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·DETERMINANTS
FOR
HONOURS AND POST-GRADUATE STUDENTS
OF
All Indian Universities and for various Competitive Examinations
By
Prof. M, L. KHANNA
Associate Professor of Mathematics,
Meerut College, MEERUT.
AUTHOR OF
A series of sixteen books for Post-graduate classes: A series of
fourteen books for Degree classes; A series of six books for
Intermediate and Higher Secondary Examinations in
Hindi and English; and a series of four
books for Roorkee and Kharagpur
Entrance Examinations,
Published by
JAI PRAKASH NATI
Fducational & Law Publishers
MEERUT CITY.
CO.

·BOOKS BY THE SAME AUTHOR For Post-Graduate Classes 1. Co-ordinate Solid Geometry 9. Integral Calculus, 2 Theory of Equations. 10. Differential Equations. 3 Vector Algebra. 11. Dynamics of a particle. 4. Vector Calculus. 12. Partial Dill. Equations. 5. Determinants. 13. Statics. 6. Spherical Trigonometry 14. Spherical Harmonics. 7 Astronomy 15. Attraction Potential. S. Differential Calculus 16 Dynamics of Rigid Bodies. For Degree Classes 17 Algebra 24. Dynamics. 18. Trigonometry. 25. Hydro-Statics. 19. Co-ordinate Geometry. 26. Set Theory & Modern Algebra. 20. Differential Calculus. 27. Matrices. 21. Integral Calculus. 28. A Tert-Book of Integral Calculus. 22. Differential Equations. 29. A Tert-Book of Differential Calculus. 23. Statics. 30. A Text Book of Solid Geometry For Roorkee and Kharagpur Entrance Examinations 31. Dynamics. 33. Co-ordinate Geometry. 32. Statics. 34. Differential Calculus. For Intermediate Students (in Hindi & English both) 35. Algebra. 38. Dynamics. 36 Trigonometry. 39. Statics. 37. Co-ordinate Geometry, 40. Differential Calculus. All rights reserved by the Author Price Rs. 2.50 only. 1st Edition Feb. 1959 (Single) 2nd Edition Sept. 1960 (Double) 3rd Edition March 1962 (Double) 4th Edition Aug 1964 (Double) 3th Edition Jan 1967 (Double) 6th Edition Jan 1969 (Double) Published by K. N. Gupta for Jai Prakash Nath & Co. Meerut. Printed at Bansal Printers, Meerut ·PREFACE TO THE THIRD EDITION In this edition the subject matter has been revised and a few new questions have been added. The first chapter is followed by a miscellaneous exercise consisting of typical and important questions based on the articles given in this chapter. In the previous editions these questions were given in the second chapter. This chapter is for B. A. and B. Sc. (pass) students. The second chapter now contains questions based upon special type of determinants, and is meant for Honours and Post-graduate students. The university papers have been added upto date, Any suggestion for improvement of the book will be highly appreciated. In the end I feel it my most pleasant duty to thank one student who did not disclose his name and was good enough to send me a list of printing mistakes which occured in the last edition. The book will be free from printing mistakes as far as possible. 315, Chhipi Tank, Meerut M. L. KHANNA March 62. PREFACE TO THE SIXTH EDITION The book has been thoroughly revised and misprints removed as far as possible. The university papers have been added upto-date. Vijay' Salet Road, Near Stadium, Mecrut. M. L. KHANVA Phone 2217. Associate Professor.

·Where is What? Page 1 28 70 Chapter 1 Elementary Properties of Determinants Exercise No. 1 Miscellaneous Exercise Chapter II Determinants of special type Laplace's method of expansion Reciprocal determinants Symmetrical determinants Skew and skew-symmetric determinants Expansion of a determinant in terms of the constituents of principal diagonal Circulant determinants Exercise University Papers : Agra M. Sc. ; Calcutta Hons. and Pass; Delhi Hons. and Pass; Gauhati Hons. ; Rajputana ; Vikram; Sagar; Agra : Kerala : Mysore ; Venkateswara ; Jiwaji ; Bihar Pass and lion's. etc. 99 100 103 109 112 123 128 133 155-172

·CHANTER
ELEMENTARY PROPERTIES OF DETERMINANTS
§ 1. Consider the following two equations :
arth'=0,
ajr+bel=0.
...(2)
d,
or
Eliminating .r and y we get
bi 5,
a, b,_agb=0.
We shall express the above eliminant in the form
Q b =0.
...(A)
1
.
ng ba
i.e. suppress the letters to be eliminated in the given aqun.
tions and enclose their coefficients as above in two parallel
lines to give you the eliminant.
(A) is called a Determinant of second order and its
value as we seen is a,b,--a.hi.
X
-J,
$2. Let us now climinaic x, y, , from the following
three equations
ap*+,+6:30.
art.cts=0,
(2)
agr+1+(9:=0.

·2 Deternuinants -les-1969 Solving equations (2) and (3) by cross-multiplication, t Y' befa-hel(0,63-1962) 22b3--2gb, sk. Substituting the values of x, 1, 2, in (1), we get A la, (1967-bacs)—b, (azf3-43ca) +41 (abzaagba)]=0 the eliminant is a. (h2c3-b2c2) - 6. (ag(3--03C,)+-C (azby magba)=0. The eliminant shall be written in the form at C1 =0, ...(A) an be 3 bg C 1.e. suppress the letters to be eliminated in the given equations and enclose their coefficients as above in two parallel lines to give you the eliminant, (A) is called a determinant of 3rd order and its value as seen above is ay (69c3-b.cz)-be (azc:- azcelt-cs (@obamaghz)=0 di b, de be or "I: 11.33/10 be C3 (3 $ 3. The value of a determinant is called its expansion. We find from above that a determinant of third order has three rows and three columns It consists of 9, ie. 37 letters which are called constituents and its expansion has 6, i.e. 3! products of the form a boca, mba.c, and so on which are called clements of the determinant. Similarly a determinant of nth order shall have n columns n rons, ni constitucnts and n! clements in it.

·Clementary Properties of DelefiylUITULE. $ 4. Leading or principal termination In determinant (A) the elementaarbigeymis called the principal element or term In this element the suffixes of the constituents occur in their natural order. Multiply the constituents diagonally starting from the left hand top corner to right hand bottom corner. The term thus obtained will be the principal term. The above determinant is sometimes expressed as (aph.cz) i.e. by enclosing within brackets its principal term. $ 5. Expansion of a determinont-We have seen that a, b, c, =a(6.63-b.co)-- 1 (AC3--07:02) +92 (ligha-agh) (A) a, b02 a, b, cз b, co or ito, al my or bg Cg d3 C3 az ba The above is called expansion with 1st row. a. (determinant obtained by removing the row and column intersecting at a.) -, (determinant obtained by removing the row and column intersecting at b) +91 (determinant obtained by renoving the row and column intersecting at ci). Now by rearranging the terms of (A), we find that 01 bi C1 =0, (6,C3-baca)-02 (b.cz-bgc,)' tag (bez-baru ag :1 bg which is called expansion with Ist column. The expansion with Ist column is obtained in the same way as the expansion with Ist rou explained before.

·Similarly if uc have a determinant of fourth order.
by co do
then following the same law of crnunsion he can write it as
with 1st row,
alba ca de fob, Cathy talab. de Idag haal
b, c, d,
a, b, d,
ng by C
illicida
or expanding with ist column the above determinant is
cqual to
a, b, c, d, -A: | by c, d, 1 to be a drag b c d
bg Cg de by ryd, b, c, d,
baca da
becode b, c, d,
Note. The above rule of expansion is true when the
determinant is expanded with either first row or first column.
We will always expand the determinant with either Ist row
or 1st column. Expansion can however be made with any
other row or column provided such row of column is made
the 1st row or column as will be explained further in Art 9.
* $ 6. The value of a determinant is not altered by
changing the rows into columns, and the columns into rons.
(Proof)
(Cal. Pass 65)
Let 4=; b, c =a, (6,63-6362)-bi(0,0,--a,ca)
+ (a,bz-agha)...(A)

·Elementary Properties of Deserminants 5 Let a new determinant j' be formed by changing the rows into columns and columns into rows of 4. Let 45 a a aa 1=0 (b2cs - bạca) – a (bạc, bạcz) +-as (b2fz-bacz), ...(B) h, ъ, ь, Now after re-arranging the terms of (B) it can be put as ai (balz-b3Cs)—b(agro-agc3)+61 (agha-a3bz), which is same as A. $ 7. (a) If two adjacent rows or columns of a determinant are interchanged, the sign of the determinant is changed, whereas its numerical value remains the same. (Proon) (Agra B Sc. 58) Let 4- 104 (0,63-1962), (,-737) too (agh-19h). ...(A) Let a new determinant A be formed by interchanging the 1st and 2nd rows of 4. . Og hy Cg =-a, (b2c3-b361)--b, (acz--239) +Cq (abs-a3b). (B) Rearranging the terms of (B), it can be put in the form -{0, (bac3-6363)(azc-scal+- (abs - agb)} which is same as A except being of opposite sigo. i A-s. (b) If any two rows or two columns of a determinant are interchanged, the determinant retains its absolute valne but changes sign.

·6
Determinants
Let A=
a b c
=a (baca-baca)-bz (azca-agro)
+ci (agby-2.h). ...(A)
a, b, c,
Let a new determinant 4' be formed by interchanging
the first and third rows.
=a(6,61--6,)-0, (a.cı-aca)
+(a,b,-46.). ...(B)
a, b, c,
a, b, c,
Re-arranging the terms of (B) it can be put in the form
- {ax (6,C3-b3ca)-by (a.cz --AzCa)+c, (anba-a3b.)} which is
same as (A) except being of opposite sign. :. A=-4.
$ 8. If any line of a determinant be passed over three
parallel lines, then the resulting determinant will be (1) of
the original determinant.
Let 4 dll bici di
a, b, c, d,
az b₃ c3 d₃
Let a new determinant 4' be formed by making the
fourth row of 4 the 1st row of A' i.e. by making it cross over
three parallel lines ; then we shall prove that
d'=(-1): s.
a b c d
a, b, şi de
a: hicils
12

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·8 Determinants make the third column take the position of Ist column for which we shall have to pass it over 2 columns and hence by article 8, =(-1) =(-1)[cz (agba-agb,)--,(a,by magbi)+cala, bo-azlik}} =(-1) (0, (baca-bacalb, (a.cg--2369)+cs (ant, -ashs}} Similarly expansion with 2nd column will be equal to (-1) 1. Hence we need not undergo the procedure to bring the row or column with which we want to expand, in the position of first row or column. We may expand with any row or column in accordance with the rules of article 5 but multiply with (-1)" where p is the number of crossings of lines which will make the expanding row or column 1st row or column. Thus a be a by expanding with 3rd row ay bg ca -1-1) (az (b1c2~0,(1)~13 (9,62–1964)+c: (a,ba--02b.) where 2 is the number of crossings of lines which will make the third row take the position of the 1st row. Above can also be written as (-1)! [62 (a,cg-asca)-b, (QCs-Q3c3)+b(A1Cz-0,61)] *$ 10. If any two rows or two columns of a determinant are identical the determinant Yanishes, (Delhi 53 ; Pb. 35, 39; Sagar 66)

·as raro
Elementary Propertics of Determinants
9
Let a determinant be such in which 2nd and 3rd
columns arc identical.
Let a new determinant, 4 be formed by interchanging
the identical columns.
d, Cg
C2
The value of a' by article no. 7, must be equal to -4.
But actually we find that there is no differncc between J
and J; A', but A'a-A (Art. 7).
Hence s-=-4 or 2.3=1; .. A=0.
$ 11. Minor.
The determinant that is left by cancelling the row and
column Intersecting at a particular constifuent is called the
minor of that constituent and is denoted by the corresponding
capital letter.
bi
ba
!
ba
1
Minor of by=B1
21
ra
{a

·10 Delcrminants Minor of ro--Cy=1 do b, ty 0 bg Phil C3 1 a b, OT 0, 61 hy he f Both the above forms can be put as aqd, ~b, 8, +6C, or GA - adatard, where capital letters denote the minors of the correspondin small letters, If expanded with 2nd rou or 3rd ron, + re 1*' or (-1) bi (1 atro bg Ca C2 dy by or Both these can be put as (-1) (0,12-0,B:+C.C.) (-1)*(az Az-b3Bz+C8C%), .: 0942-1,B,+c.Co -4, gAg-b3B, +3C,=, azerb,BAtc,fi=d, G-stazaz=. Hence when the constituents of any row or column are multiplied with their corresponding minors and summed or with alternatisely plus and minus, signs, the result=8 or - ·Elemantary Properties of Deterinmants 11 according as the row or column whose minors are written is made the first row or column by even or odd crossings of rows or columns. Now a 4,-6,B:+6,6= az be c a, b, c, a, b, c, .: agri-h,B2-7-c2Cq=1 a. b. Co =0 (Art. 10) Two τows identical a, b, c, Similarly 0,4,-0,42 tazza a b c ay by Ca as b, c, But b, 4.-bot9 + briga bi by cz b₂ b₂ CA 50 (Art. 10) Two colums identical ba ba fa Hence if the constituents of any row or column are multiplied with the minors of the constituents of some other row or column and summed up with alternately plus and minus sign, the result=1. (Gauhati 65 Hois.) : b4-6.C +b3C3=0 0341-b3B+6Cq=0. $ 12. Co-factor. The determinant that is left by cancelling the row and column intersecting at a particular constituent, when that

·12
Determinants
particular constituent is brought in the top left hand corner
of a determinant is called its co-factor and is denoted by
corresponding capital letter.
a, b, c,
Co-factor of (: Now cq can be brought in top left hand
corner by one movement of row and two movements of
columns i e, in all 3 movements
Corco-factor of C-(-1) a, b, a bi
2017
ag
ba
-(minor of ca)
B,=co-factor of be=(-1) | Az
3
C3
yao
=minor of be
Hence the co-factors are equal to minors in magnitudo
but are of same or of opposite sign according as the letters
whose co-factor is being calculated is brought to top left hand
corner by even or odd movements of lines :
a be a l=ad, ~a,A+QA,
in terms of minors
a, b, c.
andta,Å. +04
in terms of co-factors.
Just like in article No. 11, it is quite clear that
Ath Bit GC;=1

·Elements Properties of Determinus 13 A, B, +6.3, +1,8-1, (Note) algss+13B9---rCard, 0,12 -+ 6,84+CC= (Note) azhath, BatGC₂=0. azt,+ b3B.tr,C=0, where capitial letters denote the ch. factors of corresponding small letters. The expansion of a determinant in terms of minors and co-factors difter only in that in the former the terms are alternately plus and minus while in the latter all the terms are plus and that if the determinant be expanded with any ron or any column the result is either zero or +-4 (and not - as in minor). § 13. If cach constituent in any row or in any column be multiplied by the same factor, then the determinant is multiplied by that factor. Let A 9, b 6 FaAb, BitcC,. d2 b. C2 bg Cg Let a new determinant 4' be formed by multiplying the constituents of 1st row by a constant m maz mbi mci 1 ag be ca g аз by Now the minors of maz, mb, and mea in A are the same as those of a, b, and c, in A. . 4'=ma, As--mb,B,+mc,C.

·14 Determinants =mlagt, -h,0,HC)=m4. which proves the proposition. sobre Note 1. From above it is quite clear that A= Hence if in a 'given determinant 4 the constituents of any row or column be multiplicd by a constant m and a determinant 8' be formed, then m Note 2. From above it is quite clear that if in a determinant the constituents of any row or any column are k times the corresponding constituents of any other row or column, then the determinant vanishes, i.e. d. hy 2 by de h, 0 be ka, kb, ke a, bi C two rows are identical. *$ 14. Jf each constituent of any row or column consists of two terms, then the determinant can be expressed as the sum of two determinants. (Proof) If AJ, A, and Ag ate the minors of an, a, and a resp. in f, then | @ital bi b. f azta, ha ber azt az b C3 =(@ztas),-(aztag) $.+(43t-ag) fg **(47~0 ds+agds)+( aq-auditaziz) ·15 C1 bg terms resp., the determinant can prve Elementary Properties of Determinants G + 21 bi a2 b. C d3 bi bi Cs Similarly asta, butß fi Chetan b+B C. Azta, bg + B: fa ab,+Rio It a bith dy by+Big Ia bat piz ca L' f1 + ai 09 ha 02 dg ba C1 C2 ba C3 $ sum of four determinants. We can generalise above as folling, If the constituent of the three eVisomentet er $1 ca 01 61 2 (1 ag b, C. a3 mn p determinants, *$ 15. If to each constituer 5 AT 28- added or substracted the equi-positive of Si

·16
Determinants
constituents of one or more lines, the determinant remains
unaltered.
(Nagpur 61 ; Mysore 59 ; Punjab 34, 37, 44, 46, 51;
Agra 60 ; Delhi 57)
Let
01
bi
C1
ba
A3
C3
ba
Let another determinant A' be formed by adding p times
and subtracting 9 times the constituents of 2nd and 3rd
columos respectively from the corresponding constituents of
the first column.
az tphe-99 be
0x+pb2-202 bag
astpt-ge; to co
By $ 14 it can be broken into sum of three determinants.
a, b, c {+ pb, b, c,
a b c pbg by a
pb, by cz
c: 1
ara liga
Non by $ 13, the last tro determinants sanish.
alg

·Clementary Properties of Determmants 17 -- In practice generally we can alter several lines of the determinant without altering its value. Qy bica = +pbi+qcı bitsch 1 a, b, ca Ag+pbe+ac, b. +sca a, b, ca 23+phe+9c, bz+scs We can split the determinant in the R. H. S. into 3x2 determinants all of which shall vanish except a, b, c, a, b, c, $ 16. If any rth line has been alrered by means of sth and ith parallel lines, then sth and it cannot themselves be altered by means of that rth line, but we can use the rth line to alter all other parallel lines except rth and sth. If a, b, c, a, b, c, let 4= a +pbi+qc, be tra,t-sc, C, az+pbo+qca batraz+5C, Cz ag+pbz +963 63-t-raz+scy cz A' can be splitted into 3x3=9 determinants. Out of these nine, only seven determinants shall be easily seen to be vonishing and the remaining two will be ·18 Determinants A b, a ti pt ra, a pt, ra, c; Og b₃ ra pb₂ raq " =4-pred(1-pr). 45-4' because in altering the first column, the second column has benn used and again contrary to our article above in altering the second column, the first has been used and hence 4*4'. But A'=Agt-pc 1; +99 C1 atpea katga cz Ag+pcg fat gay cy A' can be broken into 2x2 i.e. 4 determinants all of which will vanish except. a, b, a dg bs Cs Here the 2nd column has not been used in altering the first column, though the 2nd has been altered by means of first, $ 17. If those constituents of a determinant, which involve x, are polynomials in x and if 4-0 when ta' be substituted for x, then xa is a factor of the determinant. Now since the constituents of s involving X are polyno. miais in .x, then the expansion of A will also be a polynomial in r and herice let 4, when expanded, give ·Elenientary Properties of Determinants 19 A=Det +P79-1+PzXn-st...+Pa-le+PA But (=p2" +2291-1+P24*-**...PA-ya+Pn (given). Substracting, A=Po (x"A")+P1 (x-1-an-)+P2 (.1---)+... -+Prom(x-a) which shows tliat .r-a is a factor of A. Similarly if the constituents of a determinant are in erms of a, b and c and by putting cb, the determinant ranishes (.e. any of the two parallel lines become identical) then (a-b) will be a factor of the determinant, *Example 1. Factorize 1 g? for 1 1 1 b a b с c c? a? 6 c? (Nagpur 30) Ist Method. Subtracting the constituents of 2nd row from first row, and the constituents of third row from first row, i e. R-Rg and Re-Rg, the determinant remains unchanged by $ 15, 0 a-b a²6² Now take (a−b) from 05-C6C* first row and (b-c) from 2nd row (§ 13), :: A-a-6) (6-2) 0 1 atb Expanding with 0 1 buc first column. . 1 c =(a-b) (c) 1 a+b =(amb)(b--c)(6+c-a-h) 1 6+6 =(a,b) (b_c) (ca) 2nd Method. Here if we put a=b, then the first and second rows ·20 Determinants become identical and hence (a-b) is a factor. Similarly it is easily seen that (0-c), (c-a) are also its factors. .: (amb) (6-c) (c-a) is a factor of the determinant. Now the determinaat is of 3rd degree as is judged from diagonal term, and the factors that we have found are also of 3rd degree. Hence there can be no other factor in terms of a, b and c. But a constant can be a factor. 1 K (a - b) (b-c)(-a). .. 1 1 a 6 C a b c Now in order to find k, we give a, b and c such small values so that on putting them in the determinant its expansion becomes easier. But care is to be taken in choosing the values, lest the determinant may become vero and the expression on the right also vanishes. Let=0,6-1,C=-1 1 1=X (0-1) (1+1) (-1-0) 0 0 1 2--2K; . K=1. I 1 =(a-b) (h-mc) (c--1). Hence 1 The set of values, a=0, h=1, r=1 is not applicable because both L. H. S. as well as R. H. S. vanish. Note. Above is the general method of finding the constant factor. The constant factor in the above example could however he found by comparing the co-efficients ·diagonal term and her 2 factor of first degree a factors found are of 1 hc results o! L2 Elementary Properties of Determinants 21 of bc2 (diaganal term) in both sides of A, and you will find that K=1. Note. Now see just now Q. 2, 3 and 4 of Miscellaneous Exercise at the end of the chapter Page 69-72, *Example 2. Factorize. 1 (Delhi 51: Pixday 58 ; 6 c Agra 45; Pb. 50) a> 63 c3 and find its value when a +b+c=0. 1st Methocl. Proceeding exactly as in Ex. 1, by (output A=!0 0 b-C c b3c as_-63 0 4=(a.) (b-c) 0 Export 1 ICI. =(a-6) (6--0) ntahPose =(a,h) (lc) (6-4-be more than (4-6) (6—c) [(cm) trase}} =(1-1)(0-c) (cm) og 2nd Method. Just as in Ex. 1, ( ammes a factor of above. No* 22. Se ceremonie ·22 Determinants factors found are cyclic. , the remaining factors must also be cyclic and of first degree. But a cyclic factor of first degree in terms of a, b and c can only be a+b+c. =K (a-6) (6-4) (-a) (a+6+ 1 1 a 0 c a 63 63 Now the set of values a=0, b=1,c=-1 given in previous question is not applicable because R.H.S vanishes by substituting these values. Choosing another set of values as a=0, b=1, c=2. 1-K(0-1) (1-2) (2-0) (0+1+2). 0 1 2 0 1 8 6=6K; K=1. =={a-6) (6-c)(cma) (a+b+c). Hence 1 1 1 b c Note. In case a+b+c=1), then clearly 1=0. *18. Multiplication of determinants, (Nagpur 61 ; Venkateswara 59 ; Agra 36, 45, 60; Pb. 40, 43, 46, 60) Let 45 my by C1 and A' a ßiri a, b, c, og B: Ya. ar a Bar Then 44=D where D is the determinant below. (101+bB+CYt Ajaz+baba+472 +bb, + 0172 agaz+baba+caya 0,22+babe+(272 0,23+baba+ czys 7202+hali+ray fida+by+ray's Ogas +3+373 ·Elementary Properties of Delerminants 23 Now by $ 14 the above determinant can be expressed as the sum of 27 determinants, a few of which will be A,a1 aa. alas ligan bp 173 dj@1 CX, bila 0921 1922 1923 12d1 b,B. (aya agai Cay, hapa Agan (1322 agag aza, C37, bls (1) (2) (3) (1) =0,2203 (Identical lines). (1) 0, (13 03 (2) =azbe's a b c (3) παιγαβ, ai og b, c, ſon interchanging 2nd and 3rd columns)=-4,72834. Similarly it can be seen that out of 27 component determinants D, 21 vanish whereas the rest 6 can be written as QBzY34, -ab3y,4, -a:B1734, azßzyid, azß.yo4, -23827,4. Summing up these six determinants, we get D=dſay (8.7's-B37.)-a(BiY:--Bgyi) tag (Biya--B.)} or D=4 u Br 71 -Xi'. pass In the same way we can multiply the determinants any order. Another proof is girea in 2nd chapter.

·24 Determinants or By multiplying a row of 4 with a row of 4', we mean to multiply the corresponding constituents of two rows and adding them i.e. multiplying 1st row of A with 3rd row of d'. be C1, we get a Qz+bps + Bs Ya ezag. Hence our rule of multiplication is as follows : Take the first row of A and multiply it successively with 1st, 2nd and 3rd rows of A' as explained above. The expressions tbus obtained will be constituents of the 1st row (in order) of (D). Then take up second row of 4 and multiply it successive ly with 1st, 2nd, 3rd roffs of A' as explained above. The three expressions thus obtained will be constituents of 2nd row (in order) of (D). In a similar manner 3rd row of (D) is obtained. *Ex. 4. Find the value of a +1? abtc ca-X X1 ab-ch2+1 bc+a1 a ca+babc--ar c+22 b 2 (Agra 27, 47 Supp. In accordance with the rule in the previous article, the constituents of the first row of the new determinant will be Ist constituent (n+1)+c (ab+c)-6 (ca-6X) = (a +62+8+). 2nd constituent=-c (a? +-12)+1 (ab+1)+e (ca-bX)=0 3rd constituent=b (22+2)-2 (ab+c)+ (ca-1)=0. Similarly find the constituents of 2nd and 3rd rows. DEL . (a2+b+c+1) 0 0 (a +62+8+18) 0 0 01 (a +-62 +c+1)

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·26 Determininis Eliminating r and : from above sve get grad, so 2 b , مد1,۲ ry AT - cl t's hi Or de be dy by ta ra dz Similarly = Az da and a, by di a b c az bz ca 1 as bg ca ! $ 20. Notation. We shall in the questions denote columns by Cand rows by R. Thus C,,C,, C,, stand for 1st, 2nd and 3rd columns respectively. Similarly Ry, Rz, R, stand for Ist, 2nd and 3rd rows respectively. Sometimes it is required while evaluating the determinants to alter rows and columns Suppose we are required to add twice the constituents of second row and subtract thrice the constituents of third Tow from the corresponding constituents of first row; we ·Elementary Properties of Determinants 27 shall express it as R. +2R, -3Rg. Similarly Cg-aCa+bC, would mean that subtract a times column no. 3 and add o times column no. 1 from column no. 2. *Example 6. Evaluate 265 240 219 240 225 198 219 198 181 (Agra 50, 53) If we expand here directly, the expansion will involve lengthy products and we are likely to commit niistakes. Hence efforts should always be made to reduce the numbers to lesser numbers and to bring in as many zeros as can be possible in any row or column and then expand with that row and column. Apply C, -C, and C.-C, which in accordance with our notation means that from the constituents of first column subtract the corresponding constituents of 2nd column. Similarly C-C, stands for subtracting from the constituents of 2nd column the corresponding constituents of 3rd .. 265-240 240-219 219 240-225 225-198 198 181 219- 198 198-181 25 21 219 15 27 198 21 181 We can further reduce the number in the above determiminant by applying C-C, and Cg -10C, ·28 Determinants 4 21 219-10(21) -12 27 198-10(27) 4 17 4 181 - 10(17) 21 9 -12 27 72 4 17 Apply R. +3R, and Rg-R, thus making two zeros in column one 4= 4 21 9 -4, 90 -45 0 90 -45 2 0 -4 2 =4 (180—180)=0 EXERCISE No. 1 Note. Before doing the questions, read Art. 20 and Ex. 6 very thoroughly. (1) Evaluate the following determinants : 13 43 16 Now 4= 7.6+1 1 6 35 7 4 7.4-4-7 74 17 3 2 7.2-+3 3 2 Break into sum of two determinants each of which will be zero because of identical lines. A=0 (2) where w is a cube root of unity. 1 107 w (Dehli 58 ; Allahabad 27) ·Clementary l'roperlies of Determinants 29 We know that if w is a cube root of unity, then 1wt0 and adsl and w=w.w=w. "Arrly categtry" and thus every constituent of fa becomes 1+w+w=0, J=0. (3) 1=a-bb- . C-a a-b . -aa -b har (Nagpur 33) Proceeding as in Q.2, 1-0. a btc Apply C9+rg and take out at htcand thuse; and 1 b c ta Cg of the new determinant become identical. ..4-0. catb (Pb. 35) (5) a+2b a+46 a +-6b Apply Rg-R and R,-R, and a +36 +5b a+7h thus R, and Rg of new determinant a +46 0+66 a 7-85 become identical. 1 ... or apply (+63-2c2; albc *(6) Evaluate 4-1 a ti 1 b69-ca cf-ab 1 a bc (Art. 15) b ca c 1 1 a a ? cab (Calcutta 61) i ja az abc (Art. 13 Note 1 abc cº C c2 abc ·30 Deicrminants We have multiplied Ry, R. and R, by a, b, c respectively and hence divided and determinant hy aho. Now tile abc common from, of 2nd determinant and male C, pass over two columns and thus becoming column No 1 llence and determinant becomes same as 1st, 0. (7) Evaluate je Apply C+Co+Cg and put 14-07-sur w? 0 and tut then crpand. J-3. 1 (8) Es atvate am 3 15 29 1 . we u 2 16 19 3337 8 38 (ora 30 ; 1'b. 10: Ald, 17) Arris C, -3C, C-C. C.-C, and create thu goftirea ditemm*2*1 o 3 cr.tor in azuin apply R, -R, aru 7,-F, srdcisurd with: ?! su *19) Paleste (1) 2 10 3 ·Elementary Properties of Determinants 31 -12 0-2-118-12-21=0. 6 2 0 6 -2 1 5 7 2 Because the first and third rows in the above are identical, i e. each constituent of first row is twice of the corresponding constituent of 3rd row. (b) Evaluate 3 7 9 6 8 4 5 8 6 10 3 a 2 9 S (Agra 56) 6 Applying Ro-Ro, we get 3 7 9 6 2 2 0 7 10 3 5 9 6 2 8 Again applying Co-C. Cs+2C, we 2:1 A=1 3 4 15 6 2 0 0 7 3 17 5 6 21 Expanding with the help of 2nd 10, 45-2 4 15 3 17 21 ·32 Determinants Adding R, to Rg, we get 4= -2 4 15 6 3 17 5 0 36 14 =-2 [4 (238--180)–3 (210–216)} =-2 (232+18)=-500. Q. 10. Find the value of (a) 30 0 38 29 26 22 6 3 0 9 25 31 27 11 -2 36 3 63 44 40 1 a 19 6 17 22 (Agra M.Sc. 55) (Agra M.Sc. 58) (a) Apply C.-20., CA-3C, etc; A=9. (b) Apply C-C, and Cy-C, and then in the new determinant apply Rs+2R, and Rs-3R, etc. ; 4-132. *(11) Dsalvate 4-1 a 4*+bed 1*+cda c+dah d d1 dahe (Luchnow pass 54: Gorakhpur 59 ; Agra 32 : Cal. Hons. 57; Patna 49; Rajasthan 53) a 07 bed D+D 1 c 1 1 h1,4 oda c c cº doi Id: abc Jutrinly Rs. Rz, R, and R, of De by a, b, c and and tesrectively and tier.ce divide the determinant by abcd.

·Elementary Properties of Determinants 33 a2 a3 abcd a? De 1 1 abcd b 62 63 abcd 662 abcd da d3 abcd (-1) 1 3-D, i, D+D,=0. 1 b 62 63 1 1 d d d3 Hence 4=0. (12) Evaluate 2a a+B aty ats ị Bta 28 B + B+8 gta gtß 2r rts i sta 8+8 Sty 28 (Rajputana 60) Applying C-C, and C2-C, and taking a--Band By respectively common from Cy and Cg of the new determinant we find that it has two columns identical and hence its value is zero. 8 (13) Find the value of 12 22 32 47 22 32 42 52 32 42 52 62 42 52 67 72 (Dacca 46 ; Mysore 49 ; Agra 54) ·34 Determinants Applying Co-Cg, Co-C2, C,--C], we get 1 3 5 7 Again if we apply C-Cs 4 5 7 9 and Co-Cg, then columns 9 7 9 11 No. 3 and 4 of the new 16 9 11 13 determinant are identical. =0 a ly b git xy P 9 P gr a b (Agra 29; Loch. 45) To prove this you have to change rows into columns and some more movements of lines (Art. 6, 7). (15) b c 0 b 0 a c h 0 0 с b 0 Ć с о а ь 0 b 0 с b Multiply the first and fourt Enlumns hy (-1) and divide the determinant by (-1): i.e. hy 1. 0 Now take (-)) common from R, and R,cuch.

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·36 Deterniinants *(17) J= Bya a Pub yda B 82 83 1 82 83 Bi δαβ γ γ γ3 1 2 g god αβγ δ δ 183 85 84 (Punjab 24) Proceed as for D of Q. 11 P. 33. *(18) If u-ax2 +2b89+cy? and r=a'x24-26.xy+c'); prove that 4= 12 -- XY' ? ax+by b.r+cy a'r+b'y b'rto's b a' b c' 11 - are one the arthy a'rub's Multiply c and co by x any j' resp. and hence divide ! hy ry: ty* -ry 18 Apply Cz+uc, and cat.ir b 0 0 Ecrand with 2nd column. ax+by b brtcy a's tb'y h' W'xtcy Art by hrtor o'rthy Fac ·Elementary Properties of Determinants 37 Exy+c* Now apply C+xC ax-t. by a'rtb'y art by a'x+by ax-t-by by+0'12 al: +2y tema d'12+2' ry+c'; ? ar+by a'x+b'y 11 1 11 w a'rtb'y u' We have changed two adjacent rows and hence ax thy' a'x+by -ive sisn. *(19) S=1 ai+by =-(ac--b*) (ar? +2b7}'+03). hrtcy (Sagar 19; Agra M. Sc. 52; brtcy 0 Nagpur 61) Apply Rg-(.*Rx+yR) and crpand with 3rd row of new determinant thus obtained. (20) IT !=ar&+4bx*+-6c.rº +4dx+e, 1111=0.x2 +2h1+c, 1122=bx2 +2c.r+d, 192=64*+2lt+e. prove b с a C 1 b c ( с C e 1122 1123 1112 0 Substitute the values of 1111. U12 and ugg in 4 Apply Re-(x2Rx+ 2xRg+Rg) thus getting three zeros in 4th row of the new determinant thus obtained. (21) Prove that ab is a perfect square. ab bc A bc Aora 41 : Gauhati Hons 657 ·38 Determinants Take out a, b and c common factors from (1, c, and cy respectively and then in the new determinant thus obtained make two zeros. A=a'b'c? (22) Prove bc c.tac =4abc Q?tab ca CP (Mysore 18) ab b24bc Taking a, b, c common from C, C, and Cg respectively Azabc a+c Apply C+0,-C. at6 C b a b b-tc с 4=abc 0 с a+c=2abic 0 atc 2b b b c 26 btc 1 btcc Now apply Ro-R, and expand etc. (23) Prove b c =a* +65 + c3-3abc. a 0-bb-cc- (Agra M.Sc. 57) b+ccta a+b Applying Cs+Cs+Cg, we get dz(++) 1 06-c c cta a+b :: O can be written as 0 (a+b+c). Apply Rg-2R, and expand with column No. l.

·Elementary Properties of Determinants 39 (24) Prove al +64 =4abc. c с C h? + a a ta b b (Vikram 63) Multiply Ry, R, and R, by c, a and b respectively and ience divide 4 by abc. 1 a+b% c? c* Now make abc zero in column no. a? htc* a? 2 by C-C, and then expand. b2 CP ta' one *(25) Prove Asa-b-c2a 2a =(a+b+c) 2b-c-a 25 2c 2cc-b (Mysore 65 ; Delhi 59 ; Rangoon 50 ; Alld. 37) Apply C-Cy, Cz-Cs. 4=(a+b+c) 0 24 Now apply Ri+RE+R; 26 and expand. 0 16-a40 (26) Prove |a+b+nc (11-1) a (n-1) b (n-1)¢ btetna (1-1) (n1) c (n-1) a ctatonb sna:

·40 Determinants Apply C1+CZ+Cg and take out n (a+b+c) common from Cy of the new determinant and then make two zeros 19 C, and expand. *27. (a) Prove 0 . =ax-by+c 0 C b 0 a 0 (Cal. Hons. 52 Raj. 57) 0 4= ar у Apply Co-6Cgt.com 0 с b -y-ac 0 a ab 0 Oax-by+c= y 0 c b 0 0 -a 0 0 Expanding with C, and hence -ive sign. b ax- -- by+c: c Take (-1) common front 0 C 0 Now you may expand the determinnant ar-bv+c. с ь ax-by+ca ar ac ah O a " 0 -00 0 .

·Elementary Properties of Determinants 41 Apply Ri-bRgt-CR, etc. (b) 4 5 6 5 6 7 Y =(x-2y+z) (Luck. l'ass 55; Bombay 58 ; Sagar 50) 6 7 8 1 0 Proceed exactly as in part (a) i.c. by applying C,-2Cg+C, etc. (28) Prove =abc. 1 1 1ta 1 1 1+b ܝܢ 1 1+0 Make three zeros in C, and expand re. R3-R1, Rg--R1, R₂ - R. (29) A= a a? 0 0 1 02 a 0 1 a a? 0 1 (Dacca 43) Apply C,+C2+Co+Cd and take out the common factor Itata? A=1+ata a2 0 1 1 Q2 ! 1 0 1 aa 0 1 1 ·Determinants Apply Re-R, Rg-R, and Re-Rg. 4=ltata? a d a2-a Apply R, - Rg+R =l+a+a% 1 +92- --(Itai-a) 1798-a ama Take out Itai-a common from R, A=[(1+a?,:-) 1 2 1 - u Now apply C,+C, and Co-C, and then expand. *(30) Prove 1 tap- 2ab 26 =(1-1-a? 4.6%)? 2ab 1-9% +62 2a 26 -2a (Pb. 3-1) Apply C-6C, and C, +aC, and take out (i+a**b*) each common from C and Cg of the new determinant thus obtained. .: 4=(1+92 +62! I -25 Now expand. 0 20 1 - 61 ·Elementary Properties of Determinants 43 *(31) Proye -a (62+(2-a) 263 2c3 2a3 --b(cata.ba) 2007 2a3 263 ac (a? +-6-1) abc (aº+b}+c^)* Taking a, b, c common from C1, C, and C, respectively, =abca?-17-? 26% 2c2 2a2 b2-%-a? 2c2 2a" 262 c2-a2-62 Apply R-R, and R3-R, and take (a3 +624c%) each common from R, and Rg of the new determinant thus obtained and then expand. (32) Prove a b dl={a+62.4-c%+-da), с a C d a b (Cal. Hons. 55 ; Punjab 64; Banaras 57) 6 d ; a ? с nd c -ac d -b - ad b Apply C1+Ct.cCat- I'm?+62 +6+4 a! 0 c 1 c - 0 0 b ·44 Determinants a' -' +c+d2} c Now expand. d -b *(33) Prove b = d. c (+b 2atb a+b+c a+b+c+d 30+2+0 40+36+2ctd 6a +36 toc 100+66 +3c+d1 3ath Apply Rs - Rz, R3 R3, Rg-Rq. a+b a+b+c 20th 31-+26+c 6a+ 3b-4c Sa+b Again apply Rg - Rg, R-Rind then expand. *(34) Find the value of { 3 5 3 6 10 15 1 10 20 35 5 15 35 69 (Allahalad 51: Agra 43) Apply Rs --Rs. R. - Rig Rs-R, RE-R, and expand with Ist colunin.

·Elementary Properties of Determinants 45 2 3 4 3 6 10 7 1 4 10 20 5 15 34 Again apply Re-R3, R3-R, R.-R and expand. 3 6 1 4 10 1 5 14 Again apply Re-R., Ro-Ry. 1 4 20 som 1 4 35. (a) Prove that sin? A sin A cos A cos? :-sin (A-B) sin (B-C) sin (C-A) sin? B sin B cos B cos? B where A+B+C=. (Punjab 60) sin' Csin C cos C cos? C Applying Cg+-C, we get sin sin 2A .: sinX+-cos r=1. sin B sin 2B sinC sin 2C 1 Apply R-R, and Rg-R, and note that C+D C-D sin C-sin D=2 cas sin 2 2 sin? d-sinB=sin (A+B) Sin (1-B). sind sin 24 CP and 1 sin (B+A) sin (B-d) cos (B+-) sin (B-1) 0 sin (C+) sin (C-A) cos (C+A) sin (C-A) 0 ·46 Determinants Now sin (A+B)=sin (7-C)=sin C and cos (A+B)-COS (7~C)=-cos C. d=sin (B-A) sin (C-A) sin C sin B =-sin (A-B) sin (B-C) sin (C-A). - cos C (C1 COS B (Vikram B.Sc. 59) 35. (b) Prove that nasinom. sina =-4 sin sina sin sin ß sin y cos a cos B cos y Apply C-C2, C,-C, and expand *(36) Prove the following identities :- Xxxx+y=+=) . x3 ; (Delhi 40. 58 ; Cal. 46 ; Pb. 48, 59 ; Agra 62; Luck, 49 Supp. ; Uthal 50, 63) Multiply C, C, and C, and x, y' and : respectively and hence divide the determinant by xy=. j? .. ges 23 XJ XY 1 =(-1) 1 1 1 (Art. 8)= 1 1 ·ETTE Elementuri Properties of Determinants 47 Now proceeding as in Ex. 1, 2, P. 19, 20, i e. by making two zeros in first row, 0 0 1 19-:2 13-23 =(x-y)(y-:) x toy "+: x2 + xy +12 1+yz-tem Again applying Co-Cs, we get x+) 2-* 1 **+ xy + y (7"--x2) +- (2-x) =(7--y) (y-2) (:-*) x+y *24xy-ty x+y+- =(x-Y)(y-s) (2-x) [(x+y)? ++ (x+y) - (x2 + xy +-2)] =(x-3)(3-2) (2-x) (*Y+y=+2r). (37) 4 a 6+c d]=-(a+b+c) (a-b)(bm) x(c-a). bota ca+b c? (Patna 23) Apply Cs+C and then interchange C, and C, Bm(a+b+c) Q? etc. from Ex. 1 P. 19. 65 C a 1 or DD, ·48 Determinants Now by Er. 1 and 2 pages 19 and 20 factors of D and , arc (amb) (1)-c) (c-a) (a+h+c) and (a-6}(6--c)(0-0) respectively :: D-De-a+b+c. (37) Factorize bc =abc . b C a b* -=abc (a--) (6-6)(-a) Ex. 1 Art. 17 page 19. Sote. Sec just now 1,7 of Misc. Exercise. (40) If a, , and z are all different and giren that 1.+23 -0, prove 1+:=0 1-4) . 1+: (Nagpur 30 : Annamalai 38 : Agra (1) .83 . ) y? : 1 1 (+.: 1 در Ilerce by It I Art. 17 pise 19. Jmir-y)(1.1:-x)(1+ ty:)0. But I canle ,tani: are all different. . I Citles of x-) wand:-ris rer. Mirros (11) 11., baadc are all different and given that --||, then whethviraichtbik.

·Elementary Properties of Determinants 49 c Proceed as above and use Q. 36 P. 46 and Ex. 2 Art. 17 P. 19. (42) Prove that a b c d =lathtftu) (a+d-b-C) late-hd) (a+b-c-d) Ortho-symmetric determinant. b (Lnck. 46 : Ph. 42, 46; Nag. 50 ; d с ь а Cal. Hons. 60 ; Delhi Hons. 65) Applying C,+Cg+CZ+Cg, the new column number one shall give a-tb-tet das a factor of A. Similarly applying C,+C2-C3-C4, we find that a+b--- will also be a factor. Again applying C1+Cy-C.-C, we get atc-b-d as a factor and applying Ci+Co-C,-Cs, we get a-t-d-h-cas a factor .. (a+b+c+d) (a-t-b-4-1)(a +c-b-7)(a+d--b-c) is a factor of fourth degree of 3 which is also of the 'sanie degree. Hence any other factor may be some constant K whose value is found to be unity by comparing the coefficient of af the diagonal term. *(43) (a) Prove that -2a + cta (=4 (64c) (c+a) (a+b). bta -20 btc ctn ctb 20 (Alld. 37 ; Patna 42 ; Luck. 46 Supp. ; Utkal 48) Expand directly or refer Ex. 1, 2. P. 19, 20, 2nd Method.

·50 Determinants If we put a+b=0 i.e. b=-a in the given determinant and expand it it vanishes, showing that a+b is a factor, Similarly btc and cta are also its factors. :. 45k (a+b) (b+c) (c+a). In order to find k, put azal, h=1,11 in both sides giving 8=2k; k=4. (b) Prove that btc a-6 = 3abc-q8–63.co cta b-c a+b C-a c Applying C+C, and taking atb+c common, d=(a+b+c) i a Now make to zeros by Ry-A, and RE-R, and expan. 1 b-c b Ca C - .. A=(a+b+c){-(a +54 62-ah-bc-ca)} =294-637-c-3abc) etc. *(44) :=-(x+y+z) (x+y--) (+1-3 0 x y X37:--) . 0 ' Y 2 (Agra 33 ; Rajputsoa 53) DI Proceeding exactiy as in Q. 42 P. 49, we get PK (*+*+-) +*-*) 0 X . (+1-3) (27.-2)...[4] 0 M I' 1' ·Elementary Properties of Determinants 51 It is easy to find K, by comparing the coefficients of x in both sides of (A). Expanding 4 only for x, z y +two other determinants which shall not giver P0 =-* [x (0--**+-+-)=ri+other terms which shall not be of r. .. the coefficient of this unity whereas it is -X on the R. S. of (A). X=-1. Hence proved. *(45) 0 =-(x+y+z) (r+ --) (3+x)(1+:-) 1 029 1 1 10.1? (Rajputana 19) 1 3 1 0 Applying C3-C, and C.-C., we get 0 0 0 0 ) 1 .? Again apply Re-R and RRx 2: 1+:: :-.7):- 2. 0:++212) (+8-03-2:) {(94-):-x {(-:):~,/?} ctc. Os gisen. |<<***** ·52 Determinants *(46) Prove that --B, C B, C deh, co B2 ag ha is where capital letters denote the minors of the corresponding small letters in the determinant on the right. (Agra B. Sc. 68) Denote the two determinants on the left and sight by D and D' respectively; then DD' Ay AymHq Betrza, 0, 42-6284+C,C, asymbgBestilo -per2+bB2-C4C, -2,4,+b28g-C2C, -03Ag+B3Bp=6zC; 2,Az-byB+Cs ag#g-b%B3+ caC, azAzmbyBn+0366 which by the result of article 11 on P. 10-11. D'(0 D'. or 00D ... DDD's DE D'2. Similarly with the help of article no. 11 P. 12-13 WC can prove A C b B 1 22 h, 013 CS B, CA where capital letters denote the cofactors of the corresponding small letters of the determinant on the right. (Bihar 66, Ph. 31. Madras 38. Annamalai 38, Luch. 12, Suppl, Calcutta 55, 61, Delhi llons 63. Agra 12, Dacca 65, Sagar 65, Aligarh 49, Bom 51, Nag. 51, 55, And. 52} Note. See also Q. 67 P. 64 and 28 P. 96.

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·54 Determinants 3 *1 x y which on multiplying is 2 . Ex2 Exy Exy 2 Exy Er Exy u Σχν: Σ.): Σx2 Ex=x+32+22= and Exy=x+yzx#*. (50) if wis a cube root of unity, then show that square of 1 1 1 U? (Ald. 27; Travancore 43 ; Nagpur 50) and hence find the value of the determinant on the L. II. S Replace ma by 1 and remember I+w+w=0. Let us denote the determinant in L. H. S. by 4. {X 1 cu 1 1 u 1 1 1 1 First constituent of first row in product determinant <ltwtw+1=1+w.twt.1-1-+0=1.

·Elementary Properties of Determinants 55 1 1 In a similar manner find other constituent of first row and other rows and we get --2 Applying Ro-Ri, R3+2R, and R-R, and then cx- --2 panding, we find that 4=-27, --2 .: A =3V --3. 1 1 1 1 1 1 1 1 (51) Find the value of A= 1 mi 11 1, ni ng 1, т. 13 (Pb. 60) where 12+11+18= *=1, etc. and 1/2+mm,+nin:=0, ctc. 42-1 l m 1 IX 1 m₂ n ܕܐܐ ܐܐܐܐܕܐ ܐ ܕܐܐ ܕܐܐܐ ܨܐ 1 1, т. п. 1. mg ng / Ell, II/= 0 0 /. Eli Ella 0 0 1l/, Elil, 1/3 0 0 J=1 *(52) Prove that a+ib ctid M -- M -tld a-1b can be written in the form -y--is atiß -BC-ID (Agra 28, ++; Travancore 47 ; Cal. Hons. 60) On multiplying, the first constituent of first row in ·56 Determinants determinant will be fat-ib) (am18)+(c+id) (-10) =da+bB+2y+do)~i (aß-batra-dy) B Similarly 2nd constituent is (a+b)(y--if)+(c+id) (a+ißl -ay +68-6cad-i (asf. cyc-da) BA с D Similarly the constituents of 2nd row can be easily found out to be -C-ID und A+iB atib ctid XXL a-By-is -ctid a-ib --idati where A, B, C and I have the values written above. iBC-iD 17c-bril Expanding these determinants, we get (29+62+c2+d)(a +8+y*f-8"]=A+B2.+C2+D? which shows that the product of iwo sums each of four squares can be expressed as the sum of four squares. Hence express (984-234-38+4) (5: +62+72 +84) as the sum of four squares. Let a=9, b=2, c-3, d=4 and a=5, B6, y=7, 8-8. Then sla +68+cy todo=110. Similarly B=40, C=S6 and D=48. i(92-723+34 +-4) (51+67+72+82) (11026 402+567 +484) ·Elementary Properties of Determinants 57 (53) Express as a determinant 0 cb c0 ha 0 0 Cb xoc 5 с о 6 ba 0 b+c2 ab ac 1 ab can be ac bc q* +-62 (54) Prove that 15+ y2 +a? 2axtry 2ay+.72 2ax+xy 2 X 02 +2.12 2arty 2axtxr 743 ta? 2ay *** a After multiplying by itself, we get deternii. nant on L.H.S. a I' Hence proved. (55) Prore that 2btry (1+3020,+304 264+rc:+30, 20, -3h, 2betro rst 30, 20, +-3h, ·58 Determinants is a multiple of the determinant a, b, c, and find the other factor. ag by C2 (Pb. 46) a, b, c, It is easy to see that a, b, 41 x 0 2 1 cic, 1:13: 3 01 2 3 0 Hence the other factor is 1 0 1 1 i.c. 13. 3 0 2 3 0 Solve the equations (56) 4= а а find =0. Put xua and b and you that 4 vanishes because of identical lines. Hence a and bare roots of a which is of second degree in x. br b (57) A=1 15--2x 11 10 11-30 17 16 7-7 14 13 Apply R-2R, and R-3Rg. -17 16 1 =0. -10 -25 23 14 13 ·Elementary Properties of Determinants 59 Again apply Cs-C, and then make two zeros in C, of the determinant and expanding, we get x=4. (58) .--2 2x-3 3x4 X-4 2x -9 3.r-16 X-8 2x-27 3.r-64 (Agra 35, 51; Luck. 48) Applying Ry-R, and Rg-Ry, we get X-2 2x-33X -4 =0. -2 -12 -6 -60 -2 2-3 3x-4 =0. 1 3 6 4 10 Applying Cg-2C, and Co--3C, and thus making two zeros in row No. 2 of the new determinant and expanding, We get X=4. *(59) 4.x 6.x+2 8x+1 6x +2 9x+2 12x 8x+1 12x 167+2 (Agra 33; Luck. 50) Apply C-C, and Co-2Cj. 2 1 10. Now expanding with third column, we find 6.7+2 8.x+1 - .*+2 *(60) 2x+3 3r++ 2843 3.8-44 4x+ 5 3r+5 5x+8 100+17 (rb. 37, 49) ·Determinants Applying Rz-Rz, RzRs, we get $=(x+1)(x+2) x+2 2.6 +3 3.x +4 ---- 1 2 6 Now apply C.-C. and Cg--Cetc. : x=-2,-1,-1. *(61) (a) Show that x=2 is a root of --6 2 -3. x-3 and solve it completely. 2x x+2 (Agra 48; Raj. 48) Applying Ro-R, and taking (r^2) common, we get 4=(x-2) 3 Now apply C.-30, Ca+C, and expand. 2 - 3 X-3 2x r2 3.x-6 1-1 12(x-2) d=(x-2) (x-2)(x-1) (-5.6---15)=0. 21.+98-1 .. Y=1, 2, -3. (b) Given a+b+c=1, solve d. b с c a (Burdnan 61; Karnatak 62; Agra 61; Bombay 18; b Venkateswara 59) Applying Cat Cz+C, and putting atb+c=0 and taling - common, we get 1 b I=0, Applying R-R, and Rs-R, 1 a 1

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·62 Determinants (Lucknow 65) 1 (d) Solic the following equation 2 (x-5) =0 9 (.1-5) 1 8 -27 (6-5)* 16 81 (x~5)* Apply R₃-R₂, R₃~R₂, R₃-R 2 (x-5) 0 2 12 (x–5) (x-5) 0 - 36 (1-5): (x-6) 0 8 108 (1-5): (x-6) Expanding with first column and taking out common factors we get 452x12x(x-5) (3-6) 1=0 2 -3 (1-5) 4 9 (x-5)* Apply Cz-C, and Co-C; 4–24 (x-5) (.1-06) 0 0 =0 1 2 4 S (1-7)(x-3) 4=24 (x-5) (x-6) 5 (x-7) )=0 IX-3 120 (X-5) (x-6) (x-7)(-x+3-1)=0 =-120 (x-2)(x-5)(x-6) (x-7)=0 =2, 5, 6, 7 –6x5(8571 To ·Elementary Properties of Determinang (c) Show that the equations 22x2+bx+6=0 Qgx+b,x+c=0 possess a common root only if a, b, 01 02 be by Cg Az C2 (Lucknow 65) If a be the common root then aja?+ba+c=0 Aşaat-bate=0 3,67-6902=Camacan abz-azba .: (4,02 -0,0)=(b_C2--b2cı) (a,b,--2,5) at C a bi by C a or H a, b, bg C2 62 Solve :-x+y+z=1, ax+by+cs=K, aºx+by+c%;=K?. (Delhi Hon,s 63; Pb. 61) By art. 19, * 1 1 1 I a bc b 1 a2 b2 c? K? 62 .: *(1-6) (b-c) (c-a)=(K-6) (b-c)(c-K). [Ex. 1, § 17). (K-h) (6-2) (0-K)_(K-6) (c-K) (a-b) (b-) (r-a) (a-b) (c-a) (P. 19) (K-h) (K-6) (-) (a-c) Similarly write the values of y' and z by symmetry. (63) Solve :- ar+hy+c2=K. a’x+baret-caz=X?. (by Art. no. 19) 6.**+by+rs=K?

·64 Determinants a 5 6 c Q2 be c? 62 C3 c3 or xabc (a-6) (6-c) (c-a)=Kbc (K-6) (6--6) (C-K). (Ex. 1 Art. 17) K(K-6) (K-5). a (a-) (2-0) Similarly write the values of y and : by symmetry. (64) Solve : x+y+=+=1, ax+by+c2+du=X, nr+hay+c2+?u=X3, al+b79+03:+duzx. 31.1 (by Art. 19) 1 a 6 d с X a 6 c 6% a be X 4 65 63 or or ula-b) (a-c)(a -1) (6-c)(b-d) (-) -(4-6) tamc) (a-X) (0-c) (h-K) (-4). (n-X) th-X) IC-) (X-X-b) (X-) (a-) (-) (0-0) (2-a) ( db) Til -7 Similarly write down the values of x, y' and : by symmetry (Sce Q. I, P. 69 for factors). (65) Soire : 15+:41-K. ox-+hy+r+duck's ; airtbiytc!:+vA", *.*+/+1:+Purk.

·Elementary Properties of Determinants 65 . Proceeding as above, u= X (K-2) (K-) (K-) (d-a) (d-6) (d-C) (66) Solve :- x+2y-+-33=6, 2x+4y+3=7, [x=y=:=1] 3.x+2y+9:=14 (57) If (2-bc) x+(ch-f8) y+(bg-hf)z=0, . (1) (ch--fg) x+(92-ca) y+(af --gh) =0, . (2) (bg-hf)x+(af-gh) y+(12-ab) :=0. ...(3) then show that abc-+2fgh-af2_bga--cha=0. In order that the three given equations may hold, then by artical No. 2 the eliminant fa-be ch-fg bg-hf 50. . (A) ch--fg gº-ca af -gh bg-hr af-gh hab Consider h 8 . hi b f с 8 f Minor of a= A=bc-f2. Similarly Brac--g2, C=ab-h2 Fraf-gh, G=lf-bg and H=ch-ſg. Hence determinant (4) becomes H F or HI G =O or by Q. 46, P.52. -H B -F с ·66 Determinants ኑ b } TE Expanding, abe+fgh-af-bg-ch=0. (68) Find the condition that Ix+my+n2=0 may be satisfied by (ar, br, fr) where r=1, 2, 3 and show that it is the same as the condition that three equations Qrx+boy+62 =0 may be satisfied by (l, m, n). Since (ar, br, Cr) for r=1, 2, 3 satisfy lx+my+n:=0. .. Altbem+91=0, 09/+bgm+c2n=0, 03/+bym tegn=0. Eliminating I, m, n by article No. 2, we get the required condition as 01 C Og b. dz bg Again since a,x+by+6:=where r= 1, 2, 3 are satisfied by l, m, n, .: 2,2+b,m+92n=0, al+b+czn=0, aglt- bzm+czn=0. Again eliminating 1, m, n by article No. 2, we get the condition as bi at d2 b. b C3 which is the same as condition ().

·Elenientary Properties of Determinants 67 OT (69) Find & when ax+y+s=0, hxt-by+f=0, gr+19+c=1 ax+3+=0, lix+by+f=0, Ex+f1-c-=0. Eliminating x, y from above set of cquations, we get a h 8+0 0 I b fto & fcor + 30. .hb : gf -A (abc+2fgh-af-bg-ch2) -- (ab-14)=0. abc+2fgh-of-bg-ch2 . ab12 (70) Find K where 11**my+12+ Park, 1,x+may+1:+P2=0, 13*+my+nga+Pa=0, 14x-tm'+14+Pa=0. Proceed as above. X=1 l m n Pi 1:1 l, m, na 12 m. le Pa 1, mis 113 1 mg ng Pa ·6S Determinants (Pb. 36) *(71) Eliminate a, b and c from b x-boyra . obrter, -ay-b+cp=0, az-:-(=D. Eliminating a, b, c by article 2, we get 1 1=0. Expanding, X we get 14.ry+-+-27=0. ' *(72) Having given the equations x=cy+b=, y=as+cx, =bxtay where ., J, are not all zero, prove that q+b37c+2abc=1, 12 1-a? Rewriting the above equations and eliminating *. ), 2, we get 1 (Pb, 45, 64) с b a a +62++2abc=} (on expansion) Again solving the first two equations by the method of cross-multiplication, we get -acb6caaaritet

·Elenientary Properties of Determinants
69
.
Squaring,
ac? 7-12 + 2abca+b+cx + 2abc-(1-c)2
Using relation (1), we get
712
1-a7-62-a cz1-62-6+6*c2-11-cīja
27
-2
or
(-as) (1-(*)(1--baj (1-с2) 11-cēja*
Cancelling (1-с2) throughout the denominator, we get
the required result.

·MISCELLANEOUS EXERCISE Q. 1. Prove that 1a-b) (a-c)(a)(1-c) (6-1) (c-d). b d 1 1 a c a? c2 43 c3 d3 (Agra M. Sc. 57; Pb. 39, 44; Sagar 66) Putting a=b and so on we find the following factors :- (a--b) (a-c) (a-d) (6-c) (6-4) (0-d) of sixth degree of the given determinant wnich too is of the same degree. Hence. if there can be any other factor that must be a constant K; 1 =K(a-b) (a-c) (a-d) (b- 1 1 (6-1)(-) b с a? 62 ccd2 aod Comparing the coefficiect of diagonal term bcf we find K=I. Q. 2. (a) If a, b, y, & be the roots of a S, stands for a'+8++++8+, then prove that So S, S, Sg =(a - b)2 ( ay)? (a-8)? (B-7 S, S, S, S, biquadratic X (B-S) ( 75)* S, SS. S S, SS, .S ·Aliscellaneous Exercise 71 Replacing a, b, cand d by a, B, y and 8 respectively in Q. I, we get A=(a-B)(amy) (a-8): (8-4) (-8)(y--8)* where A=I 1 1 1 1 1 a B 8 a y S a 82 72 8² g2 82 a 83 23 83 al 83 go3 83 By actual multiplication of determinants and using the given notation Sp="+8+ + y +-81 i e. Sq=c?+82 +42 +82 and So=a +-8° +2° +8°-=1+1+1+1=4 etc., we get the given determinant in L. H. S. Hence proved. (b) Prove that 3 a-t-btc a'+ba+r2 Jor so S1 S2 a+b+c a2 +63 +62 29+62 +63 Sz Sg qx+62+c 45+63 +63 af +66+ch S3 S4 =(amb)' (bmc)(c-a). (I. A. S. 58 ; Banaras 38) Si Sg 1 1 1 1 etc a も C b C 1 a 02 c? '(Refer Ex. 1 P. 191 Q. 3. (a) Prorc that 1 1 c d h* c 04 C *a-1)(a--() (-1)(bmc)(0-1) (-) (a+b+c+d).

·Determinants The first six factors are found as in Q I and the last facter (a+b+c+d) is found by symmetery. The remaining facte? which will be constant K is easily found to be unity by come paring the coefficient of diagonal term bcad etc. (b) Solve for x 1 x? .13 a a2 a? 8 B2 82 1 7 (Agra M.Sc. 40 : L.A.S 51) 4-(8-»)(y-a) (a-B) (x-a) (x-2) (1~-~)=0 :: X=a, , y, when a+b+y. (c) Show that each of the coefficient of any equation can be expressed in terms of the roots as the quotients of 1480 determinants. Consider an equations of 3rd degree .X2 1 a B* R? B } 73 7 tlo If we put tea, B. Y or or arb or y or B=y, then the rows become identical and hence the determinan! vanishes. Therefore, (x~a) (5-8) (x--y) (e-) (amy) (8-) are its factors of sixth degree. "If there be any other factor if is constant and let it be k. Comparing the cocflicient of diagonal term rapon both sides we find that I.

·Miscellaneous Exercise 73 or A(-2)(x-B) (x-7) (amp) (6-y) (B-Y) A=-(2-a) (x-6) (r-Y) (a-8) (B-7)(y-a). But -fa-)(A)(y-a). [Ex. 1 P. 191 1 1 a (x-a) (*--B) (1-Y). B2 B 1 Or 1 (+*+-P2124-Par+P3), B2B1 g2 g where 18+ni+*+par+P30 is an equation whose roots are Q, , . Now expanding 4 in terms of first row, we get 1 a + ! * B 1 * 1 B3 a d (18+Para+t+Pa). 1 ·1 2 1 cz 1 1 ab+cd ab*+c%.!? 74 Determinants Dividing throughout by a? B B 1 and comparing the coefficients of like powers of X, we get a a? A B 1 72 B P,= 1 BP 8 1 23 ² 1 72 y a3 a a ? a B2 B3 B BB 1 Above shows that the coefficients prPs. Ps can be exprersed as the quotient of two determinants. *Q. 4. Prove that 1 betad 6*c+a 2/2 1 ca+bd c*a* + b2d? 1 a a 1 (a-b) (a-c) (a--d) (6-c) (6-d)(cmd). (Indore B. Sc. 66; Agra 22; Pb. 51) Put bc+ad=d, ca+bd=Band abtodiac. 2abca 1 B B1_2abcd 1 с C1-2abcd

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·Determinants BCA Multiply R by A, Ra by Band R, BY C. CA AB -C1 ABC ABC - B2 B =(ab' -a'l) (bc-b'c) (ca'c'a). Hence Proved. 76 1=abc'2 '3bc2 ABC =a'b'2" =a^2}'Ac”? Q. 7. Proic ABC A2 =a’?b!!p'? (1-B)(B-C x(C-1) (Ex. 1 P 19) B BP ICC 1+a 04 a AL 1+04 Og as da a lta, =1+x+pa+mata (Delhi 65 ; Alld. 59 : Agra 51} Applying C+Co+Co+C, and taking out 1+etastas + common from C of now determinant, we get Ax-(1+4, +, tostar) п, 1 1 1+1 02 - ·Jfiscellaneous Exercise 77 Now apply Rz-Rz, Ro-R, and R.-R. i. 4=(1+x+22+(3+0) ag az 04 0 1 0 0 0 1 0 0 0 0 =i+a+az tagtas 1+01 az..... an Q. 8. =1+as+ag t...an 41 1-tas Azan (Raj. 60; Agra M. Sc. 66) .. 2 Ng.... Ita Just as Q. 7. Q. 9. (a) Prose 1+x 2 3 4 13(x+10) Proceed as above 1 2+x 3 4 2 3+x 4 1 2 3 (Agra 39; Patna 41; Mysore 42 ; Luck. 19) =(r+ 3a) (x-a) . a a 8 . d (Agra 65 : Bihar 66; Luck, 44; Supp., Dehli 17; Nagpur 36) Praceed exactly as in Q 7. Q. 10. (a) Find the value of determinant of nth order :

·78
Determinants
...
a
a
.
.
A
a
a
whose leading constituents are equal to x and the remaining
constituents are equal to 1.
(Agra M. Sc. 59 ; Raj. 61; I. A. S. 53 ; Cal. Ilon's 61)
Proceeding exactly as in Q 7, we get
A=(x+(1-1) a) (x-a)n-1.
(b) Find the valuc of the determinant of order .
1+* 1 x
1-X 1+* 1-X
1 - 1 1+r
...
...».
......
1-X 1-r 1-r ...
1+
(Gauhati Hons. 61 ; Raj. M. Sc. 59)
Proceeding exactly as in Q. 7, we get
4={1+x+(n-1) (1 x)] ((1+r)^(1 ---X)]*-*
=(n-in-2).x) (2x)*--.

·Miscellaneous Exercise 79 (c) Evaluate the determinant 1 1 1 1+k 1 1 1+k 1 1 1+k a determinant of nth order in which every constituent of the leading diagonal except the first is (k+1) and all other constituents are unity. Show also that if all the constituents of leading diagonal are equal to (k+1) and all other unity, the value of determinant is $9-1 (n+k). (Lucknow Prel. 61) Apply C.-C, C.-C,...CH-C1, you will find that everywhere there will be zero except the first column in which each element is unity and diagonal in which there will be 1, K, X,...K. 0 0 1 . 0 0 .. .. 1 0 0 +9 ... ... ... ... 0 Expanding with first row J=1.kn-kn-1. In the 2nd case if the top corner constituent be also 1+X, then adding ·vt will leave 15: #2 Dividing Ri, R. Rund R, by a, b, cand a respectively, 10 thrling 11 all the lurini tu lirit Culun ke wil 30 , coming (166ur ( nil watch on talen website scissmirent abog salership sconto te k**, Irroc llic and determinant it inkl. (d) Indit Tlic abore determinint can k casily shown ink (r.o., 04.0. +0: us fo.s.rid). 11. I'rovy 1 lind (Agra M. Sc. 51.64:1. A. S. 14; Calcutta 48: Sagar 66 ; P')). 18, 57. 58 ; Agra II. Sc. 53, 59. 61. 6f: (Col. lions, & Dilbi 33: trombay 3: Lucknow Fass 61) Is € narcof 14.4 1 Delhi 18. 52) Jrghed 1 1 1 3 6 + € 1 1 ·Miscellaneous Exercise 81 Now apply R,+Rg+Rg+Re and proceed as in Q. 7. Q. 12. (a) Prove ad =a2+1 52 4= a2+1 ab ac ab b2+1 bc bd +1 d? ac bc C+1 cd C2+1 da a ? ad bd cd d +-1 =1+a+b+c+d2 62 cº dº +1 (Allahabad 49) Multiplyingft. fe. to and fy by a, b, c and d respectively and hence dividing by abcd, 11 a (a2-4-1) aba ac? ad? abcd ab 6 (62.+1) hc2 bak bức c (c+1) cdº a'd bad c d d (12+1) Now a, b, cand d can be taken out from Rj. Ry, R, and Ra respectively. abcul a?+1 62 abcd a? 6+1 c? a? h2 3+1 f d2+1 Now proceed as in Q. 7, i.e. C+CE+Cg+C. etc. (6) Show that 07 + ab ad is divisible by as and find the other factor. bc ac ab b: + ac + bul ta (Raj. 49; Agra M. Sc. 65) ·82 Determinants Proceeding exactly as in Q. 12 (a) above by multiplying C1, C2, C3, C, by a, b, c, d respectively and taking out a, b, c, d common from Rg. Rg, R, and Re respectively, we find s=(1+ a2 +624.c+d!) X3. Hence it is divisible by 1 and the other factor is *+*+b+c2+d. Q. 13. (a) Prove 3x? 3.x 1 =(r-176 .x2.7 +2.7 2.8+1 X 2r-+! *-+-2 1 1 3 3 (Venkateswra 66; Jinaji 66; Agra 41; Luck. 19; Kerala 65) Applying C-C,+C.-C,, we get 4=(1-1) 7+2x 2x+1 2.x+1 x+2 1 1 3 3 Now make two zeros in C, by applying Ro-Realid Ry--R, and expand by it, (b) Prove that =(a-B) a* a 2a 0 87 1 0 3 28 Apply Re-Rigaud Rs-R, and take out (a-B) commes from cach R, and Rg ·Miscellaneous Exercise 83 :: A=(a-B)? | as+Ba+aß a+b 1 0 3 (a+B) 2 0 0 ps B 1 382 28 1 0 Expanding with 4th column and hence -ive sign, 45-la--8) a +82-aß a+B 1 3(a+b) 2 0 382 28 Now apply R-R, and again take (a-8) out and expand 4-la-)4. Q. 14. Show that -bc b2+ bc c* to bc1=(bctcatab). etc. al tac -ac c2-tac a?.tab b2-4ab - ab (Bombay 61) Multiply R1, R2, Rg by a, b and c respectively and hence dividing 4 by abc, 1 -abc ab (b+c) ac (c+b) abc abc be (cta) ac (a+b) bc (b+a) -abc Taking a, b, c common from C, C, and Cs. abc ab+ac ahc ab + bc ab (a+c) actab ac ac+ab ac+bc botica ·84 Determinants Applying Rg+RE+R, and taking ab+bet.ca commo, we get 4=(ab+bc+ca) ah-thc bertab 1 1 ac ac+bc bc+ca - ab Applying Co-C, and C.-C, we get 1=(ab+bc+ca) 0 abtbc-(ab + bc+ca) 0 ac+bc --(ab+he+ra) =(ab+bc+ca) (ab+be+ca)=(ab + bc+ca)». Q. 15. Prore that 0, az 02 a a B 1 B 8 =(r-a)(x-A) (x-7)(x-8). (Type Mycorer .. Apply C,--af:. C-BC3, C.-YCg. Co-Cs. r-a az-az-y 0g-8 I to ps--y too! 0 0 0 0 Teatrie is clearl cual to Ir-a)fr-f)(1-9120 ·Miscellaneous Exercise SS *Q. 16. (a) Prore that 4= (b+c) a 1=2abc (a+b+c)* 52 (Cta): 62 ; (a+b) (Gorakhpur 59 ; Cal. 31 ; Agra 50, 62, 65 ; Raj. 51 ; Sagnr 63 ; Aligarh B Sc. Hon's 59 ; Agra M.Sc. 56, 58 ; Bombay 58 ; Nagpur 61 ; Lucknow 64 ; Delhi 52). Applying C2-C,, Co-C1, we get 4- (6+c) p2-(b+c)? a?-(b+c)2 (c + a)-62 0 (at-ba-c2 (6+c)? (a+b+c) (a-b-c) (a+b+c) (a-b-c) h2 (€+a+b) (c-fa-b) (a+b-+-c).0 (a+b+c) 0 (a+b+c) (a+b--c) Taking out (a+b+c) common from C, and Cg, 4=(a+b+c) (6+c)2 a-- a-b-cl c? 0 c cta- 0 0 a+b-c Now applying R-R, -Rs, we get 4=(a+b+c) 2bc -26 -26 h2 ct0-6 0 11.b Applying Cot C, C+1C1 ·clb a+b (+0+01 =-2be (a+b+c)*{(a+b) (c+01-12 Ath B+C=2s-b-c-a; similarly C+2=band A+B=0 Multiplying R1, R2 and Rg by c, a and b respectives 86 Determinants 4-(a+b+c) 2bc 0 0 b4%c c? =2bc (a+b+c) eta balc c/b 2bc (a+b+c) a (a+b+c) 2abc (a+b+c). (b) If 25 a-th+c, prove that a2 (sma) (sona) (s-b)2 62 (s-b)* (snc) (s-c) Lets-2A, 5-6-B, S---C. .. dt B+C= 35-.(a+b+c)=s. ..d= (BC) B2 (C+4)2 B2 C2 (A+B) Q. 17. Prove that A= (a + b)2 CE bc ca (b + c) ab bc ab (c+a) =25* (s-a)(s~}{5-2 (I. A. S. 56 ; Cal. by part A: 1-2ABC (A+B+C)". = 2(s~a)(s--6)(sejo =2abc (a+b+c). and hence dividing the determinant by ahe, ·Miscellaneous Exercise 87 1 c(a+b): ca abc ac a(b+c)' alb 1 bia b(c + a) Take €, a and b common from C3, C, and C, respectively :: Aabc (a+b)2 abc a? (b+c) a (c + a)? which is of the same form as Q. 16 (a). +Q. 18. Prove that 4= ** *2-(y--)yz =(x-3)(--) (2-1) x(1+y+z) (x2+x2+za) y yº-(-x)* :* 72 :-(x,y) xy [Agra 45, 56 ; Alld. 55, 50, 32 ; Vikram 61 ; Nagpur 54(S); Celhi Hons. 60 ; Ph 57 (S)] Split the given determinant into two, one of which shall vanish because of identical lines. We get .22 (y? ++*)+2y: -(=++*)+221 -(x +32)+2xy Again split into two determinants one of which shall Vanish, z? 12 + y2 19"+ 2.5 2 x2 +33 XY Apply Ci+C, and take out x:+1+.

·88 Determinants = -(x2 + y2 +24) 1 ya 1 +x? x+y XY Now apply Ro-Ry, R3-Ry. :: 1=-(x+12+-) 1 y2+2 07-y? (1-3) 0 ''(x-2) Expanding with first column and taking x-y,tas common from R, and R3, A=--(x2+y+z) (x-y) (x-2) x+y Apply C1+C, and take x+y+z out. :: A=(+²+y2+*) (*~3) (3-X) (x+y+2) (x->> ***; **?\. ;) =(x3++)(-3) (v) (-X) (x+y+5). Q. 19. Prove bc-a? ca-62 abc -bc+catab bc-ca+ab bct-cawab (a+b)(6+c) (b+c)(b+a) (c+a)(c+b) 2-3 (h—c) (c-a) (a-hi (a+b+c) (be+ca+any; (Cal Hons, 52) Apply Ro-R,+R, and taking factor 3 out, be ab t'a any --bctca+ob bc-cat ah befca-ah * Sab hab + Jah ·Miscellaneous Evcrcise $9 Apply R,+2R, Rg-R, Sabsabt. hc +ca. ab ac ab+he+ca ab+bc+ca ab + bctca a (a+b+c) b(a+b+c) c(a++c) Taling ab +-bc+ca and a+b+c common from ho und Rg. 1=3(a+b+c) (ah+bc+ca) be ca ab b C Multiply Co. Cg. Cs by a, b, c respectively and as such divide A by abr. Then take out abc from first row of new 3. Reler Q. 36 P. 47. 1==3(a +64.c) (ab +be+ca) 1 b с 62 cs =3 (a+b+c) (ab+be+ca) (0-6) (6-c) (c-a) (Ex. 1 P. 19), Q. 20. Multiply b x с - b c a с ab and show that product of two expressions of thic form 23+A18+ N2-LAIN is of the same forin. (I. A. S. 57; Gauhati Ilons. 64) Product=arhy+r ay+bz+cx az+bx+c)' brut-cyta: by+cztar hi trxtay cx+ayat-bz ry+:+b.x cztort' ---

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·Miscellaneous Evercise 91 L.H.S. TE: a 5 a 6 If we multiply in the above form ; the 1st constituent of the 1st row of product determinant will be a+62+c?, where as we want it to be 2hc-a. Te-arranging the terms of 2nd determinant, we get b x! b a C с b c a ba C с сь a b с a b ix c b с a a C с 6 2bc-a2 ba 62 c2 c2 2ca62 ? с C a2 2ab- But b 2=(3abc-a3-53-c?) =(22+63 + c3 - 3abc)? b By expansion b Q. 22. fax,+byistcz,-d=axy?+-by:*+cz,? =axga +by;2+czy? and axzxz+byıya+c21:=ſ =axzxz+by',': +022g=ax/+by113 +212a, с then prove abc Y'1 *=(d-1: (d+28). ''2 za (Pb. 33) ·92 Determinants L.H.S.=abc X Y : 1x 1 i 2 ܕ ܕ ܕ* axt bys cm ܙ: iܙܬ ax, byz co X 13 ax, by's ca xg y's Zg On multiplying Sax;? Eaxixg Laxxs Σαχ,, Σαχ, Σαχηλ j Eax,x, Ear,r, Eaxz? Now proceed as in question 7 page 74. Q. 23. Fxpress 4-1 (2-x)? (a- y)2 (a-:) (1-x)2 (6-y) (6-3)? (C-x) (cy)? (0-3)* as product of two determinants and find its value. (Agra 32, 37, 49, 51; Andhra 92 ; 1.A.S.JO; Nagpur 51, 54, 61 ; Allahabead 56; 1.AS. 55 Calcntta Hons. 63) If we put x=y in 4, then Cy=C,; .. Hence --y is a factor. Similarly (--), (3-X) are also factors of A. But (x-3)(y-2) (-3)1 1 (Ex. 1 Art. 17 page 19) ·Miscellancours Exercise 93 Again if we put a=b in, 4 then R-R, ; :. Hence (a-b) is a factor. Similarly (6-0), (c-a) are also factors of A. But (a-b)(b-c)(cmc)= [Ex. I Art. 17 P. 19). 1 a b с a6 c? .. A=K! 1 1 1 x 1 1 a c a? bº c2 for some such suitable value of K so that on multiplying the two determinants the resulting determinant is A. 1 a a 1 1 1 * 1 с X Ixl 1 we have interchanged two columns. j2 y I b 6 22 1 1 с If we multiply we get Ist constituent of product determinant as a tax+ whereas we want it to be r2-2ax-ta?. K must be chosen=2. x2 - 2x 1 X a 72 -29 1 2: 1 1 c ·941 Determinants is required form, because on multiplication it becomes equal to 1. The value of A=2 X - . a - (Agra 35) =2 (x-3) (---)(-x) (a−b) (6-4) (---). Q. 25. Express (1 +48)(1+ay) (1+.az)* (1+bx) (+by)+ (1+bc)^ (1+cx) (1-4cy). (1+cz)" as product of two determinants and find its value. Proceeding as in Q. 24 and taking 4=(A) of Q. 24 when K-2, its value is 2 (x-) (-2) (:--x) (a-6) (6-0)(e-a? Q. 25. Find the condition in determinant form, in order that the general expression of the second degree in two variables f(x, y)-ax+2hry+By:+2gx+2fy+c may be resolved into two linear factors. Let the two linear factors be 1*+may+ne and lgx+m,y+na, Then S (x, y)=(117+may+ng) (1,8+m.)'+ng). Comparing the coefficients, 11/=a, mm=6,0,2=4. 1m2 +12me=21, m1g+min, -28. nzla+ng's=28. Now we know that 1 0=0 and (Raj. M. Sc. 59) 6 0ܢܐ ܕܐ mm2 0 m? 0 0 0 ·Miscellaneous Exercise 95 Multiplying these two zero determin ints, we get 21,12 lim,+l,m nola +12/2 =0 mila+miel 2m,ing min2+12n: m, na+m, 2n,"2 2a 2h 28 =0 ܢܐ,xm+ܕܐܢܐܐ or 2h 2b 2f 2g 2f 2c =0 is the required condition. or hbf On expanding, it becomes abc+2fgh-af--bg-ch2--0. i.e, the condition that general equation of 2nd degree may represent two straight lines. Q. 26. (a) Express as the product of two determinants and show that its value is zero. cos (a-B) cos (y-a) 1. cos (a-B) cos (-y) cos (y-a) cos (8-) (Punjab 62) sin a 0 x cosa sin ao COS a cos sin 8 0 11 COS B sin 80 each of which is zero. cos y sin 0 cos y siny o Hence the given determinant ·ay ba 0 On multiplying these two zero determinants, wet get 96 Determinants (h) Prove that 9,2, +b Bi azaltb281 Agai+b3B 30. 4,22 +byBz Qgaz+babu azaz+b3B, Gjag +b B3 0723+boss agas +-b3Bs Above is az by 0 1x a Bio a, b, 0 ay B, 0 аз B, о each of which is zero. Hence the given determinant is zero. (c) Prose that 2 a+B-ty-ts aptys a+B+8+8 2 (a+b)(x+8) aß (y+8)+y8(a+B) ab+y8 aß (Y+8)+75 (a top) 2αβγό We know that 0 i=0 and 0 Oyta ati 0 at 7+8 0 78 aß 0 αβ 1 1 - -- γδ which therefore is also equal to zero. *Q.97. Prove that 52.62+1 b2+1 6+6 =(be+cataby's 2+1 fa+1 02+1 6+a 1941 a?! a: +62+1 a+b hur cta A+b 3. (Raj. 303 ·Miscellaneous Exercise 97 By inspection, we find that 4= Obc X] c 1 а ос a C 1 a b 0 1 b0 1 1 1 1 0 where A' is easily seen by applying C, -6C.. Co-rCe and expanding, to be equal to (bc+catab). 0 -b 0 0 a 0 a 0 1 a 0 0 1 1 1 1 1 0 --[a (0+c)+b (a+c)]=-(ab +-bc-t-ca). .: 42=(ab+bc+ca)? *Q. 28. Show that a +xab-cx ac+bx ab-tex b2+xbc-ax c -C a ac b -a X Let the determinant on L.H.S. be a, b, a a, b, c, a, b, cg Cofactor of x is a? +=, Cofactor c is -(-ex-ab)=ah+cx=B, Cofactor of -h is ac-br=Cy.

·98
Determinants
L.H.S.= A, A2 A3
= A, B, C =A*
B1 B₂ B₃
C, C. Cz
A.
B, C, 1
by Q. 47 P..

·CHAPTER II DETERMINANTS OF SPECIAL TYPE $ 21, Second Minor. We have already defined in § 11 as to what we mean by the minor of any particular constituent in a determinant 1.e. the determinant which is left by suppressing the rows and columns that intersect at at particular constituent. Now we shall call this minor as first minor showing that one row and one column intersecting at that particular constituent have been suppressed. The determinant that is left is called minor determinant. a, b, a bg ca da dg by C5 dz a, b, c, d, If we now suppress two rows and two columns then the determinant that is left is called second minor i.e. if we suppress the first two rows and first two columns then the determinant left is C₂ do ic. Coded, which is called second minor. Complementary to and minor. We observe that the suppressed rows and columns have common constituents ·and this determinant 1 17 Similarly the second minor of a determinant is denoted $ 22. Expansion of a determinant in terms of minors of We have so far expanded a determinant in terms of constituents of one row and one column. Now we shall Rule In terms of first two rows. Make all possible determinants from these two rows by taking any two columns of these, i.e. there are four columns and by taking two columns at a time, we shall have six determinants of the 100 Determinants which form a determinant d. is called complementary of the second minor Notation. The first minor of a determinant A is denoted by say, ie the determinant which is left hy suppressing the rows and columns intersecting at az. It is also denoted by the corresponding capital letter Az. by fazba, i.e. the determinant which is left by suppresving the rows and columns which enclose a determinant whose leading term is a,b, and so on. any order. Laplace's Method. (M. Sc. 56, 50 ; Nagpor 61) expand the determinant in terms of two rows or two columos. type as be di li | a, b, bi by d d, b, og b. d, ·2 Determinants of Special Type 101 Multiply each of them by the corresponding determinant which is left by suppressing the rows and columns intersecting at them, i.e. by their minors and then add them with proper signs, the rule for which is as follows. Just as while expanding in terms of one row and one column we take alternately +ive and –ive, here we shall see that how many movements of rows or columns will make each of the above rows and columns the first two rows or columns. If it is odd, then sign to be attached is -ve and if even, then +ive ; e.g. ai C1 а. C2 Now the c column being 3rd can be made 2nd by one movement of columns ; a column is in the position of first column so that the total number of movements is one, i e. odd; hence the sign will be mive. Similarly be di bą da Now b column being and can be made first by one movement and d column being 4th can be made 2nd by two movements and hence the total movements are 2+1=3, i e. odd; therefore minus sign. Similarly C1 de Cg Now [ column being 3rd can be made first by two movernents and d column being fourth can be made 2nd by two movements.

·102 Determinants | CA C3 Hence the total movements are 2+2=4, l.e. even: therefore plus sign. Therefore in the light of above the expansion of the determinant is 41 bi d₂ di C1 bg 02 di 02 ba +1 a di ba Iti bi d3 ds ag d₂ b2 C2 bi by Cg di by d. h C4 de 24 The above expansion is known as Laplace's expansion of a determinant. The method will be of great use when there are zeros in any two rows or two columns, e.g. ai ba 0 0 I 11 11 03 C 1 = C2 Az b2 0 0 bg ds 04 be de By the above rule the determinant will be a, b, C3 d₃ b, d, ie zero a be 21 0 a, 0 ba da zero + 0,0 ag b₂ c₃ d zerot zero b, CA a de 2:11 16:|| a, 0 h 0 by 0 0 0 C3 i.e. zero- = d: hy zero 0 0 ·11 12 s Cg o 0 0 d3 determinant be A, then we may denote the reciprocal Determinants of Special Type 103 bi Cg dz be C4 de In a similar manner we can prove that be d. bg Pz b, Ps Is 0 0 0 Bu 0 B2 0 Bis 78 bi C1 Bi be C2 B Y by By Because if we expand the determinant by the help of first three columns and their minors, we shall get only product given in the R. H. S. All other determinants shall vanish because one row will be zero. From above we conclude that if a determinant of 2mth order contains in any position a square of m2 zeroes it can be expressed as the product of two determinants of the mth order. (Agra M. Sc. 50) 8:23. / Reciprocal Determinant. Definition :--If in a given determinant each constituent be replaced by its cofactor, then the determinant so formed is caled reciprocal or inverse determinant. If the original de 1 determinant by A'.

·104 Determinants Properties :-'-**-! where n is the order of the determinant is. the reciprocal determinant (original determinant)*- (Cal. IJons. 65; Agra M. Sc. 61, 66; Delhi Hons. 65) We have already proved the above property in the casc of a determinant of 3rd order, lc. B, C a B, C, ho C ba IQ. 7, P. 52 Similarly we may consider a determinant of the oth order. Let =) ag be C .. be ng ... ... .. ... .. ... ... .. boy CA ... B N B. C2 N, .. CA B NO Also we know that Q, Az+bB:+C+...Nq=s etc. and 4,42+bqBz+c2C+... N, etc. IS 12 P. 12-13] ·0 0 4 0 0 I.. 4 0 0 0 and 4,9-) is reciprocal determinant of 4, and hence product A' in terms of original determinant 4. (Karnatak 62; Punjab 60) Determinants of Special Type 105 With the help of above we have 4X4= A 0 0 0 Hence proved. Cor. The product of two reciprocal determinants is the reciprocal determinant of the product of the two orignal deter- If 42 and 4, be the original determinants then their product is 4,4, which is itself a determinant of nth order and (4,4,)n-1=4,1-14,n-1 But by property 1, 42n-1 is reciprocal determinant of Ja of two reciprocal determinants. Prøperty 2. To find the minor of a reciprocal determinant minants, its reciprocal is Let us first consider a determinant of fourth order say and A=1 A, B, C, D, a, b, c, a by C₂ d₂ A, B, C, D, a, b, c, d, A, B, C, D, A, B, C, D, 44z=first minor of A' which is of order 3.

·106 Determinants B. C D В. C, D, B. CA D. 1 0 0 0 (Note carefully) B₂ C₂ D₂ Ag B, C, D, B, C D :. 4x4 Asa a, b, c, do 1x1 10 " bg ca da A, B, C, D, a, b, cs da A, B, C, D, 0 0 A, B, C, Da 4 000 =9,3*. -- a, 4 0 0 a, 00 d. 004 01-4, 4-2,4-0,49-1. ..(0 of 3rd order is Now complementary of the minor 5A, dy and the corresponding from the original delermont 3 is@i Again let us consider a minor of 2nd order of reciprocal determinant. J'han, 2nd minor of J' which is of order 2. 11 nhtained by suppressing the loss and columns that constituis the determinint whose leading terms is doo,.

·Determinantr of Special Type 107 C, D, 1 0 0 Saty (Note carefully) C D B, C, D, B, C, D. 1 0 0 0 (Note carefully) 0 0 0 A B, C₃ B, C D :: 4x d'1,32 a, b, c, *1 1 0 0 0 a, b, c, de 0 1 0 0 a, b, ca da Ag Bg C, DA A, Bu CA D a, b, 00 a bi 7 lloa a, by 0 0 a, b, +all other determinants vanish (Laplace method). ag bg 40 ay be 04 We have expanded in terms of minors of 2nd order formed from the first two rows. a bi 14+all other determinants vanishing. a, by AXA 12B *013.- A' :=(azba-aba) =(226z--abs) 48–1.

·108 Determinants is I Now complementary of the minor d'AqO, of 2nd order is A, B, and the corresponding from the original A, B, determinant is ag b =0,bebas he Similarly if we have a determinants of 5th order, then A, B, C, D, E, 14.* A, B, C, D, E, A, B, C, D, E3 A, B, C, D, E. Then d'A1B2 A, B, C, D, Es C, D, Es CA DE. porco. C, D, E, ie, minor of 3rd order and it can be shown to be by 1-(a,b,-aba) 42. 02 ba In the light of above we have the general properly, that minor of order p of the reciprocal determinant is equal to the complementary of that minor (but taken from the original determinanı) multiplied by 10-1, i e. (original determinantjes Property 3. If a determinant vanishes then i's reciprocal determinant also vanishes A'=n-(property 1)=, ; 4= (given) Property 4. Il a determinant ranishes then each mmor of the reciprocal determinant ranishes, and the constituent of any ·Determinants of Special Type 109 * 1857 | (0.63 - ABC-7 row of the reciprocal determinanı are proportional to those of other row and similarly the columns By property 2 each minor of order p will contain 40-1 as its factor and as such it is zero because A=0 (given), Let us consider a determinant of fourth order. 4'4182=1 C, D =(4,69 -226) 42-1=0, 4=0. C₂ DA D .. C D,-CD=0 or ...(1) Da Similarly 4%B2C=1 43 D = b c 4-1=0 ; 4=0 A. Da bz ca C :: A3D.-A D3=0, i.e. from (1). AD-C Similarly we can prove that B, C, D, ie. the B, CD. elements of 3rd row of reciprocal determinant are proportional to those of fourth row or in general to any row. Similarly we may prove for columas. $ 25. Symmetrical Determinants. (Agra M. Sc. 51) Conjugate Elements. Any two constituents of a determinant are said to be conjugate if they are situated on the opposite sides of the rrincipal diagonal in a line perpendicular to it and at equal distances from it, ie, in the a, b, di the elements by, @zi (1, az; a, b, cy do determinant a, b, c, d, d1, 2; ca, ba; d., bridge ca are all conjugate elements. If in a determinant every two conjugate elements are equal ·which is a symmetrical determinant as every two conjugate No. 2. Hence we conclude that square of any determi 110 Determinants then such a determinant is called symmetrical determinant. e & ah & 1 is a symmetrical determinant. hbs Properties. It is casy to judge that the cofactors of an two conjugate clements of a symmetrical determinant ar same, 1 e. cofactor of h in first row is – (ch-/8) and of k i first colunin is –(ch--S.), i e. they are equal. Similarly tha of gin first row is +(18-bg) and of & in first column i +th--bg) 1.c. they are equal No 1. Hence ne conclude that reciprocal of a symmelni cal determinant is also a symmetrical determinant. Again X a, b, c, a, b, a, b, c. 0,*+b?+0,2 2,42+baba+Cc. 2,03 + balstars anaz+b2ba+fire 0;*+b+c? 0903+Bqhs +cols 0749+b, bx+cisg agas+bahaCoca elements are equal. a by a 11 ag:+6,*+ nant is a synimetrical determinant Again consider a symmetrical determinant of 4th order. & P b f ſ 9 d ·Determinants of Special Type 111 First minor of a is b ſa which is symmetrical. d First minor of b is a 8 p which is also symmetrical. 8 C r r р d Similarly first minors of e ond d are also symmetrical determinants. No. 3. Hence we conclude that all the leading first minors of a symmetrical determinant are symmetrical determinants. Ortho-symmetrica) Determinants. A symmetrical determinant in which all the elements of the principal diagonal are equal is said to be ortho-symmetric 'determinant, e.g. b d is ortho symmetric. a c ba c da b d c b (See Q. 42 P. 49) In order to factorize the above type of determinants as explained in Q. 42 P. 49, we add all the columns to first and find that a+b+c+d is a factor. Again we add first and 2nd columns and subtract 3rd and 4th, thus giving at-b-cd as a factor. Again add Ist and 3rd and subtract 2nd and 4th thus giving a-t---d as a factor. Similarly we add first and fourth and subtract 2nd and 3rd columns giving a+d-b-c as a factor. Having found all the factors the remaining factor which will be a constant ·112 Determinants is obtained by comparing the coefficient of at of the product of diagonal element with the corresponding coefficient of ar in the product of the factors. (See also Q. 44 P 50) $25. Skew and Skew-Symmetric Determinants. (Vikram M. Sc. 59 ; Agra M. Sc. 51, 57, 60) Skew Determinant. A determinant in which every ele. ment on one side of the principal diagonal is equal to its conjugate with sign changed is called skew determinant, e. g. h Pis skew determinant. a 0 f 9 C (Nagpur 61) d Shew. Symmetric Determinant. A determinant in which every clement (including those in the principal diagonal) ore equal to their conjugate with sign changed is called a shem sy mmetric determinant. Since every element in the principal diagonal is its conjugate, therefore by definition of skew-symmetric detero minant a should be equal to --a 200 Similarly b=0, 0-0, d=0 i.e. all the elements in principal diagonal should be zero. Hence a shew-symmetric determinant is zero arial. Property 1. A shea-symmetric determinant of odd order vanishes. (Nagpur M. Sc. 61; Nagpur B. Sc. 61; Sanhat! llons. 65; Calcutta llons. 53 ; Agra M. Sc. 43, 45, 49, 60; 1.e. or 0-0, Ph. 59) Let us consider a skew-symmetric determinant of fiſib order.

·Determinant of Special Type 113 0 11 P 0 f | Now we know that if in a determinant rows be changed to columns and columns to rows then the determinant remains unchanged. 0 n -9 0 0 . 0 h 0 - - -02 8 f 0 P 0 1 ni n 1 0 Now take out (-1) common from each row. :. 4=(-1) 0 h 0 8 P 71 0 r n -9 0 0 ... m 43-A or 24-0ic. 450. Hence proved, Property 2. The reciprocal of skew-symnietric determinant of odd order is a symmetric determinant. In order to find the reciprocal determinant we have to replace each constituent by its co-factor and in order to prove that it is symmetric we must prove co-factors corresponding to every two conjugate elements are equal and of the same sign.

·114 Determinants Consider the skew-symmetric determinant of 5th order given above. Co-factor corresponding to h in the first row is f 9 ni 0 n 0 -1 0 Co-factor corresponding to -h in the first column which is conjugate of h in the first row is h р h -f -9 0 n 8 0 -9 0 р r 0 n - 0 1 0 by changiog rows into columns and columns into rows. Take out (-1) common from each row =-(-1)' -h 9 m1 m 0 r n 0 r -P 0 0 -P- 0 0 which is same as (A).and also with the same sigri. Similarly we can prove that co-factors corresponding to any two conjugate elements are equal with the same sign. Hence the reciprocal of a skew.symmetric determinant of odd order is a symmetric determinant. Property 3. The reciprocal of a sken-symmetic detero minant of even order is a shew-symmetric determinant. (Agra M.Sc. 50) ·Determinants of Special Type 115 In order to find the reciprocal determinant, we have to replace each element by its co-factor and in order to prove that it is skew-symmetric determinant, we must prove that co-factors corresponding to every two conjugate elements arc equal and of opposite signs. Also we should show that cofactors of the terms in the principal diagonal are zero as a skew-symmetric determinant is zero axial. Let us consider a skew-symmetric determinant of even 8 order say P 0 f 9 0 r -P -9 0 Co-factor corresponding to h in the first row is 9 0 r P 0 Co-factor corresponding to -h in the first column which is conjugate of h in the first row is P h 8 0 r g 0 P 0 0 by changing rows into columns and columns into rows. Take out (-1) common from each row. =-(-1) -h s 1 . ::E:: which is the same as (A) but with opposite sign.

·116 Determinants Similarly we can prove that cofactors corresponding to any two conjugate elements are equal but with opposite signs. Again cofactor of the elements of the principal diagonal e g. of 0 in the top is 0 fa 0 -9 0 which is a skew-symmetric determinant of odd order any by Property (1) it is zero. Similarly co-factors of all the elements in the principal diagonal are zero. Hence all the above things combined together prove that reciprocal of a skewsymmetric determinant of even order is a skew-symmetric determinant. Property 4. A skew.symmetric determinant of even order is a perfect square. (Vikram M. Sc. 59 ; Agra M. Sc. 52, 55 (57, 64, 66 ; Cal. Hons's. 64 ; 1. A. S. 55 ; Nagpur 61 ; Karuatak 62) Let us consider a determinant of fourth order, 0 h 8 aud let its reciprocal determinant be f 0 r P 0 B D which is a skewsymmetric determinant by Property 3 Ag B, C2 D. B3 Cs B CS D.

·Now 4 A,B2 Determinants of Special Type 117 :: Aq=B=Cg=D=0 and every two conjugate elements are equal and of opposite signs, i.e. B, -A, C,=-Ag, D=-C-By, D,E-Ba etc Cg Dj = (Prop. 2 P. 104) C4 D. (C.De-C,D3)=h24. Now Cy=D,=0 and D, -Ca. .: Chor i.e. perfect square. Hence proved. Or =( fi Now Cd is cyfactor of -- 0 p 0 T =-[h (1+pS)--8 (hq))=-1[08–89+hr). :: = =(p8–89+hir): Rule. Write down the last three elements of first row in one line ; then last two of 2nd row and last of third row in one line, i.e. . h 8 P ja The value is written as under. Add the cross product of extremes and subiract the product of middle and then square, i.e. (hr-89+p%)*. 0 b then write a b c de f d ., 4-af-be+cd). If A=1 c 0 e 0 (Agra M. Sc. 48: 1. A, S, 52) 0 ·118 Determinants Note. Alternative method for this may be seen in Q. 27, page 40. Property 5. The square of any determinant of eren order can be expressed as a skew, symmetric determinant. Let 4= a bicy de =1 -b, az -d, a a, b, c, d, -bg ag a, b, c, d, —bg as d, c, ay be code -b og der a, b, c, d, by a d, c, a, b, c. d. -bg ag --d, Cg a, b, c, d, by a. d, f, as ba ca da -bag -da a Now multiply the determinants and you will find it to be of the form 0 B CI, 0 P 0 -4 0 which is a skew-symmetric determinant of even order whose value is (Ar-B4+Cp)*. $ 26. If a determinant A whose value is zero be bordered in any manner, the product of the determinant so formed jar A' by the leadine firsi minor of A (l.e. leading 2nd minor sp 3') is equal to the product of two homogeneous functions of addal constituents.

·Determinants of Special Type 119 Let us choose a determinant of fourth order, bi di =0 (given). d. be C2 d, ag bg Cg de a be C4 da If capital letters denote the corresponding cofactors, then it can be written as A=Q,44 +0.4,+-agAstanda Da Aq+6,B2+Ci+d,D, Its leading first minor is Cg de ...(2) Cg ds نبا 1 bi CE de Let us border this determinant as shown below: 4' do B 7 8 G' ay by sy di az b, c, d, a, b, c, d, a, b, a da Its reciprocal determinant may be wriiten as ki kg kis ki ...

·120 Determinants Now, ki is a minor of 2nd order of the reciprocal determinant. kg =42-1 multiplied by complementary of the minor but taken from A' (not from reciprocal) =4 by cgd, =4' x leading first minor of 4 (Prop. 2. P. 107). b, c, d, ba ca da :: 4' x leading first minor of 4=kiki-keks. Now we have to find the values of ki, ką, kg, k, which are cofactors of a, a, a' and a, of A. kn=cofactor of P of A'=1 a, b, c, d, 1=0 (given) a, b, co da as bg ca da Since kı=0, we need not calculaie ka. kg=cofactor of a of A'= a be a di B' by Cyd, go bg ca da d =--(Aya'-+- A,B'+ Agy' +428), from (1), ky=cofactor of or of Ary 8 a, b, a, b, c, d, =-(Aja +-B,B+CY+D8), ·and its leading 2nd minor (1.e. first ninor of A) have opposite leading first minor vanishes (i.e. 4-0), the determinant fself Dejerneinants of Special Type 121 .from (3), we get 4' x leading first minor of 4-gha =-(4107-B46+Cix+0,8) (Ana'+428'+ A y'+428) l.e. product of two linear homogeneous functions of the added constituents a, b, y, 8 and a', s', y', s'. Cor. If the given determinant be symmetrical then its reciprocal determinant is also symmetrical (Prop. P. 107) le. cofactors of every two conjugate elements are equal. Therefore By=ng, Co=Ag, D, A Now if a'ra, f'=B, Y=y, 8=8, then the two factors in (4) in the R. H S. become cqual l.e. a square. Hence we have the following: If a symmetrical determinant whose value is zero be bordered symmetrically (ie. ara'. B-B'), then the product of the determinane so formed by its leading 2nd minor (which will be leading first ninor of original given determinant, Note) is equal to square with a -ive sign of linear homogeneous function of the bordering constituents i.e. 4'xits leading 2nd minor or A' x first minor of A ---(Ayat AB+slay+448)2 Another form. Now A' is obtained by bordering a symmetrical determinant A symmetrically and as such a' itself is symmetrical and its cading first minor i.e. d is zero (given) and its leading 2nd minor is leading first minor Therefore regarding A' as a given determinant we can Say in the ligjit of above that If in any symmetrical determinant (4' is symmetrical) the of given determinant A.

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·Determinants of Special Type 123 .. 4' x leading first minor of A zm(Aza-t-B.B+Cy) (410'+4.8'+Asy') (26 P. 121) A' XL 0 a -(a'a t-abB+acy) (a'a'+abb'+cay'). or == 0 0 Cancel a2 from both sides .. '=-(aa+bB+cy) (aa'+bB'+ey'). $ 27. Expansion of a determinant in terms of the constituents of principal diagonal. Let there be a determinanti di bi C1 di Az b, C2 d, d3 bg d₃ 04 be CA de Just as we expand the determinant in terms of any row or any column but here we want to expand in terms of constituents of leading diagonal, i.e. algo ba, fan dg. For the sake of distinction and clarity let us take them to be A, B, C and D respectively, so that bi C1 di ...(1) B CZ dia . ag b₂ с dz be CE D The expansion of this determinant is written as follows: A=4+Ep.A+Eq.AB+ABCD. (Note that there is no term of the typ: Er.ABC.) de contains those therms of 4 which do not contain any of the diagonal elements and is, therefore, obtained by putting all A, B, C and D equal to zero.

·124 Determinants 0 di d 0 d, az bg 0 d a be 0 EpA contains those terms in which the leading coefficients occur one at a time. Clearly this sigma will contain four terms, one each with A, B, C and D. In order to find the coefficient of A, we put B, C, D equal to zero in the determinant that is left after crossing the rows and columns which intersect at A, thus giving the coefficient of A as 0 C2 d, 0 dg be 0 Similarly the coefficient of B is obtained by putting A, C and D equal to zero in the determinant that is left after crossing the row and column intersecting at B. coefficient of B 1s 0 .. ag 0 dg 04 C4 0 Similarly we can write down the coefficients of C and D. Thus we get all the terms of Ep.A. Eq. AB contains those terms in which the leading coefficients occur two at a time. There being four leading coefficients, their combinations taken two at a time will give 'Cg, i.e. 6 terms in the signs of the type AB, AC, AD. BC, BD and CD. In order to find the coefficient of AB we put C and D cqual to zero in the determinint that is left after crossins ·Determinants of Special Type 125 the two rows and two columns that contain A and B. Thus the coefficients of AB is į 0 d, is 0 CA Similarly coefficient of BC is 0 di etc. 04 0 We have seen that three is no terms of the type Er. ABC because for that we will have to put D-0 in the determinant that is left after crossing the rows and columns which contain A, B, and C and clearly it is zero i.e. coefficient of ABC is zero. Similarly coefficient of all other terms of type ABC, CDA and CDB will be zero when the determinant is of the fourth order, The last terms in the above expansion is clearly ABCD. If however the determinant be of fith order then it will not contain any terms of the type As.ABCD, 1.e. which contains four leading coefficients at a time. bi d, 0 C1 az 0 ag bg 0 be +A C 0 0 c, d, HB10 (de by o dz as 0 d, b, c, 0 0 +0 0 b, d+D 1 A o da C ·126 Determinants +AB Odg ;+ AC 0 d, +DA *10 1 bi 0 0 C b, 0 0 be + BC 0 de I+BC +CD 4,0 ag 0 a, 0 + ABCD. Ex. 2. Expand the skew-determinant b in powers of x. a c -a d e -bd X ſ . C 0 e Since we have to expand the determinant in powers of xi.e. the constituent of diagonal, we follow the method explained in $ 27. 4,=constant term in which no x occurs is obtaind by putting all the x's in diagonal elements equal to zero. 0 b which is skew-symmetric determinant of even order d and is therefore a perfect square --6 0 (Property 4. P. 116) 0 whose value is (af—be +cd). (See Rule P. 114) The coefficient of x3 i.e. x.x.x will be clearly zero. The coefficient of x2 i.e. x.x in the Ist and 2nd columns and rows is obtained by putting the two x's equal to zero in the determinant which is left by corssing the first and second TOWS ·Deteminants of Special Type 127 0 and columns and is therefore Lolan f? Similarly We can obtain other coefficients of x and they will be a”, b*, c, d, e. Thus the total coefficient of x is (a* +62 +62+da+es+fº) The coefficient of or in the first row and first column is obtained by putting the three x's equal to zero in the determinant which is left by crossing the first row and find column and is therefore 0 which is skew-symmetric determi. nant of odd order and is therefore 0 ſ zero (Prop. 1, P. 112) e 0 Similarly all the coefficients of x4 will be zero. Clearly the coefficient of xt is unity, .: 4****** (a24-62+c+dates +f2+(af-be+cd). Ex. 3. Expand the skew determinant 8 р in terms of elements of the principal dia- B b f gonal, (Agra M. Sc 19 50; с Nagpur 61) rz с r D S -S E Clearly coefficient of ABCDE=unity. There will be no term involving four of the five letters A, B, C, D, and E. Coefficient of ABC is obtained by putting D and E equal to zero and is -s. There will be 5C, such 0 ·128 Determinants 5! coefficients, i.e. =10 and they can be seen to be 21.3 a, b, c, f?, g? :p?;y?, r and sº. Coefficient of AB is obtained be putting C, D and E all equal to zero and is which being a skew-symmetric determinant of odd order is zero. -C 0 о с г s Therefore all other coefficients of the terms of the type AB etc. 6C, in number will be each zero. Coefficient of A is obtained by putting B, C, D and E all equal to zero and is 0 b j 2 0 which being a skew-symmetric determinant of even order is a perfect square. (Property 4 P. 116) 0 S 0 Its value is bf q, i.e. (bs -fr+c9) (See Rule P. 117). There will be *C, i.e. 5 such coefficients. A=ABCDE+SABC.a+EA (bs-fr-c9). $ 28. Circulant Determinants. Ex. 1. Explain the term circulant. Resolve the following determinant into linear factors, ·ag ds as and Consider the cquation x-1=0 and let its roots be 1, 0, 0, for 1, 2, 0%, , . Add to the first column the constituents of remaining columns multiplied by K, K'?, K??, K4. Since K has five different values and you will find that each constituent of first column in each transformation is same. hence we say adding all the columns to first column. It is casily seen that a, tag+da+de+a, is a factor. +X°C, ie. Ci+6C +°C -+-63C4+C ; the new first colun i Determinants of Special Type 129 0 1 an 04 is a; no (ig ag 02 fin 03 as (Nagpur 03 ag M. Sc. 61) Definition :-A determinant' is said to be circulant if the constituent in each row be the same occuring in a circukur order, but a different one standing first in each row. In the above row, the last quantity of first row is We start the 2nd row with as followed by az, ag...ending the tow by an, and again start the 3rd row with as followed by 5. Q,, ag, dg and so on. Rule to solve. Here the determinant is of 5th order. PR, 8x as we know that n nth roots of unity form a G: P. (Dehli Hon's, 65) Now take K for any of the above roots, i.e. K standing Thus Taking K=1, then K, K, K3, K'* arc all equal to !, Taking K=0, we say, apply Cz+KC +K%C3+X°C Also 65-1 You will have five factors. becomes ·130 Determinants 1 07+, +6*03+Hartola, as+an+82a3+8803+0424 94 Og +-tastaz+*a, +htag A3 az +804+003+6+4, +pla, , +as+*24+6%+fa, a If we devide and multiply each of the above by 1. 04, os. 6%, 0, respectively and put 96-1, 68=05.0=0, 88=85.93=49 ie, making the coefficient of a, unity we will observe that each element becomes OS .. 4, +822+03+-e'a, +6 ag showing thereby that it is a factor. ...(2) Similarly taking k=89 we say apply C,+KC2+K?Cz+KC4+KCg Cita-Cg +64C3+0Cq+0%C5 fo=95.8=9 and 48=05.03 =23 and proceeding as above we will find that each of the consti. tuents of new first column is 4+-6a2+8*ag+004+03as ....(3) and hence it is a factor, Similarly taking K=A3 and as we will find the other factors. A=L(Qz+aztaz+astag) (artea,+8-09 +030, +84as) (+62a, +0'az+Bag +8543) (ar to'a,+003+049 +0*ag) (a+ha,++0%ast-Bas) Comparing the coefficient of as in the given determi. nant (from diagonal terms) and in the product of above factors we find that L=1. Hence the factors.

·Determinants of Special Type 131 Ex. 2. Resolve into factors the circulant d 6 с с b b C a Consider the equation x-1=0 and let its roots be 1, 6,0where 1. Let X stand for each of the above roots. Taking K=1, then apply Co+KCo+K*C, i.e. Cs+Co+Cs. X=K=1, when K=1. The new first column gives a+b+c as a factor. Taking K=0. we apply Cz+KC+K?C, or C+0C, +6C, the new first column becomes a-t-604-02 c+al-t-b92 02 b+co-+12 0. Multiplying and dividing by 1, 82, 0 in order to make the coefficient of a unity, this column can be written as a+60+092 - 1 g=03.0=0 (a+60+60%) 12 1 (a+b9--c0). Thus we observe that a +68+602 is also a factor. ...(2) Similarly taking K=82 and proceeding as above we find that a-bga +60 is a factor. .: 4L (a+b+c) (a +-684-002) (a+b82.4-co). Comparing the coellicient of a we find that L=l. Ex. 3. If w is a cube root of unity, prove that atbus + cur is a factor of ·132 Determinants 6 b 1 6 (Rajasthan 65) (Agra 54) a b C h b C a b c b b We have interchanged 2nd and 3rd rows and hence changed the sign and now the determinant is circulant and its factors as shown before are (a+b+c) (a+be+c8) (a+b82+ch) :. 45-(a+b+c) (a +60+202) (a+b024.c) where 1, 0, 62 are the roots of x-1=0 which are given to be 1, w, ws, Ex 4. Prove that the product of two circulants of the same order is a circulant. (Nagpur M. Sc. 61) Consider the product of two circulants b 1x1 a C . a b Y b C a 2 axt by..cz cx+ay+bz a:+bx+cy cztax-tuhy bz+cx+ay ar+by+cz=A, bx+cy+az=B, ext-aytb:=C, ay-tb:+ex cy'+az-tbr by-+cztax baxt-cyta: If we put ·Determinants of Special Type 133 then the above determinant is B C B B с which is a circulant. Exercise 2 Q. 1. Prove that X =1*** +32+za+x2+y'2+z'2 +(xx' +Jy'+zz') 1 1 X' columns. Expand in terms of minors formed from the first two IS 22 P, 97] 1 1 -r' 1 + y 2 Ili | | | Now simplify etc. Q. 1. Prove that B 8 a a B 8 B ·134 Determinants =Q++84 +84 +84-2a282–2a?y? --2a²82--2B7% -28282-27*8* - Saßys. Expand as above in Q. 1, in terms of minors from the first two columns Q. 3. Prove the following identity and expand := 0 0 у 1 X 0 22 ya 1 - 0 x2 y 0 x 1 jua у X 0 0 =x*+y4+24 --2y2z2-22?x2–2x®y. Multiply C, by xy-, C, by X, C, by y and C, by 2 and therefore divide by T Xys. x y z 0 X4763 Xyz Z 0 ya: Xy's 0 xyz 372 0 Now take yz from R2, zx from R, and xy from Re 0. 4=*]" 12.X x R. H. S. x 0 Z y 0 0 In order to find the value, proced as in Q. 1 and 2.

·Determinants of Special Type 135 Note. The two determinants have been separately evaluated before in Q. 41. and 45 P. 48 and 49 respectively. Q. 4. Prove that a b ! 6 f c 3 0 0 0 0 =aa* +682 + cya +25By+2gya +211aß, where a=y='-y'z, B=zx'-:'x and y=xy' - x'y. (Nagpur 61) Here C. and Cg can be made C, and C, by (3+3), i.e. 6 movements of columns thereby causing no change and as such we expand the determinant as above in terms of minors of C, and Cg, and we get h bf y EN + J' ' y All other determinants vanish as each determinant of 2nd order will have a row of zeros. Now expand and put the 8 and y. > [ga-S(--B)+cy]-(--) {ha-b(-B)+fylta[da-h(-8)+gy =aq+683 +672 +2 By-+2gya+2haß. values of a ·136 Determinants 5. If fı (x)=2773 +352x +39;x+de, : (x)=22x3 +35r2+3c2x+de, f (x)=1378 + 3b3.12 +3c9r+dg, prove the identity f1(x) Si'(x) "(x) 18! S2(x) 1,'(x) $"(x) by C1 i fa(x) f'(x) f3"(x) =- 1 di 02 1 03 dz (Agra M.Sc. 53) A== 24x3 + 352x2+3678+d, 3 (a,x2+25,x+c) 6 (9.7+h) agx3 +3bºx?+36,1+d, 3 (azx2 +2b28+ca) 6 (a,x+62) Azr3+367x +3c2r+dz 3 (agxa +2b7x+3) 6 (a,x+hz) Take 3 and 6 common from C, and C, and then apply C,-XCg. 4=18 Qqx*+354 +2+36x+dbir-teartha 1 2,13 +359 +2+362r+d, bert.co Qr+ba a3+3+36372 +3c3F+d; byrtez agr+hs interchange C Now apply C,- XC7--2xC, and then and C, A=-18 ar+b, birta fortale aarth₂ borte cartda agx+bg birtcs Cartrig The deteminant on R.H.S. is of fourth order and as such I can be written as ·Determinants of Special Type 137 A=-18 1 0 0 0 a ax+b, bexto, carta, A ₂x+ba haute Cgxtil аз ах +h, bax+с, сах-+d. Now apply C, -C, and then in the new determinant thus obtained again Cz-xC, and then again in the new determinant thus obtained Co-xC, and you will get the determinant pn R.H.S. 6. (a) if f(x)=(x-a) (1–8) (x−y) (x–8), then prove Hat x =f(x)-xf'(x). (Agra M. Sc. 44, 51, 55 ; Delhi Hors. 56 a X X X 8 Taking log of both sides of the given relation, we get log S (x)=log (x-a)+log (x–8)+log (x−y)+log (x–8). f (x) f (x), f (x), f(x) Differentiating we get f'(x)= + X-0 Now applying R-Rg, Re-R3, Rg-Ron A, we get -B 0 0 + a-X 0 B-X 0 0 0 Y-X 8 =(-1)B-X X-y 0 0 0 0 Y-XX-8 B-XXーン 0 S 0 V-XIS ==(a-1) [(8-) {(Y-x) 8~* (7-8)}+x (r-Y) (--d)] -*(x-B) (x-7)(x-8).

·138 Determinants 1e. Collecting the terms of x and writing cach factor as X-cctc., a- . to be written as -(x-a) =(x-a) (x-)(x-y) (4-8-r1-* [(x-a) (6-8) (.1-8) += a) (r-Y) (7-8)+(x-8)(r-Y) (x--)! Taking (x-2) (PB)(x-7)(x--S) common. 1 1 =(x-a) (x–P) (+=») (x=0[1-4 litrato atos tips] I-a lir) fix) =f(x)-- * + + -B 8-tenge =f()-xf'(x) by (1). (h) If S (x)=(4,-2) (az --X) (ag-) (a,-x) (as-x), show that d1 ar =f (*)-- dx a. . 1 2 . X x x .24 72 . 4 . (Agra M. Sc. 61) Apply R.-R., Re-Rg, Re-R, R-Rs. az - Xa 0 0 C 0 02-XX-a0 0 0 D 0 0 04-XX-as ·Determinants of Special Type 139 Expanding with first column, A=(4,-*)| a;-* x-3 0 0 0 ag- 0 X 05 +x 0 0 0 02-X . 0 0 07-XX- 0 0 =(Q1-x) 0 04-XX- -x)} ax X-04 0 11 0 04-XX-08 0 0 Og X- 0 0 +x (x-a) (x-az) (x-as) (x - ay) (0, -x) (ay-x) (az--x) {a, (ar --X)-x (r-0)} +x (x-04) (x--as)] - (0-1) [(x--as) (x-2)(x-1)] +*(x-2)(x-0z) (x-as) (x-osl. Collecting the terms of .x and writing each factor as a-X instead of x-a and thus adjusting the sign, 4=(aq-_x) (a,-r) (az--x) (09-x) [(az-x)+ x) +x|(9,-2)(a,-*) (az*) (ag-x) +(9,- x)(az-X) (01-x) (05 -- ) +(91--*) (as-x) (04-x) (as-x) +(az-) (az-7) (@-x) (Qs --x)].

·'W.X 140 Determinants itor)] T - *** s(r) S(K) =)=-16 Nos talc (a, - *) (0, 1) (0, -1) (91-r) (ig-*) common. :: d=f.x) ++ + + + 0,- da - (1) - -xl () As in purt (a) taking log, sir) () 1'(x)= + + Q. 7. Show that the determinant 2.x+12+12 (-) (-b). (1-C) 11+as+cro-1 (1-a) : (1-1) 1: (1) : 1:* +1?tail-1 contains x (18+:)-1 as a factor and show also that the remaining factor is independent of a The given determinant can be made of fourth order without changing its value. i.e. 0 0 0 dr dia + cya (1-с) XP (1-0) a: +6:2-1 d(4-6).ry 199+ ar? +or- (1-a) y: J: (1-0) x: (-a) ^:? th? taj-2 Apply C1C, C-1C, C-:C. 1 de cy +62-1 -br: C.X az?+Cl- A: baz 11?ta!

·Determinants of Special Type 141 Now afply Rz-Ro-yRg-:Re, thus making the first row of the new determinant as I-(r2 +1242) 0 0 0 all other constituents remaining unchanged Now expanding with first row, 4=(1-1 (*2+12+=)] X1 Cy*+bz-1 -C.XY bx -cxy az+cx-1 са bxz ay? bx+ay2-1 Above form shows that A has a factor X (** +32 +-22)-1 and that the remaining factor is independent of A. Q. 8. Show that (b+c) a bc =(a?+b+c) (a+b+c) x(a-b) (6-c) (c--a).. (c +-a)2 62 (a-+-b)? c? ab (Agra 34) Proceed as in Q. 18 P. 85. Q. 9. Utilize Laplace's Method of expanding a determinant to prove that the product of two determinants can be put as a single determinant. (Agra M.Sc. 42, 53) Consider the determinant b. 0 0 0 ... (1) b, 0 0 0 a. ba 0 0 0 0 0 01 B 71 > 0 0 13 Ba 0 0 da

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·Determinants of Special Type 143 x 0 0 . (B) The latter determinant is clearly 0 0 -1. Hence we have from (A) and (B) by changing the rows into 0 columns and columns into rows. a, b, c, xa1 az 3 ag bg ca B, B2 B3 a, b, c, V1 72 73 0,91+byaz t-Cyag asfit-b18x+CB, 01817b18+6143 0:01+5999 tocza, ali+baba+caßg ag8z+5,72+0,45 Ugant bgag tegas afi+b38a+ c3Bs agyi+6372+-Czys Q. 10. Evaluate .82 29-1 . ... ht .88 ... ... U. .. ... ... ra .- Arrly C. - XC9-2, Comp-xCongo Cros-xCn-s ...Com', Cy-20, 0 0 0 0 0 0 0 0x(1-toys 0 0 . . I. . 0 $ ·144 Deternunants EX 0 0 r(I-X") 5 (1–7") 0 0 0 0 ... 0 0 .. (1-1) 0 .. ... ... ... ... 0 (1 -"} 0 0 0 (1-*") 0 0 0 which is a determinant of (n-1)th order. Now expand it with the last column i.e. (n-1)th which can be made first column by (n--2) movements and hence we will have to attach (-1)- Again we will be left with a determinant of (1-2)th order which has got to be expanded again with its last 1e. (n)th column. This last can he made first by (n-3) movements and hence we will have to attach (-1)-3 and so on. :: SEX(-1)"-(*(1-x)](-1)-3 (v (1~*")) x(-1)"-{v(I-)...(n-1) times (-1)(n-2)+(n-1)+(x-4 + ..3+4+1x. [x(1-x"}}"-". =(-1)(n-2) (n-1)/2 X". (1-x*j-1, Enzen (n+1)/2. Q. 11. (a) Evaluate ba 0 0 0 0 0 An1 baun 0 0 0 0 0 anne in a 0 .. 0 0 0 0 an-s bres 0 0 ... .. ... .. . ... ... . 0 D 0 0 0 og by 0 0 0 ... -1 01 ·Determinants of Special Type 145 The above determinant is called continuant determinant. All the diagonal elements are a's, the line above the diagonal is of b's and that below the diagonal is of — 1. Let us denote the determinant of nth order by 4. Expanding with first row (not with first column), we get An=on4n-1-bm-{-1 (41-2)+0}+0 An=andr-x+bring. =antom or An-1 An-1 Or Sant A-1 ba barn (by (A)] limit Ans b [by (4) annit annot Jn. Now d bg =219+b, and 4,. 여 ai 4,4,4, the Ilence continuing the above process till at last we reach to the calculation or a which is azt a TIence in terms of Contiguou fraction we can write fr-17 Crott... +0,+ ·146 Determinants 1 1 Note. If all the b's be replaced by +1, then the above determinant will be called simple continuant determinant and in that case 1 =ant annat an-at...at a (Agra M.Sc. 48) (b) Prove that the value of the determinant 1+ 0 0 0 Ani 17. . 0 0 .. 0 X 1+.13 Y 0 .. 0 0 0 0 of mth order is 1 +r+...+?m. 4,=1+x?, 4.= 14** . X 1 +.rs (I. A. $. 59) i+***** 1- 0 0 =(1+r) 4,-27 |-+*%*-7 1+1 1+* 0 1+ =(1+x)(1+7+4)-1.x (1+-+*) *(1+**) (14.7*++*-.**] =(1+x)(1+1)=l+rtx+x. Again A.=(1+x) 4.-1- 0 0 1+*** .. 0 0 1 (m-}th order ·Determinants of Special Type 147 Or or Am=(1+x2) An-1-X.c.4m- An=(1+.x*) 410-1-x*.4m-2. Putting m=4, we get 4x=(1+x) 43--13.12. =(1+x)(1+x2 + x4 +16)---(1+2+x+) =(1+x) **+(1+x8*-**) (1+x_x)=1+*+.80 +764-18. Similarly we can show that 45=1+x*-+*++6+7.710 and 4,4=1+x*--** +++ .. +.vn. Proved. Q. 11. (c) Prove that the determinant of nth order. cos a 0 =COS no. 0 0 ... 2 cos 0 1 0 0 ... 0 1 2 cos e 0 . ... ... ... 0 0 0 2 cos o 1 0 0 0 0 2 cos 4,=cos 0, 4,= COS O 1 2 cos 0 or 1.= cos: 8–1*cos 29 Cos 0 1 2 cos 0 1 0 1 2 cos e **COS 0 (4 cos? 0-1)-i (2 cos ') 4 cos' 0-3 cos acos 3..

·148 Determinants COSO 1 0 0 2 cos o 1 0 0 2 cos 1 0 0 1 2 cos e Expanding with last column, cos a 1 0 - 2 cos o cos o 0 1 2 cos 0 1 1 2 cos a 1 2 cos @ 0 0 0 -1/1 (2 cos8-1)]+2 cos 0.43. Now put 4z=cos 30. 4.=--(2 cos? 6-1)+2 cos 6 cos 30 =-cos 26+cos 49+cos 20 = cos 48. Similarly we can prove that Asrcos 50. Hence An=cos no. Q. 12. Evaluate x * ra ... 1 0 0 0 0 1 0 0 . 23 0 1 0 ... ... . ... 1 In-10 0 0 expanding with first column or first row, (Vikram M.Sc. 59 ; Agra M.Sc. 1936, 60) 4-1.(a determinant whose value is clearly unity) -X (a determinant whose value is cleariy x) ·Determinants of Special Type 149 x? .73 +*% * 1 0 0 0 0 0 0 0 ... i.e. tra (10)-1 .13 le..(-x)=-** 0 0 0 ... 0 0 1.c. ic.--.** (**)=-* {recos" 1 0 and so on, .. A=-*-* ...xn-1 1-***+***.78+ ...(-1) terms} = sum of a G. P. 1-1-Xn+. 2r:yeror Alternatire nrethod. In case there be no restriction to expand the determinant either with first column or first row we can evaluate the above determinant as below.

·150 Determinants Multiply C. Cs Cby -X. -r..-**- respectively, . .x) 7-**)... (1) X 0 ... ... ... .. ... r-1 0 0 0 0 Add all the columns to fi st column and you will find that new first column will have zero cverywhere except the first constituent which will be 1-73-..--. Expanding with this first column, 1- pr...--- (-)1-4)...(-1) 0 ... 0 .. .. . 0 0 The abuve determinant of (n-1)th order and is equal to product of the terms in principal diagonal. 1-1,-1,29–3 (-x)(x+jord (-10*(-x) (~.12...(---1} =1-1-...--721-9 ·Determinonis of Special Type =1> {(4)*-1-1 (1*-1) summing a G.P. 78-1-***+. 21-38-1 Proved. X-1 Q. 13. Find the condition that a biquadratic should be capable of being put as sum of two fourth powers and expressing it in the form ax +46x* +601? +4dxte=A (x+)4+B (x++)". find the quadratic whose roots are and $. (Agra M. Sc. 42) On comparing the coefficients, we have A+B=a, ...(1) A8+ Bø=b, ...(2) 462+B?=C, ...(3) BAS - B$=d, 404 + B*=e, ..(5) Now assume that 8 and $ are the roots of x2+2x+y=0 - (6) so that 02-7-28+4=0 ...(7) and $+$+p=0. ...(8) Now multiply (3) by 1, (2) by, and (1) by ji and add A (42+19+*+B ($2+16+x)=c+ba+aj. .. c+biltag=0 from (7) and (8) Similarly d+c+b=0 ...(10) and e+da+H=0. Eliminating and it between (9), (10) and (11), we have the required condition as b 30 or b co or J=0. € b ed e C c e ·152 Determinants Also the quadratic cquation whose roots are and $ is obtained by climinating, and I bciwccn (6) and (9) and (30) and 15 225 =0 1 or 1 c b c - 1 d b b Note: - If the cubic were expressed as sum of two cubes as above, then, also proceeding as above we would obtain the some quadratic giving the values of 0.$. (Sec Author's book on Theory of Equations P. 121) Q. 14. It b 0 0 . с s' 0 0 c' 0 0 0 0 c d 0 0 c' d' 0 0 C prove that cd EEE BOL (Ayra M. Sc. 39) Expand the given determinant in terms of minors formi ed from Ist three rows and you will get so many determinants the first of which will be b 0 which is zero. Similarly you will find that in all the ·Determinants of Special Type 153 factors three will be a determinant with one line of zeros, and hence zero except in the terms where given determinants occur, etc., Q. 15. h & a f C & f If capital letters stand for the cofactors of the corresponding small letters in the above determinant, then prove that h 8 a h b ſ m n M - 0 (Rajputana 60) [A12-+- Bm2 +-cn2+2Fmm+ 2Gnl+2Hlm) (Refer Anthor's Solid Geometry P. 239) Expanding with respect to last column or last row, we get 4=(-1) h bf iton ah & 1-0 * h b f mn m Now h! =anthH+8G=hH+bB+-fF. =&G+fF+cc. hhf 11 the mn =IA+mH+nC i.e. we have replaced a, h, 8 hbf by l, m, n, the other rows giving cofactors remain unchanged and hence capital letters remain unchanged.

·154
Determinants
=-(1H+-mB+nF)
as above.
1:172:1
3
h
8
=IG+mF+nc.
1 m 2
:: 4=(-1) [1A+m+ng)+m (IH+mB+nF)
ton (1G+F+nC)]
*-(A12+ Bm+Cn2+2Fmn+2Gnl+2Him).
Hence proved.

·AGRA UNIVERSITY M. A. & M. Sc. PAPERS 1960 (a) Define skew-symmetric and skew determinants and establish the following properties : (i) A skew-symmetric determinant of odd order vanishes. (ii) The reciprocal of a skew-symmetric determinant of even order is a skew-symmetric determinant. (b) Evaluate 1 X 1 0 0 0 . 0 1 0 0 x3 0 0 1 0 b. . х”- 0 1 0 0 1961 1. (a) Iff(x)=(a2-x) (a2-x) (az-x) (02-x) (as-x), show that az x X x |-} (x)=xLx of x x X dy X X 1 ag (b) Prove that if in any symmetrical determinant, the leading first minor vanishes, the determinant itself and its leading second minor have opposite signs.

·156 Determinants 1962 (C) Define skew-symmetric and skew determinants. Prove that every skew-symmetric determinant of even order is a perfect square. (b) Prove the identity. (b+c) a =2abc (a+b+c) 62 (c+a) c2 (a+b) 1963 (c) Prove that B y =(2+B+») (8-) (y-a) (a-B) a? B2 B+y yta ats! (d) Define a symmetrical determinant, a skew determinant and a skew-symmetric determinant. Show that a skew-symmetrical determinant of odd order has the value zero. 1964 (a) Prove that the reciprocal of any determinant is equal to the (n-1)th power of the given determinant. (b) Prove that a skew-symmetric determinant of even order is a perfect square. (c) Prove the expansion 1+a 1 1 + 04 14a 1 1 d) per 1 1 1 taz 1 1 1 1 1+04 1965 (a) Show that a skew symmetric determinant of an odd order vanishes.

·Calcutta University Hons. Papers 157 (b) Prove that ata ab ac ad ab 62 +1 bc bd dc be c²+1 od ad bd cdd2+2 is divisible by / and find the other factor. 1966 (a) Prove that the reciprocal of a determinant of the nth order is An-l. (b) Prove that a skew-symmetric determinant of even order is a perfect square. (c) Evaluate 1+1 ay 03...... On 01 1+0, Us ..... Az a1 I tag..... QR ......... AL ag Ca...... Item CALCUTTA UNIVERSITY Hons. PAPERS 1961 1. Show that =(x-2)*-* {x+(11-1) a}, . a x 4 a 2 ·158 Determinants where the determinant is of nih order with diagonal elements all r and other elements all a. 2. Resolve into factors the determinant (a-a')' (a-8')? (a-y') (B-a')? (B-8). (8-9) (v-a')? (y-8°) (y-y') 1962 1. Show that the determinant 0 (a-B) (a-Y) is equal to the product of two determinants (B-a) 0 (8-Y) (y --a)? (y-B)? 0 - 2a a? and 1 a -1 -23 Be 1 B B* -2x goz Hence or otherwise, show that 4=2 (a--)? (-y) (y-a)?. 1963 Prove that a? =(a-B) (B-y)(y-a) - a 1 68% 1 Deduce that B2y? ta’82 By+ad 1=(-y) (y-a) (a-B) x(a-8) (-5) (-8). y%22+8462 ya+ge 1 a 82 +7?5° aß+yd 1 ·Calcutta Unirersity Pass. Papers 159 1965 (a) Define the adjoint of a determinant and prove that if D is a determinant of order n and D' is its adjoint, then D'DR-1 (b) Prove that xta 5 с d =x*(x+a+b+c+d). a *+6 c d & b x+c d a b cxtd CALCUTTA UNIVERSITY PASS PAPERS 1965 (a) Show that the value of a determinant of the third order remains unaltered if the rows are changed into columns and the columns into rows. (b) Show that a2+-2ab-cx ac+bx b C 2 ab-t-cx -+ bc-ax - CH X a ac-bx bctax c*+x2 b -a DEHLI UNIVERSITY Hons. PAPERS 1963 (a) If A is a determinant of third order and A' the reciprocal determinant of A, show that 14'=42. (b) Show that a skew-symmetric determinant of odd order is zero. (c) Solve the equations *+y+z=1, ar+by+c:=k, a’x+b4y+ca:=h.

·160 Determinants 1965 (a) Show that a b C d a b c C d a b c 6 d has a +bx+c^2+d' as a factor, where a a root of r=1. Hence or otherwise show that the determinant is equal to (a+b+c+d) (a–b+c-d) {(a—c)2+1b-d)). (b) If Arlaust, A'=Adl be two determinants of order n such that All is the cofactor of as in A, prove that A'=An- What happens when A=0 ? DELHI UNIVERSITY PASS PAPERS 1965 (a) Explain the use of determinants in solving linear simultancous equations. (b) Calculate the value of the determinant an ag 1+1 1+g dy a1 dg 1 tag di ng 1.tes GAUHATI UNIVERSITY Hons. PAPERS 1963 1. Prove the rule for the multiplication of two determinants of the same order. If r, is the sum of the rth powers of the roots of an equation of degree n, then the determinant ·Gauhati University Ilons. Papers 161 So $1 * Sn-1 S2 53 ... 5, 9. .. II. ... Sn-1 Sn Sn+ 590-2 is equal to the square of the product of the differences of the roots taken in pairs. 1964 1. Write down as a determinant the product of ab c 2 b с Y C 2 Hence show that the product of two expressions of the form (p?+93 +78-3pgr) is of the same form. 2. Prove that if 1+1 1-1 1-r 1+2 1-X 1-X 1-+* ... ... ... .. ... 1+ be a determinant or tlie nth order, then D=(2x)*-} {n-(11--2)r}.

·162 Deterniinants 1965 (a) Prove that the determinant ab ac ab bc ac bc is a perfect square, and find its value. (b) Prove that if the cofactors of the elements of a certain row (or column) are multiplied, in order, by the elements of another row (or column), show that the sum of the products is zero (c) Show that a skew-symmetric determinant of odd order vanishes. RAJASTHAN UNIVERSISY B.Sc. PAPERS 1962 1. Prove that abc bº - ca cº-ab Jx b c c2 — ab a² — be 62-ac 21: : :1 62.ca (%-ab al-bc b c =(a + b3 + c3--3abc) 1963 1. Prove that (b+c)a bc (ta): bi ca (a + b) c2 ab =(a2-4-62+c)(a+b+c) (a-b) (6-c) (c-a). 1965 (b) If w is the cube root of unity, prove that athwtowa is a factor of ·Agra Unicrsity B.Sc. Papers 163 a b C b c a с a b and hence find the value of the determinant, 1966 (a) Prove that x=-I is a root of the equation 2-X 3 3 3 4-Y 5 5 4-X 3 and find the other roots. AGRA UNIVERSITY B.Sc. PAPERS 1960 1. (a) Prove that the product of two determinants each of the third order is a determinant of third order. (b) Express 2bc- 62 2ac-62 b2 2ab-c? as a product of two determinants, and find its value. 1961 1. (a) If a+b+c=0, solve the equation X =0 C c ba 6 (6) Use the properties of determinants to solve the equations Arr+by+03+d,=0, r=1, 2, 3.

·164 Determinants 1962 1. Show that (b+c)2 22 a? 2abc (a+b+c). (c+a): b? c? (a +b)? 2, Find the value of j? 1'- .8 Xy' 1963 1. Show that bº+c ah =4a?o?c?. ba c2+ a2 bc са 0? +62 2. Solve the equation a-, =0. a-X a. 2-3 4-1 + x 1964 1. Find the value of Ita 1 1 1 1+b 1+0 1+d ·165 Agra Unilersity B. Sc. Papers 2. If . 1+28 20, prove that wyzw-1 y 3.2 1+13 1+2 1965 1. (a) Evaluate 0 a a a . a а аа Х (b) Show that (6-+c) al (=2abc(a+b+c) (c+a)? b. 2 (a+b)* 1966 2. (a) Show that А, - B, C, A, B, --C a, b, c, A — Bg Ca the capital letters denoting the minors of the corresponding small letters in the determinant on the right. (b) Evaluate 1 + a 1 1-4b 1 1+c 1 1 1+0 ·166 Determinants VIKRAM UNIVERSITY B. Sc. PAPERS 1962 1. Find the value of q* +-12 ab+cx ca--bX X 2 -6 ab- 62+4* bctal d ca+b, bc-ac cºtA b 1963 1. Show that +43 =4abc. с c с 52 +c? a catal b b 1961 1. Show that (b+c)* 23 =2abc (a+b+c)* bº (c + a)' c (a+b): cº 1965 2. (a) Prove that 2bc-a? ? 63 c 208-02 b c 6 at Zab-e? (a* ++-3abc).

·Sagar Unirersity B. Sc. F'apers
167
1966
1. (a) Prove that
1 betad b*c+a'd?
I ca+hd raa+hdl
1 ahta a*b*+-
--hm-c)(-a) (a−b) (a-d) (6-d) (c-d).
SAGAR UNIVERSITY B. Sc. PAPERS
1962
1. Prove that 4 5 6 =(x-2y+z).
5 6 7
6 7 8
1963
1. Calculate the value of the determinant
21 17 7 10
24 22 6 10
6
8 2
3
1
5 7 2
2. Prove that
(6+-c)2 02
a ?
=2abc (a+b+c)?
bz (c+a)
(a+b)

·168 Determinants 1964 (a) Prove that if two rows or two columns of a determinant are identical, the determinant vanishes. (b) Express the following determinant as a product of two determinants and find its value : 2bc-a? 62 20c-6 q 52 al 2ab-c2 1965 5. (a) Explain the use of determinants in solving linear simultaneous equations. (b) Calculate the value of the determinant: 21 17 7 10 24 22 6 10 6 8 2 3 5 7 1 2 1966 1. (a) Prove that if two rows of two columns of a determinant are identical, the determinants vanishes. (b) Evaluate. a® Q2 a C 1 ·169 Niliar University Papers BINTAR UNIVERSITY PAPERS 1965 2. (a) Find the value of 14 J: rt: ***;? xy X-+y 1966 Prave that . U1+(+3)(-1). 1 x 1 1 1 1 BIHAR UNIVERSITY IIONS, PAPERS 1966 1. (a) Show that yoz-a-x8 - EX-8 xy-- E xy-22 -** EX-} where per+12+ and u=Pz+*x+xy. (b) If A, B, C ...arc the cofactors of or, bu, C...in the determinant. a be a A B, C, 14. ay by Cg A, B, C, then show that ·170 Determinants KERALA UNIVERSITY 1965 VENKATESWARA AND JIWAJI 1966 Show that 3a 3a 1 l=(a-1) al 02+2a 2a +1 2a+1 a+2 1 1 3 3 1 1965 MYSORE UNIVERSITY PAPERS (a) Prove that a-b-c 29 2a 2b b-c-a 25 2a C-9-b 2c is perfect cube. 1966 (a) Show that I (980 B I 5 X =(x-a) (x-6) (x-c) I a I b c Gorakhpur 67. Meerut 67, Rajasthan 67 11 =xyzt(1-+*+ 3 1+1 ·Lucknow (b) Prove that 4 5 6 7 5 6 7 y =(1-2y+:-)* 6 7 8 : Lucknow 67 2. (a) Prove that U =art+421-460.x44dx+c U,=ar3+2bx+c Vizb.x2+2crta Ug=0x2+2dx-te, b Uit a b c h U, 6 U, U, U Us 0 с с c c Calcutta 67 4. (a) Factorisc a, B, into four linear factors. a, 87, 72 Aty, y-ta, «+B1 Mysore 67 5 (a) Show that a determinant is unaltered by changing its rows into columns and its columns into rows. (b) Show that: at-6 b+c cta 5 ic b+c' c' ta' a'+b' =2 b' E c"+a" a*+b", +c" c" ·172 Determinants Ic) Evaluate 3 7 9 6 8 4 5 8 7 10 3 5 6 2 98 Allahabad 67, 7. (a) Prove that btc a 6 ca b 4abc a+b (b) Solve the following by Cramer's rule ! *+*+:=11, 2x--6--=0, ' +22=0. Jovan Vlie equation =0. ra C Westion that equations 0px2 + b2x+6=0 and agx?t ber+1=0 possess a common root only if HOSHOOT bg.ca 02 Agra 68. 4. Prove that btc a-cach h-ccta b-a =8abc. c-b C-a a+b

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