LA2.1.3

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If A=\begin{bmatrix} 2 & 3 \\ -1 & 2 \end{bmatrix}\, Show that A^2-4A+7I=O\,

A^2=\begin{bmatrix} 2 & 3 \\ -1 & 2 \end{bmatrix} \begin{bmatrix} 2 & 3 \\ -1 & 2 \end{bmatrix}\,

\begin{bmatrix} 4-3 & 6+6 \\ -2-2 & -3+4 \end{bmatrix}=\begin{bmatrix} 1 & 12 \\ -4 & 1 \end{bmatrix}\,

-4A=-4\begin{bmatrix} 2 & 3 \\ -1 & 2 \end{bmatrix}=\begin{bmatrix} -8 & -12 \\ 4 & -8 \end{bmatrix}\,

Therefore A^2-4A+7I=\begin{bmatrix} 1 & 12 \\ -4 & 1 \end{bmatrix}+\begin{bmatrix} -8 & -12 \\ 4 & -8 \end{bmatrix}+\begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix}\,

\begin{bmatrix} 1-8+7 & 12-12+0 \\ -4+4+0 & 1-8+7 \end{bmatrix}=\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}=O\,

Main Page:Linear Algebra

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