Given:

To show: Matrix A satisfies the equation x² + 4x – 42 = 0

If Matrix A satisfies the given equation then

A² + 4A – 42 = 0

Firstly, we find the A²

Taking LHS of the given equation .i.e.

A² + 4A – 42

= O

= RHS

∴ LHS = RHS

Hence matrix A satisfies the given equation x² + 4x – 42 = 0

Now, we have to find A^-1

Finding A^-1 using given equation

A² + 4A – 42 = O

Post multiplying by A^-1 both sides, we get

(A² + 4A – 42)A^-1 = OA^-1

⇒ A².A^-1 + 4A.A^-1 – 42.A^-1 = O [OA^-1 = O]

⇒ A.(AA^-1) + 4I – 42A^-1 = O [AA^-1 = I]

⇒ A(I) + 4I – 42A^-1 = O

⇒ A + 4I – 42A^-1 = O

⇒ A + 4I – O = 42A^-1

Ans. .