If a = [ ccc 2&-1&1 -1&2&-1 1&-1&2 ]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


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