All Mthemticl truths re reltive nd conditionl. C.P. STEINMETZ 4. Introduction DETERMINANTS In the previous chpter, we hve studied bout mtrices nd lgebr of mtrices. We hve lso lernt tht system of lgebric equtions cn be epressed in the form of mtrices. This mens, system of liner equtions like + b y c + b y c b c cn be represented s b y c. Now, this system of equtions hs unique solution or not, is determined by the number b b. (Recll tht if b or, b b b 0, then the system of liner equtions hs unique solution. The number b b Chpter 4 P.S. Lplce (749-87 b which determines uniqueness of solution is ssocited with the mtri 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 study determinnts up to order three only with rel entries. Also, we will study vrious properties of determinnts, minors, cofctors nd pplictions of determinnts in finding the re of tringle, djoint nd inverse of squre mtri, consistency nd inconsistency of system of liner equtions nd solution of liner equtions in two or three vribles using inverse of mtri. 4. Determinnt To every squre mtri A [ ij ] of order n, we cn ssocite number (rel or comple clled determinnt of the squre mtri A, where ij (i, j th element of A.
04 MATHEMATICS This my be thought of s function which ssocites ech squre mtri with unique number (rel or comple. If M is the set of squre mtrices, K is the set of numbers (rel or comple nd f : M K is defined by f (A k, where A M nd k K, then f (A is clled the determinnt of A. It is lso denoted by A or det A or Δ. b b If A c d, then determinnt of A is written s A c d Remrks det (A (i For mtri A, A is red s determinnt of A nd not modulus of A. (ii Only squre mtrices hve determinnts. 4.. Determinnt of mtri of order one Let A [ ] be the mtri of order, then determinnt of A is defined to be equl to 4.. Determinnt of mtri of order two Let A then the determinnt of A is defined s: be mtri of order, det (A A Δ Emple Evlute Solution We hve Emple Evlute Solution We hve + 4. 4 ( 4( 4 + 4 8. + ( ( + ( ( + 4..3 Determinnt of mtri of order 3 3 Determinnt of mtri of order three cn be determined by epressing it in terms of second order determinnts. This is known s epnsion of determinnt long row (or column. There re si wys of epnding 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 mtri A [ ij ] 3 3 3 i.e., A 3 Epnsion long first Row (R 3 3 33 Step Multiply first element of R by ( ( + [( sum of suffies in ] nd with the second order determinnt obtined by 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 Multiply nd element of R by ( + [( sum of suffies in ] nd the second order determinnt obtined by 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 Multiply third element 3 of R by ( + 3 [( sum of suffies in 3] nd the second order determinnt obtined by 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 epnsion of determinnt of A, tht is, A written s sum of ll three terms obtined in steps, nd 3 bove is given by 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 pply ll four steps together. Epnsion long second row (R Epnding 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... ( Epnsion long first Column (C A By epnding long C, we get 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 Clerly, vlues of A in (, ( nd (3 re equl. It is left s n eercise to the reder to verify tht the vlues of A by epnding long R 3, C nd C 3 re equl to the vlue of A obtined in (, ( or (3. Hence, epnding determinnt long ny row or column gives sme vlue. Remrks (i For esier clcultions, we shll epnd the determinnt long tht row or column which contins mimum number of zeros. (ii While epnding, insted of multiplying by ( i + j, we cn multiply by + or ccording s (i + j is even or odd. (iii Let A 4 0 nd B 0. Then, it is esy to verify 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 Emple 3 Evlute the determinnt Δ 4 3 0. 4 0 Solution Note tht in the third column, two entries re zero. So epnding long third column (C 3, we get Emple 4 Evlute Δ 3 Δ 4 0 + 0 4 4 3 4 ( 0 + 0 5 0 sin α cos α sin α 0 sin β. cos α sin β 0
08 MATHEMATICS Solution Epnding long R, we get 0 sin β sin α sin β sin α 0 Δ 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 Emple 5 Find vlues of for which Solution We hve 3 3 4 i.e. 3 3 8 i.e. 8 Hence ± 3 3. 4 EXERCISE 4. Evlute the determinnts in Eercises nd... (i 4 5 cos θ sin θ 3. If A 4. If A sin θ cos θ (ii 4, then show tht A 4 A + + + 0 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 0 (iii 0 3 (iv 0 3 0 3 5 0 6. If A 3, find A 5 4 9 7. Find vlues of, if (i 4 4 (ii 5 6 3 3 4 5 5 6 8. If 8, then 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 epnd the determinnts. In this section, we will study some properties of determinnts which simplifies its evlution by obtining mimum number of zeros in row or column. These properties re true for determinnts of ny order. However, we shll restrict ourselves upto determinnts of order 3 only. Property The vlue of the determinnt remins unchnged if its rows nd columns re interchnged. Verifiction Let Δ 3 b b b 3 c c c 3 Epnding long first row, we get Δ 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 By interchnging the rows nd columns of Δ, we get the determinnt Δ b c b c b c 3 3 3
0 MATHEMATICS Epnding Δ long first column, we get Δ (b c 3 c b 3 (b c 3 b 3 c + 3 (b c b c Hence Δ Δ Remrk It follows from bove property tht if A is squre mtri, 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 symboliclly write C i R i Let us verify the bove property by emple. Emple 6 Verify Property for Δ 3 5 6 0 4 5 7 Solution Epnding the determinnt long first row, we hve Δ 0 4 6 4 6 0 ( 3 + 5 5 7 7 5 (0 0 + 3 ( 4 4 + 5 (30 0 40 38 + 50 8 By interchnging rows nd columns, we get 6 Δ 3 0 5 (Epnding long first column 5 4 7 0 5 6 6 ( 3 + 5 4 7 4 7 0 5 (0 0 + 3 ( 4 4 + 5 (30 0 40 38 + 50 8 Clerly Δ Δ Hence, Property is verified. Property If ny two rows (or columns of determinnt re interchnged, then sign of determinnt chnges. 3 Verifiction Let Δ b b b3 c c c 3
DETERMINANTS Epnding long first row, we get Δ (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 by Δ c c c 3 b b b 3 3 Epnding long third row, we get Clerly Δ Δ Δ (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 ] Similrly, we cn verify the result by interchnging ny two columns. Note We cn denote the interchnge of rows by R R nd interchnge of i j columns by C i C j. Emple 7 Verify Property for Δ Solution Δ 3 5 6 0 4 5 7 3 5 6 0 4 5 7 8 (See Emple 6 Interchnging rows R nd R 3 i.e., R R 3, we hve Δ 3 5 5 7 6 0 4 Epnding the determinnt Δ long first row, we hve Δ. 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 Clerly Hence, Property is verified. Δ Δ Property 3 If ny 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 Δ, then Δ does not chnge. However, by Property, it follows tht Δ hs chnged its sign Therefore Δ Δ or Δ 0 Let us verify the bove property by n emple. Emple 8 Evlute Δ 3 3 3 3 3 Solution Epnding long first row, we get Δ 3 (6 6 (6 9 + 3 (4 6 0 ( 3 + 3 ( 6 6 0 Here R nd R 3 re identicl. Property 4 If ech element of row (or column of determinnt is multiplied by constnt k, then its vlue gets multiplied by k. Verifiction Let Δ b c b c b c 3 3 3 nd Δ be the determinnt obtined by multiplying the elements of the first row by k. Then Δ k kb kc b c b c 3 3 3 Epnding long first row, we get Δ 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 Δ
DETERMINANTS 3 Hence Remrks (i (ii k kb kc b c b c 3 3 3 k b c b c b c 3 3 3 By this property, we cn tke out ny common fctor from ny one row or ny one column of given determinnt. If corresponding elements of ny two rows (or columns of determinnt re proportionl (in the sme rtio, then its vlue is zero. For emple Δ Emple 9 Evlute Solution Note tht 3 b b b 3 k k k 3 0 8 36 3 4 7 3 6 0 8 36 0 (rows R nd R re proportionl 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 Property 5 If some or ll elements of row or column of determinnt re epressed s sum of two (or more terms, then the determinnt cn be epressed s sum of two (or more determinnts. For emple, Verifiction L.H.S. +λ +λ +λ 3 3 b b b 3 c c c 3 +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
4 MATHEMATICS Epnding the determinnts long the first row, we get Δ ( + 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 (by rerrnging terms λ λ λ 3 3 b b b3 + b b b3 R.H.S. c c c c c c 3 3 Similrly, we my verify Property 5 for other rows or columns. b c Emple 0 Show tht + b + y c + z 0 y z b c b c b c Solution We hve + b+ y c+ z b c + y z y z y z y z (by Property 5 0 + 0 0 (Using Property 3 nd Property 4 Property 6 If, to ech element of ny 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 pply the opertion R i R i + kr j or C i C i + k C j. Verifiction Let Δ b b b c c c 3 3 3 nd Δ + kc + kc + kc b b b3 c c c 3 3 3 where Δ is obtined by the opertion R R + kr 3. Here, we hve multiplied the elements of the third row (R 3 by constnt k nd dded them to the corresponding elements of the first row (R. Symboliclly, we write this opertion s R R + k R 3.,
DETERMINANTS 5 Now, gin Δ Hence Δ Δ Remrks (i (ii kc kc kc 3 3 b b b3 + b b b3 (Using Property 5 c c c c c c 3 3 Δ + 0 (since R nd R 3 re proportionl If Δ is the determinnt obtined by pplying R i kr i or C i kc i to the determinnt Δ, then Δ kδ. 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. Emple Prove tht + b + b + c 3 + b 4 + 3b + c. 3 6 + 3b 0 + 6b + 3c Solution Applying opertions R R R nd R 3 R 3 3R to the given determinnt Δ, we hve Now pplying R 3 R 3 3R, we get Epnding long C, we obtin Δ + b 0 + b + b + c Δ 0 + b 0 3 7 + 3b + b + b + c Δ 0 + b 0 0 + 0 + 0 ( 0 ( 3 3
6 MATHEMATICS Emple Without epnding, prove tht + y y + z z + Δ z y 0 Solution Applying R R + R to Δ, we get + y + z + y + z + y + z Δ z y Since the elements of R nd R 3 re proportionl, Δ 0. Emple 3 Evlute Δ b c bc c b Solution Applying R R R nd R 3 R 3 R, we get Δ bc 0 b c ( b 0 c b ( c Tking fctors (b nd (c common from R nd R 3, respectively, we get Δ ( b ( c 0 c bc 0 b (b (c [( b + c] (Epnding long first column ( b (b c (c b + c Emple 4 Prove tht b c + b 4bc Solution Let Δ c c + b b + c b c + b c c + b
DETERMINANTS 7 Applying R R R R 3 to Δ, we get Δ 0 c b b c + b c c + b Epnding long R, we obtin c + b b b Δ 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 3 + Emple 5 If, y, z re different nd Δ y y 3 + y 0, then z z 3 + z show tht + yz 0 Solution We hve Δ 3 + y y + y z z + z 3 3 3 3 y y + y y y (Using Property 5 3 z z z z z ( y y + yz y y (Using C 3 C nd then C C z z z z y y ( + yz z z
8 MATHEMATICS ( + yz 0 y y 0 z z (Using R R R nd R 3 R 3 R Tking out common fctor (y from R nd (z from R 3, we get Δ (+ yz ( y ( z 0 y + 0 z+ ( + yz (y (z (z y (on epnding long C Since Δ 0 nd, y, z re ll different, i.e., y 0, y z 0, z 0, we get + yz 0 Emple 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 Applying R R + R + R 3, we hve Δ + + + + + + + + + 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 pplying C C C, C 3 C 3 C, we get 0 0 Δ bc + + + b c b 0 c 0 + + + b c bc ( 0 bc + + + bc + bc + c + b R.H.S. b c Note Alterntely try by pplying C C C nd C C C, then pply 3 3 C C C 3. EXERCISE 4. Using the property of determinnts nd without epnding in Eercises to 7, prove tht: +. y b y+ b 0 3. z c z+ c 7 65 3 8 75 0 4. 5 9 86 b+ c q+ r y+ z p 5. c+ r+ p z+ b q y + b p+ q + y 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 By using properties of determinnts, in Eercises 8 to 4, show tht: 8. (i b b ( b( b c( c 9. c c (ii b c ( b( b c( c ( + b+ c 3 3 3 b c yz y y z z z y ( y (y z (z (y + yz + z + 4 + 4 5+ 4 4 0. (i ( ( + 4 y+k y y (ii y y+k y k ( 3 y+ k y y y+k b c. (i b b c b ( + b+ c 3 c c c b + y+ z y (ii z y+ z+ y ( + y+ z 3 z z+ + y
DETERMINANTS 3. ( + b b b 3. b + b ( + + b 4. b b + b c b b + bc + + b + c c cb c + Choose the correct nswer in Eercises 5 nd 6. 5. Let A be squre mtri 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 mtri. (B Determinnt is number ssocited to mtri. (C Determinnt is number ssocited to squre mtri. (D None of these 4.4 Are of Tringle In erlier clsses, we hve studied tht the re of tringle whose vertices re (, y, (, y nd ( 3, y 3, is given by the epression [ (y y 3 + (y 3 y + 3 (y y ]. Now this epression cn be written in the form of determinnt s Δ y y y 3 3 3... ( Remrks (i Since re is positive quntity, we lwys tke the bsolute vlue of the determinnt in (.
MATHEMATICS (ii If re is given, use both positive nd negtive vlues of the determinnt for clcultion. (iii The re of the tringle formed by three colliner points is zero. Emple 7 Find the re of the tringle whose vertices re (3, 8, ( 4, nd (5,. Solution The re of tringle is given by Δ 3 8 4 5 3 8 4 5 4 0 + ( ( ( + 6 ( 3 7 4 Emple 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 (, y be ny point on AB. Then, re of tringle ABP is zero (Why?. So 0 0 3 0 y This gives ( 3 y 0 or y 3, 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 ± 3 k 0 3k ± 3, i.e., k m. 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
. 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. DETERMINANTS 3 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 epnsion of determinnt in compct form using minors nd cofctors. Definition Minor of n element ij of determinnt is the determinnt obtined by deleting its ith row nd jth column in which element ij lies. Minor of n element ij is denoted by M ij. Remrk Minor of n element of determinnt of order n(n is determinnt of order n. Emple 9 Find the minor of element 6 in the determinnt Δ 3 4 5 6 7 8 9 Solution Since 6 lies in the second row nd third column, its minor M 3 is given by M 3 7 8 8 4 6 (obtined by deleting R nd C in Δ. 3 Definition Cofctor of n element ij, denoted by A ij is defined by A ij ( i + j M ij, where M ij is minor of ij. Emple 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 ( Emple Find minors nd cofctors of the elements, in the determinnt Δ 3 3 3 3 33 Solution By 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 Epnding the determinnt Δ, in Emple, long R, we hve Δ ( + 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 Similrly, Δ cn be clculted by other five wys of epnsion tht is long R, R 3, C, C nd C 3. Hence Δ sum of the product of elements of ny row (or column with their corresponding cofctors. Note If elements of row (or column re multiplied with cofctors of ny other row (or column, then their sum is zero. For emple,
DETERMINANTS 5 Δ 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 Similrly, we cn try for other rows nd columns. Emple Find minors nd cofctors of the elements of the determinnt 3 5 6 0 4 nd verify tht A 3 + A 3 + 3 A 33 0 5 7 Solution We hve M 0 4 5 7 M 6 4 7 3 3 0 0 0; A ( + ( 0 0 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 4 5 9; A ( + ( 9 9 M 3 3 0 + 3 3; A 5 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 0 + 8 8; A 6 0 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. (i 4 0 3 0 0 0 0 0 0 (ii (ii b c d 0 4 3 5 0 5 3 8 3. Using Cofctors of elements of second row, evlute Δ 0. 3 yz 4. Using Cofctors of elements of third column, evlute Δ y z. z y 5. If Δ 3 3 3 3 33 nd A ij is Cofctors of ij, then vlue of Δ is given by (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 Mtri In the previous chpter, we hve studied inverse of mtri. In this section, we shll discuss the condition for eistence of inverse of mtri. To find inverse of mtri A, i.e., A we shll first define djoint of mtri.
4.6. Adjoint of mtri DETERMINANTS 7 Definition 3 The djoint of squre mtri A [ ij ] n n is defined s the trnspose of the mtri [A ij ] n n, where A ij is the cofctor of the element ij. Adjoint of the mtri A is denoted by dj A. Let Then A 3 3 3 3 33 A A A dj ATrnsposeof A A A A A A 3 3 3 3 33 3 Emple 3 Find dj A for A 4 Solution We hve A 4, A, A 3, A A A A A A A A A A 3 3 3 3 33 Hence dj A A A 4 3 A A Remrk For squre mtri of order, given by A The dj A cn lso be obtined by interchnging nd nd by chnging signs of nd, i.e., We stte the following theorem without proof. Theorem If A be ny given squre mtri of order n, then where I is the identity mtri of order n A(dj A (dj A A AI,
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 0 0 A (dj A 0 A 0 A 0 0 A Similrly, we cn show (dj A A A I Hence A (dj A (dj A A A I 0 0 0 0 A I 0 0 Definition 4 A squre mtri A is sid to be singulr if A 0. For emple, the determinnt of mtri A Hence A is singulr mtri. 4 8 is zero Definition 5 A squre mtri 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 mtri 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 A 0 0 ( dj AA i.e. (dj A A 0 A 0 0 0 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 mtri of order n, then dj(a A n. (Why? Theorem 4 A squre mtri A is invertible if nd only if A is nonsingulr mtri. Proof Let A be invertible mtri of order n nd I be the identity mtri of order n. Then, there eists squre mtri 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. Conversely, 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 Thus A is invertible nd A Emple 4 If A 3 3 4 3 3 4 A A dj A A dj, then verify tht A dj A A I. Also find A. 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 Now A (dj A Also A A A dj 7 3 3 0 0 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 7 3 3 0 0 0 0 0 0 A. I 0 0 7 3 3 0 0 Emple 5 If A 3 nd B 4 3, then verify tht (AB B A. Solution We hve AB Since, 3 5 4 3 5 4 AB 0, (AB eists nd is given by (AB 4 5 dj (AB AB 5 4 5 5 Further, A 0 nd B 0. Therefore, A nd B both eist nd re given by A 4 3 3,B
DETERMINANTS 3 3 4 3 4 5 4 5 Therefore BA 5 5 Hence (AB B A Emple 6 Show tht the mtri A 3 stisfies the eqution A 4A + I O, where I is identity mtri nd O is zero mtri. Using this eqution, find A. Solution We hve Hence Now Therefore 3 3 7 A A.A 4 7 7 8 0 A 4A+ I 4 7 + 4 8 0 A 4A + I O A A 4A I 0 0 O 0 0 or A A (A 4 A A I A (Post multiplying by A becuse A 0 or A (A A 4I A or AI 4I A or A 4I A Hence 3 A 4 0 3 3 0 4 EXERCISE 4.5 Find djoint of ech of the mtrices in Eercises nd.. 3 4. 3 5 0 Verify A (dj A (dj A A A I in Eercises 3 nd 4 3. 3 4 6 4. 3 0 0 3
3 MATHEMATICS Find the inverse of ech of the mtrices (if it eists given in Eercises 5 to. 5. 5 6. 4 3 3 7. 0 0 3 8. 3 3 0 9. 4 0 0. 5 7 0 0. 0 cos sin α α 0 sinα cosα 3 7. Let A nd B 5 3. If A 4. For the mtri A 5. For the mtri A 3 0 4 0 0 5 0 3 3 4 6 8 7 9. Verify tht (AB B A. 3, show tht A 5A + 7I O. Hence find A. 3, find the numbers nd b such tht A + A + bi O. 3 3 Show tht A 3 6A + 5A + I O. Hence, find A. 6. If A Verify tht A 3 6A + 9A 4I O nd hence find A 7. Let A be nonsingulr squre mtri 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 mtri 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 system of liner equtions in two or three vribles nd for checking the consistency of the system of liner equtions. Consistent system A system of equtions is sid to be consistent if its solution (one or more eists. Inconsistent system A system of equtions is sid to be inconsistent if its solution does not eist. Note In this chpter, we restrict ourselves to the system of liner equtions hving unique solutions only. 4.7. Solution of system of liner equtions using inverse of mtri Let us epress the system of liner equtions s mtri equtions nd solve them using inverse of the coefficient mtri. Consider the system of equtions + b y + c z d + b y + c z d 3 + b 3 y + c 3 z d 3 Let b c d A b c,x y ndb d 3 b3 c 3 z d 3 Then, the system of equtions cn be written s, AX B, i.e., b c d b c y d 3 b3 c 3 z d 3 Cse I If A is nonsingulr mtri, then its inverse eists. Now AX B or A (AX A B (premultiplying by A or (A A X A B (by ssocitive property or I X A B or X A B This mtri eqution provides unique solution for the given system of equtions s inverse of mtri is unique. This method of solving system of equtions is known s Mtri Method.
34 MATHEMATICS Cse II If A is singulr mtri, then A 0. In this cse, we clculte (dj A B. If (dj A B O, (O being zero mtri, then solution does not eist nd the system of equtions is clled inconsistent. If (dj A B O, then system my be either consistent or inconsistent ccording s the system hve either infinitely mny solutions or no solution. Emple 7 Solve the system of equtions + 5y 3 + y 7 Solution The system of equtions cn be written in the form AX B, where A 5,X nd B 3 y 7 Now, A 0, Hence, A is nonsingulr mtri nd so hs unique solution. Note tht A Therefore X A B 5 3 i.e. y 33 3 Hence 3, y 5 3 7 Emple 8 Solve the following system of equtions by mtri method. 3 y + 3z 8 + y z 4 3y + z 4 Solution The system of equtions cn be written in the form AX B, where We see tht 3 3 8 A, X y 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 eists. 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 i.e. y z 5 8 6 9 7 0 7 5 8 7 0 7 4 A B 8 6 9 7 34 7 5 3 Hence, y nd z 3. Emple 9 The sum of three numbers is 6. If we multiply third number by 3 nd dd second number to it, we get. By dding first nd third numbers, we get double of the second number. Represent it lgebriclly nd find the numbers using mtri method. Solution Let first, second nd third numbers be denoted by, y nd z, respectively. Then, ccording to given conditions, we hve + y + z 6 y + 3z + z y or y + z 0 This system cn be written s A X B, where A 0 3, X Here A 6 (0 3 0 9 0. Now we find dj A y z nd B 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 6 0
36 MATHEMATICS 7 3 Hence dj A 3 0 3 3 Thus A 7 3 A dj (A 3 0 3 9 3 Since X A B X 7 3 6 3 0 3 9 3 0 or 4 33 + 0 y 8 0 0 + + z 9 6+ 33+ 0 9 Thus, y, z 3 EXERCISE 4.6 Emine the consistency of the system of equtions in Eercises to 6.. + y. y 5 3. + 3y 5 + 3y 3 + y 4 + 6y 8 4. + y + z 5. 3 y z 6. 5 y + 4z 5 + 3y + z y z + 3y + 5z + y + z 4 3 5y 3 5 y + 6z Solve system of liner equtions, using mtri method, in Eercises 7 to 4. 7. 5 + y 4 8. y 9. 4 3y 3 7 + 3y 5 3 + 4y 3 3 5y 7 0. 5 + y 3. + y + z. y + z 4 3 + y 5 y z 3 + y 3z 0 3y 5z 9 + y + z 3. + 3y +3 z 5 4. y + z 7 y + z 4 3 + 4y 5z 5 3 y z 3 y + 3z 9 8 7 3
DETERMINANTS 37 3 5 5. If A 3 4, find A. Using A solve the system of equtions 3y + 5z 3 + y 4z 5 + y 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 by mtri method. Miscellneous Emples Emple 30 If, b, c re positive nd unequl, show tht vlue of the determinnt Δ b c b c c b is negtive. Solution Applying C C + C + C 3 to the given determinnt, we get Δ + b+ c b c + b+ c c + b+ c b ( + b + c ( + b + c b c b c 0 c b c (Applying R R R,ndR 3 R 3 R 0 b b c ( + b + c [(c b (b c ( c ( b] (Epnding 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 c b
38 MATHEMATICS Emple 3 If, b, c, re in A.P, find vlue of y+ 4 5y+ 7 8y+ 3y+ 5 6y+ 8 9y+ b 4y+ 6 7y+ 9 0y+ c Solution Applying R R + R 3 R to the given determinnt, we obtin Emple 3 Show tht Δ 0 0 0 3y+ 5 6y+ 8 9y+ b 4y+ 6 7y+ 9 0y+ c ( + y z y z ( + y z yz ( + z yz y 0 (Since b + c yz ( + y + z 3 Solution Applying R R, R yr, R zr to Δ nd dividing by yz, we get 3 3 Δ yz y z y z y y z y z z yz z y Tking common fctors, y, z from C C nd C 3, respectively, we get Δ yz yz ( + y z ( + y z y ( + z z y Applying C C C, C 3 C 3 C, we hve Δ ( y+ z ( y+ z ( y+ z y ( + z y 0 z 0 ( + y z
DETERMINANTS 39 Tking common fctor ( + y + z from C nd C 3, we hve Δ ( + y + z Applying R R (R + R 3, we hve Δ ( + y + z ( y + z ( y+ z ( y+ z y ( + z y 0 z 0 ( + y z yz z y y y z + 0 z + y 0 z Applying C (C + y C nd C C C 3 3 z Δ ( + y + z yz 0 0 y z z y y z z y, we get Finlly epnding long R, we hve Δ ( + y + z (yz [( + z ( + y yz] ( + y + z (yz ( + y + z ( + y + z 3 (yz Emple 33 Use product Solution Consider the product 0 0 3 9 3 3 4 6 y + z y 3z 3 y + 4z 0 0 3 9 3 3 4 6 to solve the system of equtions
40 MATHEMATICS Hence or 9+ 0 + + 3 4 0 8 8 0 4 3 0 6 6 + + + 6 8+ 4 0 4+ 4 3+ 6 8 0 0 3 9 3 3 4 6 0 0 0 0 0 0 Now, given system of equtions cn be written, in mtri form, s follows 0 3 y 3 4 z y z 0 3 3 4 + 0+ 0 9 6 5 + 6+ 4 3 Hence 0, y 5 nd z 3 Emple 34 Prove tht Δ + b c+ d p+ q c p + b c + d p + q ( b d q u v w u v w Solution Applying R R R to Δ, we get Δ ( c( p( + b c + d p + q u v w c p ( + b c + d p + q u v w 0 9 3 6
DETERMINANTS 4 Applying R R R, we get c p Δ ( b d q u v w Miscellneous Eercises on Chpter 4 sin θ cosθ. Prove tht the determinnt sin θ cosθ is independent of θ. bc 3. Without epnding the determinnt, prove tht b b c b b cosα cosβ cosα sin β sin α 3. Evlute sinβ cosβ 0 sin α cosβ sin α sinβ cosα 4. If, b nd c re rel numbers, nd b+ c c+ + b Δ c+ + b b+ c 0, + b b+ c c+ Show tht either + b + c 0 or b c. + 5. Solve the eqution + 0, 0 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 3 c c b c c 7. If A ( 3.
4 MATHEMATICS 8. Let A 3. Verify tht 5 (i [dj A] dj (A (ii (A A y + y 9. Evlute 0. Evlute y + y + y y y + y y + y Using properties of determinnts in Eercises to 5, prove tht:.. 3. 4. α α β+γ β β γ+α γ γ α+β p 3 y y py 3 z z pz 3 3 +b +c b+ 3b b+ c c + c+b (β γ (γ α (α β (α + β + γ ( + pyz ( y (y z (z, where p is ny sclr. 3c + p + p+ q 3+ p 4+ 3p+ q 3 6+ 3p 0+ 6p+ 3q 6. Solve the system of equtions 3 0 y z 4 3( + b + c (b + bc + c 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 y z 6 9 0 y z Choose the correct nswer in Eercise 7 to 9. 7. If, b, c, re in A.P, then the determinnt + + 3 + + 3 + 4 + b + 4 + 5 + c is (A 0 (B (C (D 0 0 8. If, y, z re nonzero rel numbers, then the inverse of mtri A 0 y 0 is 0 0 z (A (C 9. Let A 0 0 0 y 0 0 0 z 0 0 0 y 0 yz 0 0 z (B (D 0 0 yz 0 y 0 0 0 z 0 0 0 0 yz 0 0 sinθ sin sin θ θ, where 0 θ π. Then sinθ (A Det (A 0 (B Det (A (, (C Det (A (, 4 (D Det (A [, 4]
44 MATHEMATICS Summry Determinnt of mtri A [ ] is given by Determinnt of mtri A Determinnt of mtri A b c is given by A b c b c b c 3 3 3 is given by (epnding long R b c c b b3 c3 3 c3 3 b3 3 3 3 A b c b + c b c For ny squre mtri A, the A stisfy following properties. A A, where A trnspose of A. If we interchnge ny two rows (or columns, then sign of determinnt chnges. If ny two rows or ny two columns re identicl or proportionl, then vlue of determinnt is zero. If we multiply ech element of row or column of determinnt by constnt k, then vlue of determinnt is multiplied by k. Multiplying determinnt by k mens multiply elements of only one row (or one column by k. 3 If A [ ij ] 3 3,then k.a k A If elements of row or column in determinnt cn be epressed s sum of two or more elements, then the given determinnt cn be epressed 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 (, y, (, y nd ( 3, y 3 is given by Δ y y y 3 3 Minor of n element ij of the determinnt of mtri A is the determinnt obtined by deleting i th row nd j th column nd denoted by M ij. Cofctor of ij of given by A ij ( i + j M ij Vlue of determinnt of mtri A is obtined by sum of product of elements of row (or column with corresponding cofctors. For emple, A A + A + 3 A 3. If elements of one row (or column re multiplied with cofctors of elements of ny other row (or column, then their sum is zero. For emple, A + A + 3 A 3 0 3 A A A3 If A 3, then dj A A A A 3, where A is ij 3 3 33 A3 A3 A 33 cofctor of ij A (dj A (dj A A A I, where A is squre mtri of order n. A squre mtri A is sid to be singulr or non-singulr ccording s A 0 or A 0. If AB BA I, where B is squre mtri, then B is clled inverse of A. Also A B or B A nd hence (A A. A squre mtri A hs inverse if nd only if A is non-singulr. A ( A A dj If + b y + c z d + b y + c z d 3 + b y + c z d, 3 3 3 then these equtions cn be written s A X B, where b c d A,X ndb b c y d 3 b3 c 3 z d 3
46 MATHEMATICS Unique solution of eqution AX B is given by X A B, where A 0. A system of eqution is consistent or inconsistent ccording s its solution eists or not. For squre mtri A in mtri eqution AX B (i A 0, there eists unique solution (ii A 0 nd (dj A B 0, then there eists no solution (iii A 0 nd (dj A B 0, then system my or my not be consistent. Historicl Note The Chinese method of representing the coefficients of the unknowns of severl liner equtions by using rods on clculting bord nturlly led to the discovery of simple method of elimintion. The rrngement of rods ws precisely tht of the numbers in determinnt. The Chinese, therefore, erly 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 century in his work Ki Fukudi no Ho in 683 showed tht he hd the ide of determinnts nd of their epnsion. But he used this device only in eliminting quntity from two equtions nd not directly in the solution of set of simultneous liner equtions. T. Hyshi, The Fkudoi nd Determinnts in Jpnese Mthemtics, in the proc. of the Tokyo Mth. Soc., V. Vendermonde ws the first to recognise determinnts s independent functions. He my be clled the forml founder. Lplce (77, gve generl method of epnding determinnt in terms of its complementry 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 theory of numbers. The net 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 dy, Cuchy (8 presented one on the sme subject. He used the word determinnt in its present sense. He gve the proof of multipliction theorem more stisfctory thn Binet s. The gretest contributor to the theory ws Crl Gustv Jcob Jcobi, fter this the word determinnt received its finl cceptnce.