If Verify that A³ – 6A² + 9A – 4I = O and hence find A^-1.

Here A² = A.A =

And hence A³ = A. A² =

∴ A³– 6A² + 9A -4I

Thus, A³– 6A² + 9A -4I = 0

Now, A³– 6A² + 9A -4I = 0,

→ (A.A.A)- 6 (A.A) +9A = 4I

Post-multiply with A^-1 on both sides-

→ (A.A.A.A^-1)- 6 (A.A.A^-1) +9A.A^-1 = 4I. A^-1

→ (A.A.I) – 6(A.I) + 9I = 4I. A^-1 {since A.A^-1 = I}

→ (A.A) – 6A +9I = 4A^-1 {since X.I = X}

Hence