Answer to Question #162956 in Linear Algebra for Ra

Question

To verify Cayley-Hamilton theorem for

1 2 3

M = 2 -1 4

3 1 1

Answer 1

$\mid A- \lambda I=0\\ equation 1 - \lambda^3+\lambda^2+18\lambda+30=0\\ - \lambda^3+\lambda^2+18\lambda+30I=0\\ Calculate \, A^2=\begin{bmatrix} 14 & 3&14\\ 12 & 9&6\\ 8&6&14\\ \end{bmatrix}\\ Calculate \, A^3= \begin{bmatrix} 62& 39&68\\ 48 & 21&78\\ 62&24&62\\ \end{bmatrix} \\Put \, all \, values \, in \, equation \, 1\\\\= \begin{bmatrix} -62& -39&-68\\ -48 & -21&-78\\ -62&-24&-62 \end{bmatrix}+ \begin{bmatrix} 14& 3&14\\ 12 & 9&6\\ 8&6&14\\ \end{bmatrix}+18 \begin{bmatrix} 1& 2&3 \\ 2& -1&4\\ 3&1&1\\ \end{bmatrix}+30 \begin{bmatrix} 1& 0&0 \\ 0& 1&0\\ 0&0&1\\ \end{bmatrix}\\ = \begin{bmatrix} 0&0&0\\ 0&0&0\\ 0&0&0 \end{bmatrix}\\ L.H.s=R.H.s\\ So\, this\, matrix\, meets \,Caley \, Hamilton\, Theorem$