DETERMINANTS Chpter 4 All Mthemticl truths re reltive nd conditionl. C.P. STEINMETZ 4. Introduction In the previous chpter, we hve studied bout mtrices nd lgebr of mtrices. We hve lso lernt tht sstem of lgebric equtions cn be expressed in the form of mtrices. This mens, sstem of liner equtions like x + b c x + b c b x c cn be represented s b c. Now, this sstem of equtions hs unique solution or not, is determined b the number b b. (Recll tht if b P.S. Lplce or, b b b 0, then the sstem of liner (749-87 equtions hs unique solution. The number b b b which determines uniqueness of solution is ssocited with the mtrix A b nd is clled the determinnt of A or det A. Determinnts hve wide pplictions in Engineering, Science, Economics, Socil Science, etc. In this chpter, we shll stud determinnts up to order three onl with rel entries. Also, we will stud vrious properties of determinnts, minors, cofctors nd pplictions of determinnts in finding the re of tringle, djoint nd inverse of squre mtrix, consistenc nd inconsistenc of sstem of liner equtions nd solution of liner equtions in two or three vribles using inverse of mtrix. 4. Determinnt To ever squre mtrix A [ ij ] of order n, we cn ssocite number (rel or complex clled determinnt of the squre mtrix A, where ij (i, j th element of A.
04 MATHEMATICS This m be thought of s function which ssocites ech squre mtrix with unique number (rel or complex. If M is the set of squre mtrices, K is the set of numbers (rel or complex nd f : M K is defined b f(a k, where A M nd k K, then f(a is clled the determinnt of A. It is lso denoted b A or det A or. b If A c d Remrks, then determinnt of A is written s A b c d det (A (i For mtrix A, A is red s determinnt of A nd not modulus of A. (ii Onl squre mtrices hve determinnts. 4.. Determinnt of mtrix of order one Let A [] be the mtrix of order, then determinnt of A is defined to be equl to 4.. Determinnt of mtrix of order two Let A then the determinnt of A is defined s: be mtrix of order, det (A A Exmple Evlute 4. Solution We hve 4 ( 4( 4 + 4 8. Exmple Evlute Solution We hve x x x + x x x x + x x (x (x + (x x (x x x + 4..3 Determinnt of mtrix of order 3 3 Determinnt of mtrix of order three cn be determined b expressing it in terms of second order determinnts. This is known s expnsion of determinnt long row (or column. There re six ws of expnding determinnt of order
DETERMINANTS 05 3 corresponding to ech of three rows (R, R nd R 3 nd three columns (C, C nd C 3 giving the sme vlue s shown below. Consider the determinnt of squre mtrix A [ ij ] 3 3 3 i.e., A 3 Expnsion long first Row (R 3 3 33 Step Multipl first element of R b ( ( + [( sum of suffixes in ] nd with the second order determinnt obtined b deleting the elements of first row (R nd first column (C of A s lies in R nd C, i.e., ( + 3 3 33 Step Multipl nd element of R b ( + [( sum of suffixes in ] nd the second order determinnt obtined b deleting elements of first row (R nd nd column (C of A s lies in R nd C, i.e., ( + 3 3 33 Step 3 Multipl third element 3 of R b ( + 3 [( sum of suffixes in 3] nd the second order determinnt obtined b deleting elements of first row (R nd third column (C 3 of A s 3 lies in R nd C 3, i.e., ( + 3 3 3 3 Step 4 Now the expnsion of determinnt of A, tht is, A written s sum of ll three terms obtined in steps, nd 3 bove is given b det A A ( + + ( + 3 3 3 + 3 + ( 3 33 3 33 3 3 or A ( 33 3 3 ( 33 3 3 + 3 ( 3 3
06 MATHEMATICS 33 3 3 33 + 3 3 + 3 3 3 3... ( Note We shll ppl ll four steps together. Expnsion long second row (R Expnding long R, we get A A 3 3 3 3 33 ( ( + ( + 3 + 3 + 3 33 3 33 + 3 3 3 3 ( 33 3 3 + ( 33 3 3 3 ( 3 3 A 33 + 3 3 + 33 3 3 3 3 + 3 3 33 3 3 33 + 3 3 + 3 3 3 3... ( Expnsion long first Column (C B expnding long C, we get A A 3 3 3 3 33 3 3 ( + + ( + 3 33 3 33 + 3 ( 3 + 3 3 ( 33 3 3 ( 33 3 3 + 3 ( 3 3
DETERMINANTS 07 A 33 3 3 33 + 3 3 + 3 3 3 3 33 3 3 33 + 3 3 + 3 3 3 3... (3 Clerl, vlues of A in (, ( nd (3 re equl. It is left s n exercise to the reder to verif tht the vlues of A b expnding long R 3, C nd C 3 re equl to the vlue of A obtined in (, ( or (3. Hence, expnding determinnt long n row or column gives sme vlue. Remrks (i For esier clcultions, we shll expnd the determinnt long tht row or column which contins mximum number of zeros. (ii While expnding, insted of multipling b ( i + j, we cn multipl b + or ccording s (i + j is even or odd. (iii Let A 4 0 nd B 0. Then, it is es to verif tht A B. Also A 0 8 8 nd B 0. Observe tht, A 4( B or A n B, where n is the order of squre mtrices A nd B. In generl, if A kb where A nd B re squre mtrices of order n, then A k n B, where n,, 3 Exmple 3 Evlute the determinnt 4 3 0 4 0 Solution Note tht in the third column, two entries re zero. So expnding long third column (C 3, we get. 3 4 0 + 0 4 4 3 4 ( 0 + 0 5 Exmple 4 Evlute 0 sin α cos α sin α 0 sin β cos α sin β 0.
08 MATHEMATICS Solution Expnding long R, we get 0 sin β sin α sin β sin α 0 0 sin α cos α sin β 0 cos α 0 cos α sin β 0 sin α (0 sin β cos α cos α (sin α sin β 0 sin α sin β cos α cos α sin α sin β 0 Exmple 5 Find vlues of x for which 3 x 3. x 4 Solution We hve 3 x 3 x 4 i.e. 3 x 3 8 i.e. x 8 Hence x ± EXERCISE 4. Evlute the determinnts in Exercises nd.. 4 5. (i cos θ sin θ sin θ cos θ (ii x x + x x + x + 3. If A 4, then show tht A 4 A 0 4. If A 0, then show tht 3 A 7 A 0 0 4 5. Evlute the determinnts (i 3 0 0 3 5 0 (ii 3 4 5 3
DETERMINANTS 09 (iii 0 0 3 3 0 (iv 0 3 5 0 6. If A 3, find A 5 4 9 7. Find vlues of x, if (i 4 x 4 (ii 5 6 x 3 x 3 4 5 x 5 x 6 8. If 8 x, then x is equl to 8 6 (A 6 (B ± 6 (C 6 (D 0 4.3 Properties of Determinnts In the previous section, we hve lernt how to expnd the determinnts. In this section, we will stud some properties of determinnts which simplifies its evlution b obtining mximum number of zeros in row or column. These properties re true for determinnts of n order. However, we shll restrict ourselves upto determinnts of order 3 onl. Propert The vlue of the determinnt remins unchnged if its rows nd columns re interchnged. Verifiction Let 3 b b b 3 c c c 3 Expnding long first row, we get b b b b b b 3 3 + 3 c c3 c c3 c c (b c 3 b 3 c (b c 3 b 3 c + 3 (b c b c B interchnging the rows nd columns of, we get the determinnt b c b c b c 3 3 3
0 MATHEMATICS Expnding long first column, we get (b c 3 c b 3 (b c 3 b 3 c + 3 (b c b c Hence Remrk It follows from bove propert tht if A is squre mtrix, then det (A det (A, where A trnspose of A. Note If R ith row nd C ith column, then for interchnge of row nd i i columns, we will smbolicll write C i R i Let us verif the bove propert b exmple. 3 5 Exmple 6 Verif Propert for 6 0 4 5 7 Solution Expnding the determinnt long first row, we hve 0 4 6 4 6 0 ( 3 + 5 5 7 7 5 (0 0 + 3 ( 4 4 + 5 (30 0 40 38 + 50 8 B interchnging rows nd columns, we get 6 3 0 5 5 4 7 (Expnding long first column 0 5 6 6 ( 3 + 5 4 7 4 7 0 5 (0 0 + 3 ( 4 4 + 5 (30 0 40 38 + 50 8 Clerl Hence, Propert is verified. Propert If n two rows (or columns of determinnt re interchnged, then sign of determinnt chnges. Verifiction Let 3 b b b 3 c c c 3
DETERMINANTS Expnding long first row, we get (b c 3 b 3 c (b c 3 b 3 c + 3 (b c b c Interchnging first nd third rows, the new determinnt obtined is given b c c c 3 b b b 3 3 Expnding long third row, we get (c b 3 b c 3 (c b 3 c 3 b + 3 (b c b c [ (b c 3 b 3 c (b c 3 b 3 c + 3 (b c b c ] Clerl Similrl, we cn verif the result b interchnging n two columns. Note We cn denote the interchnge of rows b R R nd interchnge of i j columns b C i C j. Exmple 7 Verif Propert for 3 5 6 0 4 5 7. Solution 3 5 6 0 4 5 7 8 (See Exmple 6 Interchnging rows R nd R 3 i.e., R R 3, we hve 3 5 5 7 6 0 4 Expnding the determinnt long first row, we hve 5 7 7 5 ( 3 + 5 0 4 6 4 6 0 (0 0 + 3 (4 + 4 + 5 (0 30 40 + 38 50 8
MATHEMATICS Clerl Hence, Propert is verified. Propert 3 If n two rows (or columns of determinnt re identicl (ll corresponding elements re sme, then vlue of determinnt is zero. Proof If we interchnge the identicl rows (or columns of the determinnt, then does not chnge. However, b Propert, it follows tht hs chnged its sign Therefore or 0 Let us verif the bove propert b n exmple. Exmple 8 Evlute 3 3 3 3 3 Solution Expnding long first row, we get Here R nd R 3 re identicl. 3 (6 6 (6 9 + 3 (4 6 0 ( 3 + 3 ( 6 6 0 Propert 4 If ech element of row (or column of determinnt is multiplied b constnt k, then its vlue gets multiplied b k. Verifiction Let b c b c b c 3 3 3 nd be the determinnt obtined b multipling the elements of the first row b k. Then k k b k c b c b c 3 3 3 Expnding long first row, we get k (b c 3 b 3 c k b ( c 3 c 3 + k c ( b 3 b 3 k [ (b c 3 b 3 c b ( c 3 c 3 + c ( b 3 b 3 ] k
DETERMINANTS 3 Hence Remrks (i (ii k k b k c b c b c 3 3 3 k b c b c b c 3 3 3 B this propert, we cn tke out n common fctor from n one row or n one column of given determinnt. If corresponding elements of n two rows (or columns of determinnt re proportionl (in the sme rtio, then its vlue is zero. For exmple 3 b b b 3 k k k 3 0 (rows R nd R re proportionl Exmple 9 Evlute 0 8 36 3 4 7 3 6 Solution Note tht 0 8 36 6(7 6(3 6(6 7 3 6 3 4 3 4 6 3 4 0 7 3 6 7 3 6 7 3 6 (Using Properties 3 nd 4 Propert 5 If some or ll elements of row or column of determinnt re expressed s sum of two (or more terms, then the determinnt cn be expressed s sum of two (or more determinnts. For exmple, + λ + λ + λ 3 3 b b b 3 c c c 3 λ λ λ 3 3 b b b + b b b 3 3 c c c c c c 3 3 Verifiction L.H.S. + λ + λ + λ 3 3 b b b 3 c c c 3
4 MATHEMATICS Expnding the determinnts long the first row, we get ( + λ (b c 3 c b 3 ( + λ (b c 3 b 3 c + ( 3 + λ 3 (b c b c (b c 3 c b 3 (b c 3 b 3 c + 3 (b c b c + λ (b c 3 c b 3 λ (b c 3 b 3 c + λ 3 (b c b c (b rerrnging terms λ λ λ 3 3 b b b3 + b b b3 R.H.S. c c c c c c 3 3 Similrl, we m verif Propert 5 for other rows or columns. b c Exmple 0 Show tht + x b + c + z 0 x z b c b c b c Solution We hve + x b + c + z b c + x z x z x z x z (b Propert 5 0 + 0 0 (Using Propert 3 nd Propert 4 Propert 6 If, to ech element of n row or column of determinnt, the equimultiples of corresponding elements of other row (or column re dded, then vlue of determinnt remins the sme, i.e., the vlue of determinnt remin sme if we ppl the opertion R i R i + kr j or C i C i + kc j. Verifiction Let 3 b b b 3 c c c 3 nd where is obtined b the opertion R R + kr 3. + k c + k c + k c 3 3 b b b 3 c c c 3 Here, we hve multiplied the elements of the third row (R 3 b constnt k nd dded them to the corresponding elements of the first row (R. Smbolicll, we write this opertion s R R + k R 3.,
DETERMINANTS 5 Now, gin k c k c k c 3 3 b b b3 + b b b3 (Using Propert 5 c c c c c c 3 3 Hence Remrks (i (ii + 0 (since R nd R 3 re proportionl If is the determinnt obtined b ppling R i kr i or C i kc i to the determinnt, then k. If more thn one opertion like R i R i + kr j is done in one step, cre should be tken to see tht row tht is ffected in one opertion should not be used in nother opertion. A similr remrk pplies to column opertions. Exmple Prove tht + b + b + c 3 + b 4 + 3b + c. 3 6 + 3b 0 + 6b + 3c 3 Solution Appling opertions R R R nd R 3 R 3 3R to the given determinnt, we hve Now ppling R 3 R 3 3R, we get Expnding long C, we obtin + b + b + c 0 + b 0 3 7 + 3b + b + b + c 0 + b 0 0 + b 0 + 0 + 0 ( 0 ( 3
6 MATHEMATICS Exmple Without expnding, prove tht x + + z z + x z x 0 Solution Appling R R + R to, we get x + + z x + + z x + + z z x Since the elements of R nd R 3 re proportionl, 0. Exmple 3 Evlute b c bc c b Solution Appling R R R nd R 3 R 3 R, we get bc 0 b c ( b 0 c b ( c Tking fctors (b nd (c common from R nd R 3, respectivel, we get ( b ( c 0 c bc 0 b (b (c [( b + c] (Expnding long first column ( b (b c (c b + c Exmple 4 Prove tht b c + b 4bc c c + b Solution Let b + c b c + b c c + b
DETERMINANTS 7 Appling R R R R 3 to, we get 0 c b b c + b c c + b Expnding long R, we obtin c + b b b 0 ( c c + b c + b + ( b c + b c c c ( b + b bc b (b c c c b c + cb bc b c + bc + bc 4 bc Exmple 5 If x,, z re different nd show tht + xz 0 Solution We hve x x + x 3 0, then 3 + z z + z 3 3 x x + x 3 + 3 z z + z 3 x x x x x 3 + (Using Propert 5 3 z z z z z x x x x ( + xz (Using C 3 C nd then C C x z z z z ( + xz z x z
8 MATHEMATICS ( x + xz 0 x x 0 x z x z x (Using R R R nd R 3 R 3 R Tking out common fctor ( x from R nd (z x from R 3, we get x (+ xz ( x ( z x 0 + x 0 z + x x ( + xz ( x (z x (z (on expnding long C Since 0 nd x,, z re ll different, i.e., x 0, z 0, z x 0, we get + xz 0 Exmple 6 Show tht + + b bc + + + bc + bc + c + b b c + c Solution Tking out fctors,b,c common from R, R nd R 3, we get L.H.S. + bc + b b b + c c c Appling R R + R + R 3, we hve + + + + + + + + + b c b c b c bc + b b b + c c c
DETERMINANTS 9 bc + + + b c + b b b + c c c Now ppling C C C, C 3 C 3 C, we get 0 0 bc + + + b c b 0 c 0 + + + b c bc ( 0 bc + + + bc + bc + c + b R.H.S. b c Note Alterntel tr b ppling C C C nd C C C, then ppl 3 3 C C C 3. EXERCISE 4. Using the propert of determinnts nd without expnding in Exercises to 7, prove tht: x x +. b + b 0 3. z c z + c 7 65 3 8 75 0 4. 5 9 86 b + c q + r + z p x 5. c + r + p z + x b q + b p + q x + c r z b b c c. b c c b 0 c b b c ( + ( ( + c b c + 0 bc b c b c b
0 MATHEMATICS 0 b 6. 0 c 0 7. b c 0 b c b b bc 4 b c c cb c B using properties of determinnts, in Exercises 8 to 4, show tht: 8. (i b b ( b( b c( c c c (ii b c ( b( b c( c ( + b + c 3 3 3 b c 9. x x z zx z z x (x ( z (z x (x + z + zx x + 4 x x x x + 4 x 5x + 4 4 x 0. (i ( ( x x x + 4 + k (ii + k k ( 3 + k + k b c. (i b b c b ( + b + c 3 c c c b x + + z x (ii z + z + x ( x + + z 3 z x z + x +
DETERMINANTS x 3. x x ( x x x x + b b b 3. b + b ( + + b b b 3 4. + b c b b + bc + + b + c c cb c + Choose the correct nswer in Exercises 5 nd 6. 5. Let A be squre mtrix of order 3 3, then ka is equl to (A k A (B k A (C k 3 A (D 3k A 6. Which of the following is correct (A Determinnt is squre mtrix. (B Determinnt is number ssocited to mtrix. (C Determinnt is number ssocited to squre mtrix. (D None of these 4.4 Are of Tringle In erlier clsses, we hve studied tht the re of tringle whose vertices re (x,, (x, nd (x 3, 3, is given b the expression [x ( 3 + x ( 3 + x 3 ( ]. Now this expression cn be written in the form of determinnt s x x x 3 3... ( Remrks (i Since re is positive quntit, we lws tke the bsolute vlue of the determinnt in (.
MATHEMATICS (ii (iii If re is given, use both positive nd negtive vlues of the determinnt for clcultion. The re of the tringle formed b three colliner points is zero. Exmple 7 Find the re of the tringle whose vertices re (3, 8, ( 4, nd (5,. Solution The re of tringle is given b 3 8 4 5 3 8 4 5 4 0 + ( ( ( + 6 ( 3 7 4 Exmple 8 Find the eqution of the line joining A(, 3 nd B (0, 0 using determinnts nd find k if D(k, 0 is point such tht re of tringle ABD is 3sq units. Solution Let P (x, be n point on AB. Then, re of tringle ABP is zero (Wh?. So 0 0 3 x 0 This gives ( 3 x 0 or 3x, which is the eqution of required line AB. Also, since the re of the tringle ABD is 3 sq. units, we hve This gives, 3 0 0 k 0 3k ± 3, i.e., k. ± 3 EXERCISE 4.3. Find re of the tringle with vertices t the point given in ech of the following : (i (, 0, (6, 0, (4, 3 (ii (, 7, (,, (0, 8 (iii (, 3, (3,, (, 8
DETERMINANTS 3. Show tht points A (, b + c, B (b, c +, C (c, + b re colliner. 3. Find vlues of k if re of tringle is 4 sq. units nd vertices re (i (k, 0, (4, 0, (0, (ii (, 0, (0, 4, (0, k 4. (i Find eqution of line joining (, nd (3, 6 using determinnts. (ii Find eqution of line joining (3, nd (9, 3 using determinnts. 5. If re of tringle is 35 sq units with vertices (, 6, (5, 4 nd (k, 4. Then k is (A (B (C, (D, 4.5 Minors nd Cofctors In this section, we will lern to write the expnsion of determinnt in compct form using minors nd cofctors. Definition Minor of n element ij of determinnt is the determinnt obtined b deleting its ith row nd jth column in which element ij lies. Minor of n element ij is denoted b M ij. Remrk Minor of n element of determinnt of order n(n is determinnt of order n. Exmple 9 Find the minor of element 6 in the determinnt 3 4 5 6 7 8 9 Solution Since 6 lies in the second row nd third column, its minor M 3 is given b M 3 7 8 8 4 6 (obtined b deleting R nd C in. 3 Definition Cofctor of n element ij, denoted b A ij is defined b A ij ( i + j M ij, where M ij is minor of ij. Exmple 0 Find minors nd cofctors of ll the elements of the determinnt 4 3 Solution Minor of the element ij is M ij Here. So M Minor of 3 M Minor of the element 4 M Minor of the element
4 MATHEMATICS M Minor of the element Now, cofctor of ij is A ij. So A ( + M ( (3 3 A ( + M ( 3 (4 4 A ( + M ( 3 ( A ( + M ( 4 ( Exmple Find minors nd cofctors of the elements, in the determinnt 3 3 3 3 33 Solution B definition of minors nd cofctors, we hve Minor of M 3 3 33 33 3 3 Cofctor of A ( + M 33 3 3 Minor of M 3 3 33 33 3 3 Cofctor of A ( + M ( ( 33 3 3 33 + 3 3 Remrk Expnding the determinnt, in Exmple, long R, we hve ( + 3 3 33 + ( + 3 3 33 + ( +3 3 3 3 A + A + 3 A 3, where A ij is cofctor of ij sum of product of elements of R with their corresponding cofctors Similrl, cn be clculted b other five ws of expnsion tht is long R, R 3, C, C nd C 3. Hence sum of the product of elements of n row (or column with their corresponding cofctors. Note If elements of row (or column re multiplied with cofctors of n other row (or column, then their sum is zero. For exmple,
DETERMINANTS 5 A + A + 3 A 3 ( + 3 3 3 3 33 3 3 33 + ( + 3 + 3 3 ( +3 33 0 (since R nd R re identicl 3 3 Similrl, we cn tr for other rows nd columns. Exmple Find minors nd cofctors of the elements of the determinnt 3 5 6 0 4 nd verif tht A 3 + A 3 + 3 A 33 0 5 7 Solution We hve M 0 4 5 7 0 0 0; A ( + ( 0 0 M 6 4 7 4 4 46; A ( + ( 46 46 M 3 6 0 5 30 0 30; A 3 ( +3 (30 30 M 3 5 5 7 5 4; A ( + ( 4 4 M 5 7 M 3 3 5 4 5 9; A ( + ( 9 9 0 + 3 3; A 3 ( +3 (3 3 M 3 3 5 0 4 0 ; A 3 ( 3+ (
6 MATHEMATICS M 3 5 6 4 8 30 ; A 3 ( 3+ ( nd M 33 3 6 0 0 + 8 8; A 33 ( 3+3 (8 8 Now, 3, 3 5; A 3, A 3, A 33 8 So A 3 + A 3 + 3 A 33 ( + ( 3 ( + 5 (8 4 66 + 90 0 EXERCISE 4.4 Write Minors nd Cofctors of the elements of following determinnts:. (i 4 0 3 (ii b c d. (i 0 0 0 0 0 0 (ii 0 4 3 5 0 3. Using Cofctors of elements of second row, evlute 4. Using Cofctors of elements of third column, evlute 5 3 8 0. 3 x z zx z x. 5. If 3 3 3 3 33 nd A ij is Cofctors of ij, then vlue of is given b (A A 3 + A 3 + 3 A 33 (B A + A + 3 A 3 (C A + A + 3 A 3 (D A + A + 3 A 3 4.6 Adjoint nd Inverse of Mtrix In the previous chpter, we hve studied inverse of mtrix. In this section, we shll discuss the condition for existence of inverse of mtrix. To find inverse of mtrix A, i.e., A we shll first define djoint of mtrix.
4.6. Adjoint of mtrix DETERMINANTS 7 Definition 3 The djoint of squre mtrix A [ ij ] n n is defined s the trnspose of the mtrix [A ij ] n n, where A ij is the cofctor of the element ij. Adjoint of the mtrix A is denoted b dj A. Let Then A 3 3 3 3 33 A A A dj A Trnspose of A A A A A A 3 3 3 3 33 A A A A A A A A A 3 3 3 3 33 3 Exmple 3 Find dj A for A 4 Solution We hve A 4, A, A 3, A Hence dj A A A 4 3 A A Remrk For squre mtrix of order, given b A The dj A cn lso be obtined b interchnging nd nd b chnging signs of nd, i.e., We stte the following theorem without proof. Theorem If A be n given squre mtrix of order n, then A(dj A (dj A A A I, where I is the identit mtrix of order n
8 MATHEMATICS Verifiction 3 A A A3 Let A 3, then dj A A A A 3 3 3 33 A3 A3 A 33 Since sum of product of elements of row (or column with corresponding cofctors is equl to A nd otherwise zero, we hve A (dj A A 0 0 0 A 0 0 0 A A 0 0 0 0 0 0 A I Similrl, we cn show (dj A A A I Hence A (dj A (dj A A A I Definition 4 A squre mtrix A is sid to be singulr if A 0. For exmple, the determinnt of mtrix A 4 8 is zero Hence A is singulr mtrix. Definition 5 A squre mtrix A is sid to be non-singulr if A 0 Let A 3 4. Then A 3 4 4 6 0. Hence A is nonsingulr mtrix We stte the following theorems without proof. Theorem If A nd B re nonsingulr mtrices of the sme order, then AB nd BA re lso nonsingulr mtrices of the sme order. Theorem 3 The determinnt of the product of mtrices is equl to product of their respective determinnts, tht is, AB A B, where A nd B re squre mtrices of the sme order Remrk We know tht (dj A A A I A 0 0 0 A 0 0 0 A
DETERMINANTS 9 Writing determinnts of mtrices on both sides, we hve ( dj A A A 0 0 0 A 0 0 0 A i.e. (dj A A 3 0 0 A 0 0 0 0 i.e. (dj A A A 3 ( i.e. (dj A A In generl, if A is squre mtrix of order n, then dj(a A n. (Wh? Theorem 4 A squre mtrix A is invertible if nd onl if A is nonsingulr mtrix. Proof Let A be invertible mtrix of order n nd I be the identit mtrix of order n. Then, there exists squre mtrix B of order n such tht AB BA I Now AB I. So AB I or A B (since I, AB A B This gives A 0. Hence A is nonsingulr. Conversel, let A be nonsingulr. Then A 0 Now A (dj A (dj A A A I (Theorem or A dj A dj A A I A A or AB BA I, where B A A dj Thus A is invertible nd A A A dj Exmple 4 If A 3 3 4 3, then verif tht A dj A A I. Also find A. 3 4 Solution We hve A (6 9 3 (4 3 + 3 (3 4 0
30 MATHEMATICS Now A 7, A, A 3, A 3, A,A 3 0, A 3 3, A 3 0, A 33 Therefore dj A 7 3 3 0 0 Now A (dj A 3 3 7 3 3 4 3 0 3 4 0 7 3 3 3 + 3 + 0 3 + 0 + 3 7 4 3 3 4 0 3 0 3 + + + + 7 3 4 3 + 3 + 0 3 + 0 + 4 0 0 0 0 0 0 ( 0 0 0 0 0 0 A. I 7 3 3 Also A A A d j 0 0 7 3 3 0 0 Exmple 5 If A 3 nd B 4 3, then verif tht (AB B A. 3 5 Solution We hve AB 4 3 5 4 Since, AB 0, (AB exists nd is given b 4 5 4 5 (AB dj (AB AB 5 5 Further, A 0 nd B 0. Therefore, A nd B both exist nd re given b A 4 3 3,B
DETERMINANTS 3 3 4 3 Therefore B A Hence (AB B A 4 5 5 4 5 5 Exmple 6 Show tht the mtrix A 3 stisfies the eqution A 4A + I O, where I is identit mtrix nd O is zero mtrix. Using this eqution, find A. 3 3 7 Solution We hve A A.A 4 7 Hence 7 8 0 A 4A + I 4 7 + 4 8 0 Now A 4A + I O Therefore A A 4A I 0 0 O 0 0 or A A (A 4 A A I A (Post multipling b A becuse A 0 or A (A A 4I A or AI 4I A or 4 0 3 3 A 4I A 0 4 Hence 3 A EXERCISE 4.5 Find djoint of ech of the mtrices in Exercises nd.. 3 4. 3 5 0 Verif A (dj A (dj A A A I in Exercises 3 nd 4 3. 3 4 6 4. 3 0 0 3
3 MATHEMATICS Find the inverse of ech of the mtrices (if it exists given in Exercises 5 to. 5. 4 3 6. 5 3 7. 3 0 4 0 0 5 8. 0 0 3 3 0 5 9. 3 4 0 7 0. 0 3 3 4. 0 0 0 cosα sin α 0 sin α cos α. 3 7 Let A 5 nd B 6 8 7 9. Verif tht (AB B A. 3. If A 3, show tht A 5A + 7I O. Hence find A. 3 4. For the mtrix A, find the numbers nd b such tht A + A + bi O. 5. For the mtrix A 3 3 Show tht A 3 6A + 5A + I O. Hence, find A. 6. If A Verif tht A 3 6A + 9A 4I O nd hence find A 7. Let A be nonsingulr squre mtrix of order 3 3. Then dj A is equl to (A A (B A (C A 3 (D 3 A 8. If A is n invertible mtrix of order, then det (A is equl to (A det (A (B det (A (C (D 0
DETERMINANTS 33 4.7 Applictions of Determinnts nd Mtrices In this section, we shll discuss ppliction of determinnts nd mtrices for solving the sstem of liner equtions in two or three vribles nd for checking the consistenc of the sstem of liner equtions. Consistent sstem A sstem of equtions is sid to be consistent if its solution (one or more exists. Inconsistent sstem A sstem of equtions is sid to be inconsistent if its solution does not exist. Note In this chpter, we restrict ourselves to the sstem of liner equtions hving unique solutions onl. 4.7. Solution of sstem of liner equtions using inverse of mtrix Let us express the sstem of liner equtions s mtrix equtions nd solve them using inverse of the coefficient mtrix. Consider the sstem of equtions x + b + c z d x + b + c z d 3 x + b 3 + c 3 z d 3 b c x d Let A b c, X nd B d 3 b3 c 3 z d 3 Then, the sstem of equtions cn be written s, AX B, i.e., b c x b c 3 b3 c 3 z d d d 3 Cse I If A is nonsingulr mtrix, then its inverse exists. Now AX B or A (AX A B (premultipling b A or (A A X A B (b ssocitive propert or I X A B or X A B This mtrix eqution provides unique solution for the given sstem of equtions s inverse of mtrix is unique. This method of solving sstem of equtions is known s Mtrix Method.
34 MATHEMATICS Cse II If A is singulr mtrix, then A 0. In this cse, we clculte (dj A B. If (dj A B O, (O being zero mtrix, then solution does not exist nd the sstem of equtions is clled inconsistent. If (dj A B O, then sstem m be either consistent or inconsistent ccording s the sstem hve either infinitel mn solutions or no solution. Exmple 7 Solve the sstem of equtions x + 5 3x + 7 Solution The sstem of equtions cn be written in the form AX B, where A 5 x,x nd B 3 7 Now, A 0, Hence, A is nonsingulr mtrix nd so hs unique solution. Note tht A 5 3 Therefore X A B 5 3 7 i.e. x 33 3 Hence x 3, Exmple 8 Solve the following sstem of equtions b mtrix method. 3x + 3z 8 x + z 4x 3 + z 4 Solution The sstem of equtions cn be written in the form AX B, where We see tht 3 3 x 8 A, X nd B 4 3 z 4 A 3 ( 3 + (4 + 4 + 3 ( 6 4 7 0
DETERMINANTS 35 Hence, A is nonsingulr nd so its inverse exists. Now A, A 8, A 3 0 A 5, A 6, A 3 A 3, A 3 9, A 33 7 Therefore A So X 5 8 6 9 7 0 7 5 8 7 0 7 4 A B 8 6 9 i.e. x z 7 34 7 5 3 Hence x, nd z 3. Exmple 9 The sum of three numbers is 6. If we multipl third number b 3 nd dd second number to it, we get. B dding first nd third numbers, we get double of the second number. Represent it lgebricll nd find the numbers using mtrix method. Solution Let first, second nd third numbers be denoted b x, nd z, respectivel. Then, ccording to given conditions, we hve x + + z 6 + 3z x + z or x + z 0 This sstem cn be written s A X B, where x A 0 3, X nd B z Here A ( + 6 (0 3 + ( 0 9 0. Now we find dj A 6 0 A ( + 6 7, A (0 3 3, A 3 A ( + 3, A 0, A 3 ( 3 A 3 (3, A 3 (3 0 3, A 33 ( 0
36 MATHEMATICS Hence dj A 7 3 3 0 3 3 Thus A 7 3 A dj (A 3 0 3 9 3 Since X A B 7 3 6 X 3 0 3 9 3 0 or x 4 33 + 0 8 0 0 + + z 9 6 + 33 + 0 Thus x,, z 3 9 9 8 7 3 EXERCISE 4.6 Exmine the consistenc of the sstem of equtions in Exercises to 6.. x +. x 5 3. x + 3 5 x + 3 3 x + 4 x + 6 8 4. x + + z 5. 3x z 6. 5x + 4z 5 x + 3 + z z x + 3 + 5z x + + z 4 3x 5 3 5x + 6z Solve sstem of liner equtions, using mtrix method, in Exercises 7 to 4. 7. 5x + 4 8. x 9. 4x 3 3 7x + 3 5 3x + 4 3 3x 5 7 0. 5x + 3. x + + z. x + z 4 3x + 5 x z 3 x + 3z 0 3 5z 9 x + + z 3. x + 3 +3 z 5 4. x + z 7 x + z 4 3x + 4 5z 5 3x z 3 x + 3z
DETERMINANTS 37 5. If A 3 5 3 4, find A. Using A solve the sstem of equtions x 3 + 5z 3x + 4z 5 x + z 3 6. The cost of 4 kg onion, 3 kg whet nd kg rice is Rs 60. The cost of kg onion, 4 kg whet nd 6 kg rice is Rs 90. The cost of 6 kg onion kg whet nd 3 kg rice is Rs 70. Find cost of ech item per kg b mtrix method. Miscellneous Exmples Exmple 30 If, b, c re positive nd unequl, show tht vlue of the determinnt b c b c c b is negtive. Solution Appling C C + C + C 3 to the given determinnt, we get + b + c b c + b + c c + b + c b ( + b + c b c c b b c ( + b + c 0 c b c (ApplingR R R,ndR 3 R 3 R 0 b b c ( + b + c [(c b (b c ( c ( b] (Expnding long C ( + b + c( b c + b + bc + c ( + b + c ( + b + c b bc c ( + b + c [( b + (b c + (c ] which is negtive (since + b + c > 0 nd ( b + (b c + (c > 0
38 MATHEMATICS Exmple 3 If, b, c, re in A.P, find vlue of + 4 5 + 7 8 + 3 + 5 6 + 8 9 + b 4 + 6 7 + 9 0 + c Solution Appling R R + R 3 R to the given determinnt, we obtin 0 0 0 3 + 5 6 + 8 9 + b 4 + 6 7 + 9 0 + c Exmple 3 Show tht 0 (Since b + c ( + z x zx ( + x x z z ( + xz z x xz (x + + z 3 Solution Appling R xr, R R, R zr to nd dividing b xz, we get 3 3 xz ( + x z x x z ( + x x z z ( + xz z z x Tking common fctors x,, z from C C nd C 3, respectivel, we get xz xz ( + z x x ( + x z ( + z z x Appling C C C, C 3 C 3 C, we hve ( + z x ( + z x ( + z ( x + z 0 z 0 ( x + z
DETERMINANTS 39 Tking common fctor (x + + z from C nd C 3, we hve (x + + z Appling R R (R + R 3, we hve ( + z x ( + z x ( + z ( x + z 0 z 0 ( x + z (x + + z z z x z + 0 z x + 0 z Appling C (C + C nd C C C 3 + 3, we get z (x + + z z 0 0 x+ z z z z x+ Finll expnding long R, we hve (x + + z (z [(x + z (x + z] (x + + z (z (x + x + xz (x + + z 3 (xz 0 Exmple 33 Use product 0 3 9 3 to solve the sstem of equtions 3 4 6 x + z 3z 3x + 4z 0 Solution Consider the product 0 3 9 3 3 4 6
40 MATHEMATICS 9 + 0 + + 3 4 0 8 8 0 4 3 0 6 6 + + + 6 8 + 4 0 4 + 4 3 + 6 8 0 0 0 0 0 0 0 Hence 0 3 9 3 3 4 6 Now, given sstem of equtions cn be written, in mtrix form, s follows x 0 3 3 4 z x or 0 3 z 3 4 + 0 + 0 9 + 6 5 6 + 4 3 Hence x 0, 5 nd z 3 Exmple 34 Prove tht + bx c + dx p + qx c p x + b cx + d px + q ( x b d q u v w u v w 0 9 3 6 Solution Appling R R x R to, we get ( x c( x p( x x + b cx + d px + q u v w c p ( x x + b cx + d px + q u v w
DETERMINANTS 4 Appling R R x R, we get c p ( x b d q u v w Miscellneous Exercises on Chpter 4 x sin θ cos θ. Prove tht the determinnt sin θ x is independent of θ. cosθ x bc 3. Without expnding the determinnt, prove tht b b c b b 3 c c b c c 3. cosα cosβ cosα sin β sin α 3. Evlute sin β cosβ 0 sin α cosβ sin α sinβ cosα 4. If, b nd c re rel numbers, nd. b+ c c+ + b c + + b b + c 0, + b b + c c+ Show tht either + b + c 0 or b c. x + x x 5. Solve the eqution x x + x 0, 0 x x x + 6. Prove tht bc c+ c + b b c b b + bc c 4 b c 3 5 6 5 nd B 3 0, find AB 5 0 7. If A (
4 MATHEMATICS 8. Let A 3. Verif tht 5 (i [dj A] dj (A (ii (A A 9. Evlute x x+ x + x x + x 0. Evlute x x + x x+ Using properties of determinnts in Exercises to 5, prove tht:. α α β + γ β β γ + α γ γ α + β (β γ (γ α (α β (α + β + γ. x x + px 3 + p 3 z z + pz 3 ( + pxz (x ( z (z x, where p is n sclr. 3. 3 +b + c b + 3b b + c c + c+b 3c 3( + b + c (b + bc + c 4. + p + p + q 3 + p 4 + 3 p + q 3 6 + 3 p 0 + 6 p + 3q 6. Solve the sstem of equtions 3 0 + + 4 x z 5. ( ( ( sin α cosα cos α + δ sinβ cosβ cos β + δ 0 sin γ cos γ cos γ + δ
DETERMINANTS 43 4 6 5 + x z 6 9 0 + x z Choose the correct nswer in Exercise 7 to 9. 7. If, b, c, re in A.P, then the determinnt x + x + 3 x + x + 3 x + 4 x + b x + 4 x + 5 x + c is (A 0 (B (C x (D x 8. If x,, z re nonzero rel numbers, then the inverse of mtrix x 0 0 A 0 0 is 0 0 z (A x 0 0 0 0 0 0 z (B x 0 0 xz 0 0 0 0 z (C x 0 0 0 0 xz 0 0 z (D 0 0 0 0 xz 0 0 9. Let A sin θ sin sin θ θ, where 0 θ π. Then sin θ (A Det(A 0 (B Det(A (, (C Det(A (, 4 (D Det(A [, 4]
44 MATHEMATICS Summr Determinnt of mtrix A [ ] is given b Determinnt of mtrix A is given b A b c Determinnt of mtrix A b c 3 b3 c 3 is given b (expnding long R A b c b c c b b3 c3 3 c3 3 b3 3 3 3 b c b + c b c For n squre mtrix A, the A stisf following properties. A A, where A trnspose of A. If we interchnge n two rows (or columns, then sign of determinnt chnges. If n two rows or n two columns re identicl or proportionl, then vlue of determinnt is zero. If we multipl ech element of row or column of determinnt b constnt k, then vlue of determinnt is multiplied b k. Multipling determinnt b k mens multipl elements of onl one row (or one column b k. If A [ ],then k.a k A ij 3 3 3 If elements of row or column in determinnt cn be expressed s sum of two or more elements, then the given determinnt cn be expressed s sum of two or more determinnts. If to ech element of row or column of determinnt the equimultiples of corresponding elements of other rows or columns re dded, then vlue of determinnt remins sme.
DETERMINANTS 45 Are of tringle with vertices (x,, (x, nd (x 3, 3 is given b x x x 3 3 Minor of n element ij of the determinnt of mtrix A is the determinnt obtined b deleting i th row nd j th column nd denoted b M ij. Cofctor of ij of given b A ij ( i+j M ij Vlue of determinnt of mtrix A is obtined b sum of product of elements of row (or column with corresponding cofctors. For exmple, A A + A + 3 A 3. If elements of one row (or column re multiplied with cofctors of elements of n other row (or column, then their sum is zero. For exmple, A + A + 3 A 3 0 If 3 A 3, 3 3 33 then A A A dj A A A A A A A 3 3 3 3 33, where A ij is cofctor of ij A (dj A (dj A A A I, where A is squre mtrix of order n. A squre mtrix A is sid to be singulr or non-singulr ccording s A 0 or A 0. If AB BA I, where B is squre mtrix, then B is clled inverse of A. Also A B or B A nd hence (A A. A squre mtrix A hs inverse if nd onl if A is non-singulr. A ( A A dj If x + b + c z d x + b + c z d 3 x + b + c z d, 3 3 3 then these equtions cn be written s A X B, where b c x d A,X nd B b c d 3 b3 c 3 z d 3
46 MATHEMATICS Unique solution of eqution AX B is given b X A B, where A 0. A sstem of eqution is consistent or inconsistent ccording s its solution exists or not. For squre mtrix A in mtrix eqution AX B (i A 0, there exists unique solution (ii A 0 nd (dj A B 0, then there exists no solution (iii A 0 nd (dj A B 0, then sstem m or m not be consistent. Historicl Note The Chinese method of representing the coefficients of the unknowns of severl liner equtions b using rods on clculting bord nturll led to the discover of simple method of elimintion. The rrngement of rods ws precisel tht of the numbers in determinnt. The Chinese, therefore, erl developed the ide of subtrcting columns nd rows s in simplifiction of determinnt Mikmi, Chin, pp 30, 93. Seki Kow, the gretest of the Jpnese Mthemticins of seventeenth centur in his work Ki Fukudi no Ho in 683 showed tht he hd the ide of determinnts nd of their expnsion. But he used this device onl in eliminting quntit from two equtions nd not directl in the solution of set of simultneous liner equtions. T. Hshi, The Fkudoi nd Determinnts in Jpnese Mthemtics, in the proc. of the Toko Mth. Soc., V. Vendermonde ws the first to recognise determinnts s independent functions. He m be clled the forml founder. Lplce (77, gve generl method of expnding determinnt in terms of its complementr minors. In 773 Lgrnge treted determinnts of the second nd third orders nd used them for purpose other thn the solution of equtions. In 80, Guss used determinnts in his theor of numbers. The next gret contributor ws Jcques - Philippe - Mrie Binet, (8 who stted the theorem relting to the product of two mtrices of m-columns nd n- rows, which for the specil cse of m n reduces to the multipliction theorem. Also on the sme d, Cuch (8 presented one on the sme subject. He used the word determinnt in its present sense. He gve the proof of multipliction theorem more stisfctor thn Binet s. The gretest contributor to the theor ws Crl Gustv Jcob Jcobi, fter this the word determinnt received its finl cceptnce.