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Characteristic equation of A is ∣A−λI∣=0
λ3−6λ2+[(4+4+4)−(1+1+1)]λ−∣A∣=0
λ3−6λ2+(12−3)λ−4=0
λ3−6λ2+9λ−4=0
To verify Cayley-Hamilton Theorem, we need to prove that A3−6A2−4I=0
LHS=A3−6A2+9A−4I
= [22−2121 −2122−21 21−2122 ]-6[6−55−56−55−56]+9[2−11−12−11−12] - [400040004]
= [22−2121 −2122−21 21−2122 ]-6[36−3030−3036−3030−3036]+9[18−99−918−99−918]
= [22−2121 −2122−21 21−2122 ]-6[6−55−56−55−56]+9[2−11−12−11−12] - [400040004]
= [000 000 000 ]= RHS A3−6A2+9A−4I=0 (1) Cayley-Hamilton Theorem verified. Now to find A−1 and A4. multiply (1) by A−1 we get, A−1(A3−6A2+9A−4I)=A−10 A2−6A+9I−4A−1=0 4A−1=A2−6A+9I A−1=14(A2−6A+9I) A−1=[6−55−56−55−56] −6[2−11−12−11−12] + [900090009]]
[6−55−56−55−56] -[12−66−612−66−612] + [900090009]]
=[311131−113]
To find A4 multiply (1) by A
A(A3−6A2+9A−4I)=A(0)
A4−6A3+9A2−4A=0
A4=6A3−9A2+4A
=6[22−2121−2122−2121−2122] −9[6−55−56−55−56] +4 [2−11−12−11−12]
=[132−126126−126132−126126−126132] −[54−4545−4554−4545−4554] + [8−44−48−44−48]
=[86−8585−8586−8585−8586]