If A = 2 3 5 - 2 be such that A - 1 = k A , then find the value of k

Given that A = 2 3 5 - 2 and A - 1 = k A ⇒ | A | = 2 3 5 - 2 = - 14 - 15 = - 19 The value is non-zero, so A - 1 exists A - 1 A = I k A · A = I Substituting A - 1 = k A k 2 3 5 - 2 2 3 5 - 2 = 1 0 0 1 k 4 + 15 6 - 6 10 - 10 15 + 4 = 1 0 0 1 k 19 0 0 19 = 1 0 0 1 Therefore, k = 1 19

If A = 2 3 5 - 2 be such that A - 1 = k A , then find the value of k

Given that A = 2 3 5 - 2 and A - 1 = k A ⇒ | A | = 2 3 5 - 2 = - 14 - 15 = - 19 The value is non-zero, so A - 1 exists A - 1 A = I k A · A = I Substituting A - 1 = k A k 2 3 5 - 2 2 3 5 - 2 = 1 0 0 1 k 4 + 15 6 - 6 10 - 10 15 + 4 = 1 0 0 1 k 19 0 0 19 = 1 0 0 1 Therefore, k = 1 19

Find the inverse of the matrix 3 - 2 - 7 5

Given that inverse of the matrix is 3 - 2 - 7 5 Let A = 3 - 2 - 7 5 ⇒ | A | = 3 - 2 - 7 5 = 1 ≠ 0 ⇒ A is a singular matrix ⇒ it is invertible. ⇒ inverse exist We know that c o f a c t o r o f A i j o f a i j ⇒ A i j = ( - 1 ) i + j d e t M i j w h e r e d e t M i j i s m i n o r o f a i j ⇒ Let C i j be a cofactor of a i j in A then the cofactors of element A ⇒ C 11 = 5 , C 12 = 7 , C 21 = 2 , C 22 = 3 We know that A - 1 = a d j ( A ) A ⇒ A - 1 = 1 | A | 5 7 2 3 T = 5 2 7 3

Find the inverse of the matrix cos θ sin θ - sin θ cos θ

Given that A = cos θ sin θ - sin θ cos θ ⇒ | A | = cos 2 θ + sin 2 θ = 1 ≠ 0 ⇒ A is singular matrix ⇒ It is invertible ⇒ inverse exist We know that c o f a c t o r o f A i j o f a i j ⇒ A i j = ( - 1 ) i + j d e t M i j w h e r e d e t M i j i s m i n o r o f a i j ⇒ Let C i j be a cofactor of a i j in A ⇒ C 11 = cos θ , C 12 = sin θ , C 21 = - sin θ and C 22 = cos θ ⇒ adj A = cos θ sin θ - sin θ cos θ T = cos θ - sin θ sin θ cos θ ⇒ A - 1 = 1 | A | adj A ⇒ cos θ - sin θ sin θ cos θ

If A = 1 - 3 2 0 write a d j A

Given that A = 1 - 3 2 0 ⇒ | A | = 1 - 3 2 0 = 6 ≠ 0 ⇒ A is a non-singular matrix. ⇒ it is invertible ⇒ inverse exist We know that c o f a c t o r o f A i j o f a i j ⇒ A i j = ( - 1 ) i + j d e t M i j w h e r e d e t M i j i s m i n o r o f a i j Let C i j be a cofactor of a i j in A ⇒ C 11 = 0 , C 12 = - 2 , C 21 = 3 , C 22 = 1 ⇒ adj A = C 11 C 12 C 21 C 22 T = 0 - 2 3 1 T = 0 3 - 2 1

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