Verify Cayley – Hamilton theorm for A = $\begin{bmatrix} 1& 2&-2 \\ -1& 3& 0 \\ 0&-2& 1 \end{bmatrix} $. Also find A$^{-1}$

Subject: Applied Mathematics 4

Topic: Matrix Theory

Difficulty: Medium

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

For characteristic equation |A - $\lambda$ I| = 0

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

$\lambda^3$ - (Sum of diagonal elements)$\lambda^2$ + (Sum of Principal minors)$\lambda$ - |A| = 0

$ \lambda^3 - (1+3+1)\lambda^2 + (3+1+5)\lambda - 1 = 0 \\ \lambda^3 - 5 \lambda^2 + 9 \lambda - 1 = 0 $

By Cayley Hamilton theorem, $ A^3 - 5 A^2 + 9 A - I = 0 $ ...(1)

Consider, $ A^3 - 5 A^2 + 9 A - I $

$ = \begin{bmatrix} -13 & 42 & -2 \\ -11 & 9 & 10 \\ 10 & -22 & -3 \end{bmatrix} -5 \begin{bmatrix} -1 & 12 & -4 \\ -4 & 7 & 2 \\ 2 & -8 & 1 \end{bmatrix} +9 \begin{bmatrix} 1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1 \end{bmatrix} - \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} = 0 $

Hence, Cayley Hamilton theorem is verified

Pre-multiply both sides of (1) by A$^{-1}$

$ A^{-1} (A^3 - 5 A^2 + 9 A - I) = 0 \\ A^2 -5A +9I - A^{-1} = 0 \\ A^{-1} = A^2 -5A +9I \\ = \begin{bmatrix} -1 & 12 & -4 \\ -4 & 7 & 2 \\ 2 & -8 & 1 \end{bmatrix} -5 \begin{bmatrix} 1& 2&-2 \\ -1& 3& 0 \\ 0&-2& 1 \end{bmatrix} + \begin{bmatrix} 9 & 0 & 0 \\ 0 & 9 & 0 \\ 0 & 0 & 9 \end{bmatrix} \\ = \begin{bmatrix} 3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5 \end{bmatrix} $