Given:

To verify: A² + 3A + 4I = 0

Firstly, we find the A²

Taking LHS of the given equation .i.e.

A² + 3A + 4I

= O

= RHS

∴ LHS = RHS

Hence verified

Now, we have to find A^-1

Finding A^-1 using given equation

A² + 3A + 4I = O

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

(A² + 3A + 4I)A^-1 = OA^-1

⇒ A².A^-1 + 3A.A^-1 + 4I.A^-1 = O [OA^-1 = O]

⇒ A.(AA^-1) + 3I + 4A^-1 = O [AA^-1 = I]

⇒ A(I) + 3I + 4A^-1 = O

⇒ A + 3I + 4A^-1 = O

⇒ 4A^-1 = – A – 3I + O

Ans.