Inverse of matrix using Cayley Hamilton method Example [[2,3,1],[0,5,6],[1,1,2]]

Find Inverse of matrix using Cayley Hamilton method
A=[72-2-6-1262-1] $A=[[7,2,-2],[-6,-1,2],[6,2,-1]]$

Solution:
To apply the Cayley-Hamilton theorem, we first determine the characteristic polynomial p(t) of the matrix A.

|A-tI| $|A-tI|$

(7-t) $(7-t)$	2 $2$	-2 $-2$
-6 $-6$	(-1-t) $(-1-t)$	2 $2$
6 $6$	2 $2$	(-1-t) $(-1-t)$

=(7-t)((-1-t)×(-1-t)-2×2)-2((-6)×(-1-t)-2×6)+(-2)((-6)×2-(-1-t)×6) $=(7-t)((-1-t) × (-1-t) - 2 × 2)-2((-6) × (-1-t) - 2 × 6)+(-2)((-6) × 2 - (-1-t) × 6)$

=(7-t)((1+2t+t2)-4)-2((6+6t)-12)-2((-12)-(-6-6t)) $=(7-t)((1+2t+t^2)-4)-2((6+6t)-12)-2((-12)-(-6-6t))$

=(7-t)(-3+2t+t2)-2(-6+6t)-2(-6+6t) $=(7-t)(-3+2t+t^2)-2(-6+6t)-2(-6+6t)$

=(-21+17t+5t2-t3)-(-12+12t)-(-12+12t) $= (-21+17t+5t^2-t^3)-(-12+12t)-(-12+12t)$

=-t3+5t2-7t+3 $=-t^3+5t^2-7t+3$

p(t)=-t3+5t2-7t+3 $p(t)=-t^3+5t^2-7t+3$

The Cayley-Hamilton theorem yields that

O=p(A)=-A3+5A2-7A+3I $O = p(A)=-A^3+5A^2-7A+3I$

Rearranging terms, we have

$:. 3I = A^3-5A^2+7A$

$:. 3I = A(A^2-5A+7I)$

$:. A^-1 = 1/3(A^2-5A+7I)$

Now, first we find

$A^2-5A+7I$

$A^2$

$A×A$

$7$	$2$	$-2$
$-6$	$-1$	$2$
$6$	$2$	$-1$

$7$	$2$	$-2$
$-6$	$-1$	$2$
$6$	$2$	$-1$

$7×7+2×-6-2×6$	$7×2+2×-1-2×2$	$7×-2+2×2-2×-1$
$-6×7-1×-6+2×6$	$-6×2-1×-1+2×2$	$-6×-2-1×2+2×-1$
$6×7+2×-6-1×6$	$6×2+2×-1-1×2$	$6×-2+2×2-1×-1$

$49-12-12$	$14-2-4$	$-14+4+2$
$-42+6+12$	$-12+1+4$	$12-2-2$
$42-12-6$	$12-2-2$	$-12+4+1$

$25$	$8$	$-8$
$-24$	$-7$	$8$
$24$	$8$	$-7$

$A^2$

$7$	$2$	$-2$
$-6$	$-1$	$2$
$6$	$2$	$-1$

$25$	$8$	$-8$
$-24$	$-7$	$8$
$24$	$8$	$-7$

$5 × A$

$5$

$7$	$2$	$-2$
$-6$	$-1$	$2$
$6$	$2$	$-1$

$35$	$10$	$-10$
$-30$	$-5$	$10$
$30$	$10$	$-5$

$A^2 - 5 × A$

$25$	$8$	$-8$
$-24$	$-7$	$8$
$24$	$8$	$-7$

$35$	$10$	$-10$
$-30$	$-5$	$10$
$30$	$10$	$-5$

$25+35$	$8+10$	$-8-10$
$-24-30$	$-7-5$	$8+10$
$24+30$	$8+10$	$-7-5$

$-10$	$-2$	$2$
$6$	$-2$	$-2$
$-6$	$-2$	$-2$

$7 × I$

$7$

$1$	$0$	$0$
$0$	$1$	$0$
$0$	$0$	$1$

$7$	$0$	$0$
$0$	$7$	$0$
$0$	$0$	$7$

$A^2 - 5 × A + 7 × I$

$-10$	$-2$	$2$
$6$	$-2$	$-2$
$-6$	$-2$	$-2$

$7$	$0$	$0$
$0$	$7$	$0$
$0$	$0$	$7$

$-10+7$	$-2+0$	$2+0$
$6+0$	$-2+7$	$-2+0$
$-6+0$	$-2+0$	$-2+7$

$-3$	$-2$	$2$
$6$	$5$	$-2$
$-6$	$-2$	$5$

Now,

$A^-1 = 1/3(A^2-5A+7I)$

$:. A^-1 =$

$1/(3)$

$-3$	$-2$	$2$
$6$	$5$	$-2$
$-6$	$-2$	$5$

This material is intended as a summary. Use your textbook for detail explanation.
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2. Example [231056112][[2,3,1],[0,5,6],[1,1,2]]

2. Example [231056112] $[[2,3,1],[0,5,6],[1,1,2]]$