If A = 3 - 5 4 2 , find A ( a d j A ) .

We have to find the value of A ( a d j A ) . Given matrix A = 3 - 5 4 2 Let C ij be cofactor of a ij in A . Then, the cofactors of elements of A are given by C 11 = 2 , C 12 = - ( 4 ) = - 4 , C 21 = - ( - 5 ) = 5 , C 22 = 3 ∴ adjA = C 11 C 12 C 21 C 22 T = 2 - 4 5 3 T = 2 5 - 4 3 Now, A ( adj A ) = 3 - 5 4 2 2 5 - 4 3 = 6 + 20 15 - 15 8 - 8 20 + 6 = 26 0 0 26 This is the required value of A ( a d j A ) .

If the matrix A = x + 4 x x x x + 4 x x x x + 4 is singular, find x .

Given, A = x + 4 x x x x + 4 x x x x + 4 If the matrix is singular, its determinant has to be zero ∴ A = 0 x + 4 x x x x + 4 x x x x + 4 = 0 Applying C 1 → C 1 + C 2 + C 3 , ⇒ x + 4 + x + x x x x + x + 4 + x x + 4 x x + x + x + 4 x x + 4 = 0 Taking out ( 3 x + 4 ) common from C 1 , we get ⇒ ( 3 x + 4 ) 1 x x 1 x + 4 x 1 x x + 4 = 0 Applying R 2 → R 2 - R 1 and R 3 → R 3 - R 1 [to create two zeros in the first column] ⇒ ( 3 x + 4 ) 1 x x 0 4 0 0 0 4 = 0 Expanding by first column, we get ⇒ ( 3 x + 4 ) 1 4 0 0 4 = 0 ⇒ ( 3 x + 4 ) 1 ( 4 2 - 0 ) = 0 ⇒ 16 ( 3 x + 4 ) = 0 ⇒ 3 x = - 4 ⇒ x = - 4 3 .

Find the adjoint of the matrix 1 2 3 4 and also verify that A ( a d j A ) = A I 2 = ( a d j A ) A .

Given matrix A = 1 2 3 4 Let C ij be cofactor of a ij in A . Then, the cofactors of elements of A are given by C 11 = 4 , C 12 = - ( 3 ) = - 3 , C 21 = - 2 , C 22 = 1 ∴ adjA = C 11 C 12 C 21 C 22 T = 4 - 3 - 2 1 T = 4 - 2 - 3 1 Now A = 1 2 3 4 = 4 - 6 = - 2 ≠ 0 A ( adj A ) = 1 2 3 4 4 - 2 - 3 1 = 4 - 6 - 2 + 2 12 - 12 - 6 + 4 = - 2 0 0 - 2 = - 2 1 0 0 1 = - 2 I 2 = A I 2 . . . . . . . . . . . . ( i ) ( a d j A ) A = 4 - 2 - 3 1 1 2 3 4 = 4 - 6 8 - 8 - 3 + 3 - 6 + 4 = - 2 0 0 - 2 = - 2 1 0 0 1 = A I 2 . . . . . . . . . . . . ( ii ) From equation ( i ) and ( ii ) , It is proved that A ( adj A ) = A I 2 = ( adj A ) A .

Find the adjoint of the matrix 2 3 - 4 - 6 and also verify that A ( a d j A ) = A I 2 = ( a d j A ) A .

Given matrix A = 2 3 - 4 - 6 Let C ij be cofactor of a ij in A . Then, the cofactors of elements of A are given by C 11 = - 6 , C 12 = - ( - 4 ) = 4 , C 21 = - 3 , C 22 = 2 ∴ adjA = C 11 C 12 C 21 C 22 T = - 6 4 - 3 2 T = - 6 - 3 4 2 Now A = 2 3 - 4 - 6 = - 12 + 12 = 0 A ( adj A ) = 2 3 - 4 - 6 - 6 - 3 4 2 = - 12 + 12 - 6 + 6 24 - 24 12 - 12 = 0 0 0 0 = 0 1 0 0 1 = 0 I 2 = A I 2 . . . . . . . . . . . . ( i ) ( a d j A ) A = - 6 - 3 4 2 2 3 - 4 - 6 = - 12 + 12 - 18 + 18 8 - 8 12 - 12 = 0 0 0 0 = 0 1 0 0 1 = A I 2 . . . . . . . . . . . . ( ii ) From equation ( i ) and ( ii ) , It is proved that A ( adj A ) = A I 2 = ( adj A ) A .

If A = 3 1 2 2 - 3 - 1 1 2 1 find A ( a d j A ) without calculating a d j A .

Given matrix A = 3 1 2 2 - 3 - 1 1 2 1 ∴ A = 3 1 2 2 - 3 - 1 1 2 1 Expanding along R 1 , = 3 - 3 - 1 2 1 - 1 2 - 1 1 1 + 2 2 - 3 1 2 = 3 ( - 3 + 2 ) - 1 ( 2 + 1 ) + 2 ( 4 + 3 ) = - 3 - 3 + 14 = 8 ≠ 0 According to property of adjoint, We know that A ( adj A ) = A I 3 = 8 1 0 0 0 1 0 0 0 1 = 8 0 0 0 8 0 0 0 8 .

Find the adjoint of the matrix A = 1 - 1 2 2 3 5 - 2 0 1 and also verify that A ( a d j A ) = A I 3 = ( a d j A ) A .

Given matrix A = 1 - 1 2 2 3 5 - 2 0 1 We have A = 1 3 - 0 + 1 2 + 10 + 2 0 + 6 = 3 + 12 + 12 = 27 ≠ 0 Co-factors of elements of coefficient matrix A are as follows: C 11 = 3 5 0 1 = 3 - 0 = 3 , C 12 = - 2 5 - 2 1 = - ( 2 + 10 ) = - 12 , C 13 = 2 3 - 2 0 = 0 + 6 = 6 , C 21 = - - 1 2 0 1 = - ( - 1 - 0 ) = 1 , C 22 = 1 2 - 2 1 = 1 + 4 = 5 , C 23 = - 1 - 1 - 2 0 = - ( 0 - 2 ) = 2 , C 31 = - 1 2 3 5 = - 5 - 6 = - 11 , C 32 = - 1 2 2 5 = - ( 5 - 4 ) = - 1 , C 33 = 1 - 1 2 3 = 3 + 2 = 5 Therefore, the Co-factor matrix C of the matrix A is C = 3 - 12 6 1 5 2 - 11 - 1 5 Adj A = 3 - 12 6 1 5 2 - 11 - 1 5 ' = 3 1 - 11 - 12 5 - 1 6 2 5 Now A ( adjA ) = 1 - 1 2 2 3 5 - 2 0 1 3 1 - 11 - 12 5 - 1 6 2 5 = 3 + 12 + 12 1 - 5 + 4 - 11 + 1 + 10 6 - 36 + 30 2 + 15 + 10 - 22 - 3 + 25 - 6 + 0 + 6 - 2 + 0 + 2 22 + 0 + 5 = 27 0 0 0 27 0 0 0 27 = 27 1 0 0 0 1 0 0 0 1 = A I 3 . . . . . . . . . . . . . . ( i ) and ( adjA ) A = 3 1 - 11 - 12 5 - 1 6 2 5 1 - 1 2 2 3 5 - 2 0 1 = 3 + 2 + 22 - 3 + 3 + 0 6 + 5 - 11 - 12 + 10 + 2 12 + 15 + 0 - 24 + 25 - 1 6 + 4 - 10 - 6 + 6 + 0 12 + 10 + 5 = 27 0 0 0 27 0 0 0 27 = 27 1 0 0 0 1 0 0 0 1 = A I 3 . . . . . . . . . . . . . . ( ii ) From equation ( i ) and ( ii ) , It is proved that A ( adj A ) = A I 3 = ( adj A ) A .