Here A² = A.A =

And hence A³ = A. A² =

∴ A³– 6A² + 5A + 11 I =

Thus, A³– 6A² + 5A + 11 I = 0

Now, A³– 6A² + 5A + 11 I = 0,

→ (A.A.A)- 6 (A.A) +5A = -11I

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

→ (A.A.A.A^-1)- 6 (A.A.A^-1) +5A.A^-1 = -11I. A^-1

→ (A.A.I) – 6(A.I) + 5I = -11I. A^-1 {since A.A^-1 = I}

→ (A.A) – 6A +5I = -11A^-1 {since X.I = X}

Hence