5

Method I

We evaluate A 2 and A 3 and write the given equation as AA -1 = I = [A 3 + cA 2 + dA].

Comparing the corresponding elements on both the sides, we get c = - 6, d = 11.

Method II

Characteristic equation for A is |A – xI|= 0 =

(1 - x) [(1 - x) (4 - x) + 2] = 0

Hence characteristic equation is x 3 - 6x 2 + 11x - 6 = 0

Then by Caley Hamilton theorem

A 3 - 6A 2 + 11A – 6I – 0

multiply by A -1 both the sides,

we get (A 2 -6A + 117) = A -1 ...(i)

given A -1 = 1 (A 2 + cA + d) ...(ii)

then from equation (i) and (ii)

we get c = -6 d = 11

then c + d = 5