Given:

We have to show that matrix A satisfies the equation A³ – A² – 3A – I = O

Firstly, we find the A²

Now, we have to calculate A³

Taking LHS of the given equation .i.e.

A³ – A² – 3A – I

Putting the values, we get

= O

= RHS

∴ LHS = RHS

Hence, the given matrix A satisfies the equation A³ – A² – 3A – I

Now, we have to find A^-1

Finding A^-1 using given equation

A³ – A² – 3A – I

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

(A³ – A² – 3A – I)A^-1 = OA^-1

⇒ A³.A^-1 – A².A^-1 – 3A.A^-1 – I.A^-1 = O [OA^-1 = O]

⇒ A².(AA^-1) – A.(AA^-1) – 3I – A^-1 = O

⇒ A²(I) – A(I) – 3I – A^-1 = O [AA^-1 = I]

⇒ A² – A – 3I – A^-1 = O

⇒ O + A^-1 = A² – A – 3I

⇒ A^-1 = A² – A – 3I

Ans. .