If A = 3 - 1 4 5 , then find a d j ( a d j A ) .

We have to find a d j ( a d j A ) . Given that A = 3 - 1 4 5 We know that for a non-singular matrix B of order n , we have B a d j B = B I . . . . . . . . . . . . . ( 1 ) replacing B by a d j A in eqn. ( 1 ) ; we have a d j A a d j ( a d j A ) = adj A I Pre-multiplying by A both sides, ( A a d j A ) a d j ( a d j A ) = A adj A [ ∵ A · I = A ] ⇒ A I a d j ( a d j A ) = A adj A [ ∵ ( Aadj A ) = A I ] ⇒ a d j ( a d j A ) = A adj A A ⇒ a d j ( a d j A ) = A A n - 1 A [ ∵ A ≠ 0 , adj A = A n - 1 ] ⇒ a d j ( a d j A ) = A n - 2 A . . . . . . . . . . . . . . . . . ( 2 ) And A = 3 - 1 4 5 = 15 + 4 = 19 ≠ 0 From ( 2 ) , we have ⇒ a d j ( a d j A ) = ( 19 ) 2 - 2 3 - 1 4 5 = 3 - 1 4 5 .

If A is a non-singular matrix, prove that A - 1 = 1 A .

Since it is given A is a non-singular matrix, therefore A ≠ 0 And according to property of inverse of matrix, AA - 1 = I = A - 1 A Taking determinant both sides, ⇒ AA - 1 = I ⇒ A A - 1 = 1 [ ∵ AB = A B and I = 1 ] ⇒ A - 1 = 1 A Hence, it is proved that If A is a non-singular matrix, then A - 1 = 1 A .

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 .