Question

Verify Cayley-Hamilton theorem of A= ( 1 2 2 3 1 0 1 1 1 ) 

Answer 1

$$A=\begin{pmatrix} 1 & 2 & 2 \\ 3 & 1 & 0 \\ 1 & 1 & 1 \\ \end{pmatrix}$$

$$P_A(\lambda)=\det(A-\lambda I)=\begin{vmatrix} 1-\lambda & 2 & 2 \\ 3 & 1-\lambda & 0 \\ 1 & 1 & 1-\lambda \\ \end{vmatrix}$$

$$=(1-\lambda)\begin{vmatrix} 1-\lambda & 0 \\ 1 & \lambda-1 \end{vmatrix}-2\begin{vmatrix} 3 & 0 \\ 1 & 1-\lambda \end{vmatrix}$$

$$+2\begin{vmatrix} 3 & 1-\lambda \\ 1 & 1 \end{vmatrix}=(1-\lambda)^3-6(1-\lambda)+6-2(1-\lambda)$$

$$=1-3\lambda+3\lambda^2-\lambda^3-8+8\lambda+6$$

$$=-\lambda^3+3\lambda^2+5\lambda-1$$

$$A^2=\begin{pmatrix} 1 & 2 & 2 \\ 3 & 1 & 0 \\ 1 & 1 & 1 \\ \end{pmatrix}\begin{pmatrix} 1 & 2 & 2 \\ 3 & 1 & 0 \\ 1 & 1 & 1 \\ \end{pmatrix}$$

$$=\begin{pmatrix} 1+6+2 & 2+2+2 & 2+0+2 \\ 3+3+0 & 6+1+0 & 6+0+0 \\ 1+3+1 & 2+1+1 & 2+0+1 \\ \end{pmatrix}$$

$$=\begin{pmatrix} 9 & 6 & 4 \\ 6 & 7 & 6 \\ 5 & 4 & 3 \\ \end{pmatrix}$$

$$A^3=\begin{pmatrix} 9 & 6 & 4 \\ 6 & 7 & 6 \\ 5 & 4 & 3 \\ \end{pmatrix}\begin{pmatrix} 1 & 2 & 2 \\ 3 & 1 & 0 \\ 1 & 1 & 1 \\ \end{pmatrix}$$

$$=\begin{pmatrix} 9+18+4 & 18+6+4 & 18+0+4 \\ 6+21+6 & 12+7+6 & 12+0+6 \\ 5+12+3 & 10+4+3 & 10+0+3 \\ \end{pmatrix}$$

$$=\begin{pmatrix} 31 & 28 & 22 \\ 33 & 25 & 18 \\ 20 & 17 & 13 \\ \end{pmatrix}$$

$$-A^3+3A^2+5A-I=-\begin{pmatrix} 31 & 28 & 22 \\ 33 & 25 & 18 \\ 20 & 17 & 13 \\ \end{pmatrix}$$

$$+3\begin{pmatrix} 9 & 6 & 4 \\ 6 & 7 & 6 \\ 5 & 4 & 3 \\ \end{pmatrix}+5\begin{pmatrix} 1 & 2 & 2 \\ 3 & 1 & 0 \\ 1 & 1 & 1 \\ \end{pmatrix}-\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{pmatrix}$$

$$=\begin{pmatrix} -31+27+5-1 & -28+18+10-0 & -22+12+10-0 \\ -33+18+15-0 & -25+21+5-1 & -18+18+0-0 \\ -20+15+5-0 & -17+12+5-0 & -13+9+5-1 \\ \end{pmatrix}$$

$$=\begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{pmatrix}$$

The matrix $A$ satisfies its own characteristic equation. 

Therefore we verify  Cayley-Hamilton theorem of $A=\begin{pmatrix} 1 & 2 & 2 \\ 3 & 1 & 0 \\ 1 & 1 & 1 \\ \end{pmatrix}.$