CHAPTER 4: DETERMINANTS

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CHAPTER 4: DETERMINANTS MARKS WEIGHTAGE 0 mrks NCERT Importnt Questions & Answers 6. If, then find the vlue of. 8 8 6 6 Given tht 8 8 6 On epnding both determinnts, we get 8 = 6 6 8 36 = 36 36 36 = 0 = 36 = ± 6. Prove tht b b 3 b 4 3b 3 6 3b 0 6b 3 Prepred b: M. S. KumrSwm, TGT(Mths) Pge - - 3 Appling opertions R R R nd R 3 R 3 3R to the given determinnt Δ, we hve b b 0 b 0 3 7 3b Now ppling R 3 R 3 3R, we get b b 0 b 0 0 Epnding long C, we obtin b 0 0 ( 0) ( ) 0 b 3 3. Prove tht b b 4b b b Let b b b Appling R R R R 3 to Δ, we get 0 b b b b Epnding long R, we obtin 0 b ( ) b b ( b) b b b

= ( b + b b) b (b ) = b + b b b + b + b = 4 b 4. If,, z re different nd We hve 3 3 z z z 3 3 3 z z z 0 then show tht + z = 0 3 Now, we know tht If some or ll elements of row or olumn of determinnt re epressed s sum of two (or more) terms, then the determinnt n be epressed s sum of two (or more) determinnts. 3 3 z z z z z 3 ( ) z (Using C 3 C nd then C C ) z z z z ( z) z ( z) 0 0 z z z (Using R R R nd R 3 R 3 R ) Tking out ommon ftor ( ) from R nd (z ) from R 3, we get ( z)( )( z ) 0 0 z = ( + z) ( ) (z ) (z ) (on epnding long C ) Sine Δ = 0 nd,, z re ll different, i.e., 0, z 0, z 0, we get + z = 0 5. Show tht b b b b b b LHS b Tking out ftors,b, ommon from R, R nd R 3, we get Prepred b: M. S. KumrSwm, TGT(Mths) Pge - -

b b b b Appling R R + R + R 3, we hve b b b b b b b Now ppling C C C, C 3 C 3 C, we get 0 0 b b b 0 0 b ( 0) b b b b b = RHS b 6. Using the propert of determinnts nd without epnding, prove tht b q r z p r p z b q b p q r z b q r z LHS r p z b p q b b q r r p p q (interhnge row nd olumn) z z b q r r p r [usingc 3 C 3 (C + C )] z z z b ( ) q r r p r (tking ommon from C 3 ) z z z Prepred b: M. S. KumrSwm, TGT(Mths) Pge - 3 -

b ( ) q p r (using C C C 3 nd C C C 3 ) z b p q r (using z C C ) p b q RHS (interhnge row nd olumn) r z 7. Using the propert of determinnts nd without epnding, prove tht b b b b 4 b b b b LHS b b b b b R 3 ] b b ( b)( b) b 0 0 0 Epnding orresponding to first row R, we get 0 b [tking out ftors from R, b from R nd from (tking out ftors from C, b from C nd from C 3 ) (using R R + R nd R R R 3 ) b (0 ) 4 b RHS 8. Using the propert of determinnts nd without epnding, prove tht b b ( b)( b )( ) LHS b b Appling R R R3 nd R R R3, we get 0 0 ( )( ) 0 b b 0 b ( b )( b ) Tking ommon ftors ( ) nd (b ) from R nd R respetivel, we get Prepred b: M. S. KumrSwm, TGT(Mths) Pge - 4 -

0 ( ) ( )( b ) 0 ( b ) Now, epnding orresponding to C, we get = ( ) (b ) (b + ) = ( b) (b ) ( ) = RHS 9. Using the propert of determinnts nd without epnding, prove tht b ( b)( b )( )( b ) b 3 3 3 LHS b b 3 3 3 Appling C C C nd C C C3, we get 0 0 b b b b 3 3 3 3 3 0 0 b b ( b)( b b ) ( b )( b b ) 3 Tking ommon ( b) from C nd (b ) from C, we get 0 0 ( b)( b ) b b ( b b ) ( b b ) 3 Now, epnding long R, we get = ( b) (b ) [ (b + b + ) ( + b + b )] = ( b) (b ) [b + b + b b ] = ( b) (b ) (b b + ) = ( b) (b ) [b( ) + ( ) ( + )] = ( b) (b ) ( ) ( + b + )= RHS. 0. Using the propert of determinnts nd without epnding, prove tht z z z z z z z z z LHS z z z ( )( )( )( ) Appling R R, R R nd R3 zr3, we hve 3 z 3 z z z z 3 z Prepred b: M. S. KumrSwm, TGT(Mths) Pge - 5 -

z z z 3 3 (tke out z ommon from C 3 ) z 3 3 3 3 z z 3 3 Epnding orresponding to C 3, we get 3 3 3 3 z z 0 0 (using R R R nd R3 R3 R ) 3 3 3 3 ( )( z ) ( z )( ) = ( + ) ( ) (z ) (z + + z) (z + ) (z ) ( ) ( + + ) = ( ) (z ) [( + ) (z + + z) (z + ) ( + + )] = ( )(z )[z + + z + z + 3 + z z z z 3 ] = ( )(z )[z z + z ] = ( )(z )[z(z ) + (z )] = ( )(z )[z(z ) + (z )(z + )] = ( ) (z ) [(z ) ( + z + z)] = ( ) ( z) (z ) ( + z + z) = RHS.. Using the propert of determinnts nd without epnding, prove tht 4 4 (5 4)(4 ) 4 4 LHS 4 4 5 4 5 4 4 5 4 4 (5 4) 4 4 (5 4) 0 4 0 0 0 4 Epnding long C, we get = (5 + 4) {(4 ) (4 )} (5 4)(4 ) = RHS. (using C C + C + C 3 ) [tke out (5 + 4) ommon from C ]. (Using R R R nd R 3 R 3 R ). Using the propert of determinnts nd without epnding, prove tht k k k (3 k) k Prepred b: M. S. KumrSwm, TGT(Mths) Pge - 6 -

k LHS k k 3 k 3 k k 3 k k (3 k) k k (using C C + C + C 3 ) [tke out (5 + 4) ommon from C ]. (3 k) 0 k 0 (Using R R R nd R 3 R 3 R ) 0 0 Epnding long C 3, we get (3 k) ( k 0) k (3 k) = RHS k 3. Using the propert of determinnts nd without epnding, prove tht b b b b ( b ) b b LHS b b b b b b b b b b (Using R R R R ) b Tke out ( + b + ) ommon from R, we get ( b ) b b b b 0 0 ( b ) b b 0 0 b Epnding long R, we get = ( + b + ) {( b ) ( b)} = ( + b + ) [ (b + + ) ( ) ( + + b)] 3 ( b )( b )( b ) ( b ) = RHS 3 3 (Using C C C nd C 3 C 3 C ) Prepred b: M. S. KumrSwm, TGT(Mths) Pge - 7 -

4. Using the propert of determinnts nd without epnding, prove tht z z z ( z) z z z LHS z z z z ( z) ( z) z (using C C + C + C 3 ) ( z) z ( z) z [tke out ( + + z) ommon from C ]. z ( z) 0 z 0 (Using R R R nd R 3 R 3 R ) 0 0 z ( z)( z)( z) 0 0 0 0 Epnding long R 3, we get ( z)( z)( z) ( 0) 3 ( z)( z)( z) ( z) =RHS 3 5. Using the propert of determinnts nd without epnding, prove tht LHS (using C C + C + C 3 ) ( ) [tke out ( ) 0 0 3 ( ) ommon from C ]. (Using R R R nd R 3 R 3 R ) Prepred b: M. S. KumrSwm, TGT(Mths) Pge - 8 -

( ) 0 ( ) 0 ( ) Tke out ( ) ommon from R nd sme from R 3, we get ( )( )( ) 0 Epnding long C, we get ( )( )( ) 0 ( )( )( )( ) 3 3 3 = RHS 6. Using the propert of determinnts nd without epnding, prove tht b b b b b ( b ) 3 b b b b b LHS b b b b 0 b b 0 b b( b ) ( b ) b 0 b ( b ) 0 b b 0 b ( b ) 0 0 0 b Epnding long R, we get ( b ) ( b ) 3 ( b ) RHS ( R R br R ) 3 3 (Using C C bc3 nd C C C3 ) 7. Using the propert of determinnts nd without epnding, prove tht b b b b b b b LHS b b b b Prepred b: M. S. KumrSwm, TGT(Mths) Pge - 9 -

Tking out ommon ftors, b nd from R, R nd R 3 respetivel, we get b b b b b b 0 (Using R R R nd R 3 R 3 R ) 0 Multipl nd divide C b, C b b nd C 3 b nd then tke ommon out from C, C nd C 3 respetivel, we get b b b 0 0 b 0 Epnding long R 3, we get ( ) ( ) ( b ) b RHS 0 8. Find vlues of k if re of tringle is 4 sq. units nd verties re (i) (k, 0), (4, 0), (0, ) (ii) (, 0), (0, 4), (0, k) k 0 (i) We hve Are of tringle = 4 0 4 0 k(0 ) + (8 0) = 8 k(0 ) + (8 0) = ± 8 On tking positive sign k + 8 = 8 k = 0 k = 0 On tking negtive sign k + 8 = 8 k = 6 k = 8 k =0, 8 0 (ii) We hve Are of tringle = 0 4 4 0 k (4 k) + (0 0) = 8 (4 k) + (0 0) = ± 8 [ 8 + k] = ± 8 On tking positive sign, k 8 = 8 k = 6 k = 8 On tking negtive sign, k 8 = 8 k = 0 k = 0 k =0, 8 Prepred b: M. S. KumrSwm, TGT(Mths) Pge - 0 -

9. If re of tringle is 35 sq units with verties (, 6), (5, 4) nd (k, 4). Then find the vlue of k. 6 We hve Are of tringle = 5 4 35 k 4 (4 4) + 6(5 k) + (0 4k) = 70 (4 4) + 6 (5 k) + (0 4k) = ± 70 30 6k + 0 4k = ± 70 On tking positive sign, 0k + 50 = 70 0k = 0 k = On tking negtive sign, 0k + 50 = 70 0k = 0 k = k =, 0. Using Coftors of elements of seond row, evlute Given tht 5 3 8 0 3 Coftors of the elements of seond row 3 8 A ( ) (9 6) 7 3 A 5 8 3 ( ) (5 8) 7 5 3 8 0 3 3 5 3 nd A3 ( ) (0 3) 7 Now, epnsion of Δ using oftors of elements of seond row is given b A A 3 A3 = 7 + 0 7 + ( 7) = 4 7 = 7 Prepred b: M. S. KumrSwm, TGT(Mths) Pge - - 3. If A =, show tht A 5A + 7I = O. Hene find A. 3 Given tht A = Now, A 5A + 7I = O 3 3 9 3 8 5 A A. A 3 4 5 3 8 5 3 0 5 7 5 3 0 8 5 5 5 7 0 5 3 5 0 0 7 8 5 7 5 5 0 0 0 O 5 5 0 30 3 0 0

A 5A + 7I = O 3 A 6 7 0 A eists. Now, A.A 5A = 7I Multipling b A on both sides, we get A.A (A ) 5A(A ) = 7I(A ) AI 5I = 7A (using AA = I nd IA = A ) A ( A 5 I) 5I A 5 0 3 7 7 7 0 5 7 3 A 7 3 3. For the mtri A =, find the numbers nd b suh tht A + A + bi = O. 3 Given tht A = 3 3 9 6 8 A A. A 3 4 3 Now, A A bi O 8 3 0 b O 4 3 0 8 3 b 0 O 4 3 0 b 3 b 8 0 0 4 3 b 0 0 If two mtries re equl, then their orresponding elements re equl. + 3 + b = 0 (i) 8 + = 0 (ii) 4 + = 0 (iii) nd 3 + + b = 0 (iv) Solving Eqs. (iii) nd (iv), we get 4 + = 0 = 4 nd 3 + + b = 0 3 4 + b = 0 b = Thus, = 4 nd b = 3. For the mtri A = 3, Show tht A 3 6A + 5A + I = O. Hene, find A. 3 Prepred b: M. S. KumrSwm, TGT(Mths) Pge - -

Given tht A = 3 3 3 3 4 A A. A 3 3 6 4 3 6 9 3 8 4 3 3 6 3 3 9 7 3 4 4 4 4 4 4 6 3 3 nd A A. A 3 8 4 3 3 8 8 3 6 4 3 4 4 7 3 4 3 7 3 8 7 6 4 7 9 4 8 7 3 7 69 3 3 58 3 A 6A 5A I 8 7 4 0 0 3 7 69 6 3 8 4 5 3 0 0 3 3 58 7 3 4 3 0 0 8 7 4 6 5 5 5 0 0 3 7 69 8 48 84 5 0 5 0 0 3 3 58 4 8 84 0 5 5 0 0 8 4 5 7 5 0 6 5 0 3 8 5 0 7 48 0 69 84 5 0 3 4 0 0 38 5 0 58 84 5 0 0 0 0 0 0 O 0 0 0 A 3 (6 3) (3 6) ( 4) 3 9 5 0 3 A eist 3 Now, A 6A 5A I O AA( AA ) 6 A( AA ) 5( AA ) ( IA ) O AAI 6AI 5I A O A 6A 5I A ( A A 6 A 5 I ) ( A A 6 A 5 I ) 4 0 0 A 3 8 4 6 3 5 0 0 7 3 4 3 0 0 Prepred b: M. S. KumrSwm, TGT(Mths) Pge - 3 -

4 6 6 6 5 0 0 A 3 8 4 6 8 0 5 0 7 3 4 6 8 0 0 5 4 6 5 6 0 6 0 A 3 6 0 8 5 4 8 0 7 0 3 6 0 4 8 5 3 4 5 A 9 4 5 3 4. Solve sstem of liner equtions, using mtri method, + + z = z = 3 3 5z = 9 The given sstem n be written s AX = B, where A 4, X nd B 3 0 3 5 z 9 A 4 (0 6) ( 0 0) (6 0) 0 3 5 = 5 + 0 + 6 = 68 0 Thus, A is non-singulr, Therefore, its inverse eists. Therefore, the given sstem is onsistent nd hs unique solution given b X = A B. Coftors of A re A = 0 + 6 = 6, A = ( 0 + 0) = 0, A 3 = 6 + 0 = 6 A = ( 5 3) = 8, A = 0 0 = 0, A 3 = (6 0) = 6 A 3 = ( + 4) =, A 3 = ( 4 ) = 6, A 33 = 8 = 0 6 0 6 6 8 dj( A) 8 0 6 0 0 6 6 0 6 6 0 6 8 A ( dja) 0 0 6 A 68 6 6 0 6 8 Now, X A B 0 0 6 3 68 z 6 6 0 9 T Prepred b: M. S. KumrSwm, TGT(Mths) Pge - 4 -

6 4 8 68 0 30 54 34 68 68 z 6 8 90 0 3 3 Hene,, nd z 5. Solve sstem of liner equtions, using mtri method, + z = 4 + 3z = 0 + + z = The given sstem n be written s AX = B, where 4 A 3, X nd B 0 z Here, A 3 = ( + 3) ( ) ( + 3) + ( ) = 4 + 5 + = 0 0 Thus, A is non-singulr, Therefore, its inverse eists. Therefore, the given sstem is onsistent nd hs unique solution given b X = A B. Coftors of A re A = + 3 = 4, A = ( + 3) = 5, A 3 = =, A = ( ) =, A = = 0, A 3 = ( + ) =, A 3 = 3 =, A 3 = ( 3 ) = 5, A 33 = + = 3 T 4 5 4 dj( A) 0 5 0 5 5 3 3 4 A ( dja) 5 0 5 A 0 3 4 4 Now, X A B 5 0 5 0 0 z 3 6 0 4 0 0 0 0 0 0 0 z 4 0 6 0 Hene, =, = nd z =. Prepred b: M. S. KumrSwm, TGT(Mths) Pge - 5 -

6. Solve sstem of liner equtions, using mtri method, + 3 +3 z = 5 + z = 4 3 z = 3 The given sstem n be written s AX = B, where 3 3 5 A, X nd B 4 3 z 3 3 3 Here, A 3 = (4 + ) 3 ( 3) + 3 ( + 6) = 0 + 5 + 5 = 40 0 Thus, A is non-singulr. Therefore, its inverse eists. Therefore, the given sstem is onsistent nd hs unique solution given b X = A B Coftors of A re A = 4 + = 5, A = ( 3) = 5, A 3 = ( + 6) = 5, A = ( 6 + 3) = 3, A = ( 4 9) = 3, A 3 = ( 9) =, A 3 = 3 + 6 = 9, A 3 = ( 3) =, A 33 = 4 3 = 7 T 5 5 5 5 3 9 dj( A) 3 3 5 3 9 7 5 7 5 3 9 A ( dja) 5 3 A 40 5 7 5 3 9 5 Now, X A B 5 3 4 40 z 5 7 3 5 7 40 5 5 3 80 40 40 z 5 44 40 Hene, =, = nd z =. 7. Solve sstem of liner equtions, using mtri method, + z = 7 3 + 4 5z = 5 + 3z = The given sstem n be written s AX = B, where Prepred b: M. S. KumrSwm, TGT(Mths) Pge - 6 -

7 A 3 4 5, X nd B 5 3 z Here, A 3 4 5 = ( 5) ( ) (9 + 0) + ( 3 8) 3 = 7 + 9 = 4 0 Thus, A is non-singulr. Therefore, its inverse eists. Therefore, the given sstem is onsistent nd hs unique solution given b X = A B Coftors of A re A = 5 = 7, A = (9 + 0) = 9, A 3 = 3 8 =, A = ( 3 + ) =, A = 3 4 =, A 3 = ( + ) =, A 3 = 5 8 = 3, A 3 = ( 5 6) =, A 33 = 4 + 3 = 7 T 7 9 7 3 dj( A) 9 3 7 7 7 3 A ( dja) 9 A 4 7 7 3 5 Now, X A B 9 4 4 z 7 3 49 5 36 8 33 5 3 4 4 4 z 77 5 84 3 Hene, =, = nd z = 3. 3 5 8. If A = 3 4 find A. Using A,Solve sstem of liner equtions: 3 + 5z = 3 + 4z = 5 + z = 3 The given sstem n be written s AX = B, where 3 5 A 3 4, X nd B 5 z 3 Prepred b: M. S. KumrSwm, TGT(Mths) Pge - 7 -

3 5 Here, A 3 4 = ( 4 + 4) ( 3) ( 6 + 4) + 5 (3 ) = 0 6 + 5 = 0 Thus, A is non-singulr. Therefore, its inverse eists. Therefore, the given sstem is onsistent nd hs unique solution given b X = A B Coftors of A re A = 4 + 4 = 0, A = ( 6 + 4) =, A 3 = 3 =, A = (6 5) =, A = 4 5 = 9, A 3 = ( + 3) = 5, A 3 = ( 0) =, A 3 = ( 8 5) = 3, A 33 = 4 + 9 = 3 T 0 0 dj( A) 9 5 9 3 3 3 5 3 0 0 A ( dja) 9 3 9 3 A 5 3 5 3 0 Now, X A B 9 3 5 z 5 3 3 0 5 6 45 69 z 5 39 3 Hene, =, = nd z = 3. 9. The ost of 4 kg onion, 3 kg whet nd kg rie is Rs 60. The ost of kg onion, 4 kg whet nd 6 kg rie is Rs 90. The ost of 6 kg onion kg whet nd 3 kg rie is Rs 70. Find ost of eh item per kg b mtri method. Let the pries (per kg) of onion, whet nd rie be Rs., Rs. nd Rs. z, respetivel then 4 + 3 + z = 60, + 4 + 6z = 90, 6 + + 3z = 70 This sstem of equtions n be written s AX = B, where 4 3 60 A 4 6, X nd B 90 6 3 z 70 4 3 Here, A 4 6 = 4( ) 3(6 36) + (4 4) 6 3 = 0 + 90 40 = 50 0 Thus, A is non-singulr. Therefore, its inverse eists. Therefore, the given sstem is onsistent nd hs unique solution given b X = A B Coftors of A re, Prepred b: M. S. KumrSwm, TGT(Mths) Pge - 8 -

A = = 0, A = (6 36) = 30, A 3 = 4 4 = 0, A = (9 4) = 5, A = = 0, A 3 = (8 8) = 0, A 3 = (8 8) = 0, A 3 = (4 4) = 0, A 33 = 6 6 = 0 T 0 30 0 0 5 0 dj( A) 5 0 0 30 0 0 0 0 0 0 0 0 0 5 0 A ( dja) 30 0 0 A 50 0 0 0 0 5 0 60 Now, X A B 30 0 0 90 50 z 0 0 0 70 0 450 700 50 5 800 0 400 400 8 50 50 z 00 900 700 400 8 = 5, = 8 nd z = 8. Hene, prie of onion per kg is Rs. 5, prie of whet per kg is Rs. 8 nd tht of rie per kg is Rs. 8. 30. Without epnding the determinnt, prove tht b LHS b b b Appling R R, R br nd R3 R3, we get 3 b 3 b b b b 3 b b b b 3 3 b [Tking out ftor b from C 3 ] 3 3 b 3 b b b b 3 b 3 ( ) b b (using C C 3 nd C C 3 ) 3 3 Prepred b: M. S. KumrSwm, TGT(Mths) Pge - 9 -

3 3 b b RHS 3 b b 3. If, b nd re rel numbers, nd b b 0. Show tht either + b + = 0 or = b =. b b b b b b ( b ) b ( b ) b b ( b ) b ( b ) b b b b ( b ) 0 b 0 b b b Epnding long C, we get ( ) b b b b ( b ) ( b )( b) ( )( b ) b b ( b ) b b b ( b b ) ( b ) b b b b b ( b ) b b b It is given tht Δ= 0, ( b ) b b b 0 Either b 0 or b b b 0 b b b 0 b b b 0 b b b 0 b b b b 0 ( b) ( b ) ( ) 0 (using C C + C + C 3 ) [tke out ( b ) ommon from C ]. (Using R R R nd R 3 R 3 R ) ( b) ( b ) ( ) 0 [sine squre of n rel number is never negtive] ( b) ( b ) ( ) 0 b, b, b Prepred b: M. S. KumrSwm, TGT(Mths) Pge - 0 -

3. Prove tht b b b 4 b b b b b LHS b b b b b Tking out from C, b from C nd from C 3, we get b b b b 0 b b b b b b b 0 b 0 b b Epnding long C,we get = (b) [ ( b) { ( ) + ( + ) } ] = (b ) () = 4 b = RHS. [Using C C + C C 3] [Using R R R 3] 33. Using properties of determinnts, prove tht LHS (using C 3 C 3 + C ) ( ) (Tking out (α + β + γ) ommon from C ) ( ) 0 0 (Using R R R nd R 3 R 3 R ) ( )( )( )( ) Epnding long C 3, we get = (α + β + γ) [(β α)(γ α ) (γ α)(β α )] = (α + β + γ) [(β α)(γ α)(γ + α) (γ α)(β α)(β + α)] = (α + β + γ) (β α)(γ α)[γ + α β α] = (α + β + γ) (β α)(γ α)(γ β) = (α + β + γ) (α β)(β γ)(γ α) = RHS Prepred b: M. S. KumrSwm, TGT(Mths) Pge - -

34. Using properties of determinnts, prove tht 3 b LHS b 3b b b 3 b b b 3b b b b 3 b ( b ) 3b b b 3 Now ppling R R R, R 3 R 3 R, we get b ( b ) 0 b b 0 3 b b 3b b 3( b )( b b ) b 3 (using C C + C + C 3 ) (Tking out ( + b + ) ommon from C ) Epnding long C, we get = ( + b + )[(b + ) ( + ) ( b) ( )] = ( + b + )[4b + b + + + + b b] = ( + b + ) (3b + 3b + 3) = 3( + b + )(b + b + ) = RHS 35. Solve the sstem of equtions: 3 0 4 z 4 6 5 z 6 9 0 z Let p, q nd r, then the given equtions beome z p + 3q + 0r = 4, 4p 6q + 5r =, 6p + 9q 0r = This sstem n be written s AX = B, where 3 0 p 4 A 4 6 5, X q nd B 6 9 0 r 3 0 Here, A 4 6 5 (0 45) 3( 80 30) 0(36 36) 6 9 0 = 50 + 330 + 70 = 00 0 Thus, A is non-singulr. Therefore, its inverse eists. Therefore, the bove sstem is onsistent nd hs unique solution given b X = A B Coftors of A re A = 0 45 = 75, Prepred b: M. S. KumrSwm, TGT(Mths) Pge - -

A = ( 80 30) = 0, A 3 = (36 + 36) = 7, A = ( 60 90) = 50, A = ( 40 60) = 00, A 3 = (8 8) = 0, A 3 = 5 + 60 = 75, A 3 = (0 40) = 30, A 33 = = 4 75 0 7 75 50 75 dj( A) 50 00 0 0 00 30 75 30 4 7 0 4 75 50 75 A ( dja) 0 00 30 A 00 7 0 4 75 50 75 4 X A B 0 00 30 00 z 7 0 4 300 50 50 600 440 00 60 400 00 00 3 z 88 0 48 40 5 p, q, r 3 5,, 3 z 5 =, = 3 nd z = 5. 36. If, b,, re in A.P, then find the determinnt of 3 Let A 3 4 b 4 5 3 0 0 ( b ) 4 5 T 3 3 4 b 4 5 (using R R R R 3 ) But,b, re in AP. Using b = +, we get 3 A 0 0 0 0 [Sine, ll elements of R re zero] 4 5 Prepred b: M. S. KumrSwm, TGT(Mths) Pge - 3 -

3 37. Show tht the mtri A stisfies the eqution A 4A + I = O, where I is identit mtri nd O is zero mtri. Using this eqution, find A. 3 Given tht A A 3 3 7 AA 4 7 7 3 0 Hene, A 4A I 4 4 7 0 7 8 0 7 8 0 0 0 4 7 4 8 0 4 4 0 7 8 O 0 0 Now, A 4A I O AA 4A I AA( A ) 4AA IA (Post multipling b A beuse A 0) A( AA ) 4I A AI 4I A 0 3 4 0 3 3 A 4I A 4 0 0 4 3 A 38. Solve the following sstem of equtions b mtri method. 3 + 3z = 8 + z = 4 3 + z = 4 The sstem of eqution n be written s AX = B, where 3 3 8 A, X nd B 4 3 z 4 3 3 Here, A 4 3 = 3 ( 3) + (4 + 4) + 3 ( 6 4) = 7 0 Hene, 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 8 0 5 dj( A) 5 6 8 6 9 9 7 0 7 5 A ( dja) 8 6 9 A 7 0 7 T Prepred b: M. S. KumrSwm, TGT(Mths) Pge - 4 -

5 8 X A B 8 6 9 7 z 0 7 4 7 34 7 z 5 3 Hene =, = nd z = 3. Given tht ( ) z z 39. Show tht ( z) z 3 z( z) z z ( ) ( z) z ( z) z z z ( ) Appling R R, R R,R 3 zr 3 to Δ nd dividing b z, we get ( z) z ( z) z z z z z( ) Tking ommon ftors,, z from C, C nd C 3 respetivel, we get ( z) z ( z) z z z ( ) Appling C C C, C 3 C 3 C, we hve ( z) ( z) ( z) ( z) 0 z 0 ( ) z Tking ommon ftor ( + + z) from C nd C 3, we hve ( z) ( z) ( z) ( z) ( z) 0 z 0 ( ) z Appling R R (R + R 3 ), we hve z z ( z) z 0 z 0 z Appling C (C + C ) nd C 3 C 3 + z C, we get Prepred b: M. S. KumrSwm, TGT(Mths) Pge - 5 -

z 0 0 ( ) z z z z z Finll epnding long R, we hve Δ = ( + + z) (z) [( + z) ( + ) z] = ( + + z) (z) ( + + z) = ( + + z) 3 (z) 0 40. Use produt 0 3 9 3 to solve the sstem of equtions 3 4 6 + z = 3z = 3 + 4z = 0 Consider the produt 0 3 9 3 3 4 6 9 0 3 4 0 0 0 8 8 0 4 3 0 6 6 0 0 6 8 4 0 4 4 3 6 8 0 0 0 Hene, 0 3 9 3 3 4 6 Now, given sstem of equtions n be written, in mtri form, s follows 0 3 3 4 z 0 0 3 9 3 z 3 4 6 0 0 9 6 5 6 4 3 Hene = 0, = 5 nd z = 3 Prepred b: M. S. KumrSwm, TGT(Mths) Pge - 6 -

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