If A = a b b 2 - a 2 - a b , then prove that A 2 = 0 0 0 0 .

Given, A = a b b 2 - a 2 - a b Let us find A 2 . Since A 2 = A × A ⇒ a b b 2 - a 2 - a b × a b b 2 - a 2 - a b ⇒ a b × a b - a 2 b 2 a b × b 2 - b 2 × a b - a 2 × a b + - a b × - a 2 - a 2 × b 2 + - a b × - a b ⇒ a 2 b 2 - a 2 b 2 a b 3 - a b 3 - a 3 b + a 3 b - a 2 b 2 + a 2 b 2 ⇒ A 2 = 0 0 0 0 Hence, proved.

If A = a b b 2 - a 2 - a b , then prove that A 2 = 0 0 0 0 .

Given, A = a b b 2 - a 2 - a b Let us find A 2 . Since A 2 = A × A ⇒ a b b 2 - a 2 - a b × a b b 2 - a 2 - a b ⇒ a b × a b - a 2 b 2 a b × b 2 - b 2 × a b - a 2 × a b + - a b × - a 2 - a 2 × b 2 + - a b × - a b ⇒ a 2 b 2 - a 2 b 2 a b 3 - a b 3 - a 3 b + a 3 b - a 2 b 2 + a 2 b 2 ⇒ A 2 = 0 0 0 0 Hence, proved.

If A = 1 2 2 3 , B = 2 1 3 2 and C = 1 3 3 1 , then find C B - A .

Given, A = 1 2 2 3 , B = 2 1 3 2 and C = 1 3 3 1 . We have to find C B - A . First, find B - A : B - A = 2 1 3 2 - 1 2 2 3 = 2 - 1 1 - 2 3 - 2 2 - 3 = 1 - 1 1 - 1 . Now, let us find C B - A : C B - A = 1 3 3 1 × 1 - 1 1 - 1 ⇒ 1 × 1 + 3 × 1 1 × - 1 + 3 × - 1 3 × 1 + 1 × 1 3 × - 1 + 1 × - 1 ⇒ 1 + 3 - 1 - 3 3 + 1 - 3 - 1 Hence, C B - A = 4 - 4 4 - 4 .

If P = 4 1 - 1 2 , then prove that 6 P - P 2 = 9 I 2 .

Given, P = 4 1 - 1 2 . We have to prove that 6 P - P 2 = 9 I 2 . Therefore, 6 P = 6 4 1 - 1 2 = 4 × 6 1 × 6 - 1 × 6 2 × 6 = 24 6 - 6 12 P 2 = P × P = 4 1 - 1 2 × 4 1 - 1 2 ⇒ 4 × 4 + 1 × - 1 4 × 1 + 1 × 2 - 1 × 4 + 2 × - 1 - 1 × 1 + 2 × 2 ⇒ P 2 = 16 - 1 4 + 2 - 4 - 2 - 1 + 4 = 15 6 - 6 3 Now, 6 P - P 2 = 24 6 - 6 12 - 15 6 - 6 3 6 P - P 2 = 24 - 15 6 - 6 - 6 + 6 12 - 3 = 9 0 0 9 . . . . . . . . . . . . . . i We know that, Identity matrix I 2 = 1 0 0 1 Therefore, ⇒ 9 I 2 = 9 1 0 0 1 = 9 0 0 9 . . . . . . . . . . . . . ii Thus, from i and ii , we get, LHS = RHS Therefore, 6 P - P 2 = 9 I 2 Hence, proved.

If P = 2 - 2 - 3 4 , then evaluate - P 2 + 6 P .

Given, P = 2 - 2 - 3 4 . We have to find - P 2 + 6 P . First, let us find - P 2 : P 2 = P × P = 2 - 2 - 3 4 × 2 - 2 - 3 4 ⇒ P 2 = 2 × 2 + - 2 × - 3 2 × - 2 + - 2 × 4 - 3 × 2 + 4 × - 3 - 3 × - 2 + 4 × 4 = 4 + 6 - 4 - 8 - 6 - 12 6 + 16 = 10 - 12 - 18 22 ⇒ - P 2 = - 10 - 12 - 18 22 = - 10 12 18 - 22 . Now, let us find 6 P : 6 P = 6 2 - 2 - 3 4 = 2 × 6 - 2 × 6 - 3 × 6 4 × 6 = 12 - 12 - 18 24 Therefore, - P 2 + 6 P = - 10 12 18 - 22 + 12 - 12 - 18 24 = - 10 + 12 12 - 12 18 - 18 - 22 + 24 ⇒ - P 2 + 6 P = 2 0 0 2 . Hence, - P 2 + 6 P = 2 0 0 2 .

If A = 3 1 - 1 2 and I = 1 0 0 1 , then find A 2 - 5 A + 7 I .

Given, A = 3 1 - 1 2 and I = 1 0 0 1 . We have to find A 2 - 5 A + 7 I . A 2 = A × A = 3 1 - 1 2 × 3 1 - 1 2 ⇒ A 2 = 3 × 3 + 1 × - 1 3 × 1 + 1 × 2 - 1 × 3 + 2 × - 1 - 1 × 1 + 2 × 2 = 9 - 1 3 + 2 - 3 - 2 - 1 + 4 ⇒ A 2 = 8 5 - 5 3 . 5 A = 5 3 1 - 1 2 = 3 × 5 1 × 5 - 1 × 5 2 × 5 = 15 5 - 5 10 . 7 I = 7 1 0 0 1 = 7 0 0 7 Therefore, A 2 - 5 A + 7 I = 8 5 - 5 3 - 15 5 - 5 10 + 7 0 0 7 ⇒ 8 - 15 + 7 5 - 5 + 0 - 5 - - 5 + 0 3 - 10 + 7 = 0 0 0 0 Hence, A 2 - 5 A + 7 I = 0 0 0 0 .

If A = 2 3 5 7 , B = 0 4 - 1 7 and C = 1 0 - 1 4 , find A C + B 2 - 10 C .

Given, A = 2 3 5 7 , B = 0 4 - 1 7 and C = 1 0 - 1 4 . We have to find A C + B 2 - 10 C . A C = 2 3 5 7 × 1 0 - 1 4 = 2 × 1 + 3 × - 1 2 × 0 + 3 × 4 5 × 1 + 7 × - 1 5 × 0 + 7 × 4 ⇒ A C = 2 - 3 0 + 12 5 - 7 0 + 28 = - 1 12 - 2 28 B 2 = B × B = 0 4 - 1 7 × 0 4 - 1 7 = 0 × 0 + 4 × - 1 0 × 4 + 4 × 7 - 1 × 0 + 7 × - 1 - 1 × 4 + 7 × 7 ⇒ B 2 = 0 - 4 0 + 28 0 - 7 - 4 + 49 = - 4 28 - 7 45 10 C = 10 1 0 - 1 4 = 1 × 10 0 × 10 - 1 × 10 4 × 10 = 10 0 - 10 40 Therefore, A C + B 2 - 10 C = - 1 12 - 2 28 + - 4 28 - 7 45 - 10 0 - 10 40 = - 1 - 4 - 10 12 + 28 - 0 - 2 - 7 - - 10 28 + 45 - 40 = - 15 40 1 33 . Hence, A C + B 2 - 10 C = - 15 40 1 33 .

Let A = 1 0 2 1 , B = 2 3 - 1 0 , then find A 2 + A B + B 2 .

Given, A = 1 0 2 1 , B = 2 3 - 1 0 We have to find A 2 + A B + B 2 . A 2 = A × A = 1 0 2 1 × 1 0 2 1 = 1 × 1 + 0 × 2 1 × 0 + 0 × 1 2 × 1 + 1 × 2 2 × 0 + 1 × 1 A 2 = 1 + 0 0 + 0 2 + 2 0 + 1 = 1 0 4 1 A B = A × B = 1 0 2 1 × 2 3 - 1 0 = 1 × 2 + 0 × - 1 1 × 3 + 0 × 0 2 × 2 + 1 × - 1 2 × 3 + 1 × 0 ⇒ A B = 2 + 0 3 + 0 4 - 1 6 + 0 = 2 3 3 6 B 2 = B × B = 2 3 - 1 0 × 2 3 - 1 0 = 2 × 2 + 3 × - 1 2 × 3 + 3 × 0 - 1 × 2 + 0 × - 1 - 1 × 3 + 0 × 0 ⇒ B 2 = 4 - 3 6 + 0 - 2 + 0 - 3 + 0 = 1 6 - 2 - 3 Therefore, A 2 + A B + B 2 = 1 0 4 1 + 2 3 3 6 + 1 6 - 2 - 3 = 1 + 2 + 1 0 + 3 + 6 4 + 3 - 2 1 + 6 - 3 = 4 9 5 4 Hence, A 2 + A B + B 2 = 4 9 5 4 .

If A = 8 6 - 2 4 , B = - 3 5 1 0 , then solve for 2 × 2 matrix X , such that X - B = A .

Given, A = 8 6 - 2 4 , B = - 3 5 1 0 , we have to find the matrix X , if, X - B = A . ⇒ X = A + B ⇒ X = 8 6 - 2 4 + - 3 5 1 0 ⇒ X = 8 - 3 6 + 5 - 2 + 1 4 + 0 ⇒ X = 5 11 - 1 4 . Hence, X = 5 11 - 1 4 , which is a 2 × 2 matrix.

If 1 4 - 2 3 + 2 M = 3 3 2 0 - 3 , then find the matrix M .

Given, 1 4 - 2 3 + 2 M = 3 3 2 0 - 3 , then we have to find the matrix M . ⇒ 2 M = 3 3 2 0 - 3 - 1 4 - 2 3 ⇒ 2 M = 3 × 3 2 × 3 0 × 3 - 3 × 3 - 1 4 - 2 3 ⇒ 2 M = 9 6 0 - 9 - 1 4 - 2 3 ⇒ 2 M = 9 - 1 6 - 4 0 - - 2 - 9 - 3 ⇒ 2 M = 8 2 2 - 12 ⇒ M = 1 2 8 2 2 - 12 ⇒ M = 8 2 2 2 2 2 - 12 2 ⇒ M = 4 1 1 - 6 Hence, M = 4 1 1 - 6 .