If A = a b c d , B = 1 0 0 1 , find adj A B .

Given that A = a b c d , B = 1 0 0 1 A × B = a b c d A × B is a non-singular matrix. It is invertible. So inverse exist We know that cofactor A i j o f a i j ⇒ A i j = ( - 1 ) i + j M i j where M i j is minor of a i j Let C i j be a cofactor of A B i j in A B ⇒ C 11 = d , C 12 = - c , C 21 = - b , C 22 = a adj A B = C 11 C 12 C 21 C 22 T Therefore, adj A B = d - c - b a T = d - b - c a

If A = a b c d , B = 1 0 0 1 , find adj A B .

Given that A = a b c d , B = 1 0 0 1 A × B = a b c d A × B is a non-singular matrix. It is invertible. So inverse exist We know that cofactor A i j o f a i j ⇒ A i j = ( - 1 ) i + j M i j where M i j is minor of a i j Let C i j be a cofactor of A B i j in A B ⇒ C 11 = d , C 12 = - c , C 21 = - b , C 22 = a adj A B = C 11 C 12 C 21 C 22 T Therefore, adj A B = d - c - b a T = d - b - c a

If A = 2 3 5 - 2 write A - 1 in terms of A .

Given that A = 2 3 5 - 2 | A | = 2 3 5 - 2 = - 19 ≠ 0 A is a non-singular matrix. It is invertible. So inverse exist We know that cofactor A i j o f a i j ⇒ A i j = ( - 1 ) i + j M i j where M i j is minor of a i j Let C i j be a cofactor of a i j in A ⇒ C 11 = - 2 , C 12 = - 5 , C 21 = - 3 , C 22 = 2 adj A = - 2 - 5 - 3 2 T = - 2 - 3 - 5 2 A - 1 = 1 | A | adj A = 1 - 19 - 2 - 5 - 3 2 = 2 19 3 19 5 19 - 2 19 Therefore, A - 1 = 1 19 A

Write A - 1 for A = 2 5 1 3

Given that A = 2 5 1 3 | A | = 2 5 1 3 = 6 - 5 = 1 ≠ 0 A is a non-singular matrix. It is invertible. So inverse exist We know that cofactor A i j o f a i j ⇒ A i j = ( - 1 ) i + j M i j where M i j is minor of a i j Let C i j be the cofactor of a i j in A ⇒ C 11 = 3 , C 12 = - 1 , C 21 = - 5 , C 22 = 2 adj A = 3 - 1 - 5 2 T = 3 - 5 - 1 2 Therefore, A - 1 = 1 | A | adj A = 1 1 3 - 5 - 1 2