Question
If $\text{A}=\begin{bmatrix}4&2\\-1&-1 \end{bmatrix},$ prove that (A - 2I)(A - 3I) = 0
✓
Answer
Given, $\text{A}=\begin{bmatrix}4&2\\-1&-1 \end{bmatrix}$
$(\text{A}-2\text{I})(\text{A}-3\text{I})$
$=\begin{pmatrix}\begin{bmatrix}4&2\\-1&-1 \end{bmatrix}-2\begin{bmatrix}1&0\\0&1 \end{bmatrix}\end{pmatrix}\begin{pmatrix}\begin{bmatrix}4&2\\-1&-1 \end{bmatrix}-3\begin{bmatrix}1&0\\0&1 \end{bmatrix} \end{pmatrix}$
$=\begin{pmatrix}\begin{bmatrix}4&2\\-1&-1 \end{bmatrix}-\begin{bmatrix}2&0\\0&2 \end{bmatrix} \end{pmatrix}\begin{pmatrix}\begin{bmatrix}4&2\\-1&-1 \end{bmatrix}-\begin{bmatrix}3&0\\0&3 \end{bmatrix} \end{pmatrix}$
$=\begin{pmatrix}\begin{bmatrix}4-2&2-0\\-1-0&1-2 \end{bmatrix} \end{pmatrix}\begin{pmatrix}\begin{bmatrix}4-3&2-0\\-1-0&1-3 \end{bmatrix} \end{pmatrix}$
$=\begin{bmatrix}2&2\\-1&-1 \end{bmatrix}\begin{bmatrix}1&2\\-1&-2 \end{bmatrix}$
$=\begin{bmatrix}2-2&4-4\\-1+1&-2+2 \end{bmatrix}$
$=\begin{bmatrix}0 & 0 \\0 & 0 \end{bmatrix}$
$=0$
Hence,
$(\text{A}-2\text{I})(\text{A}-3\text{I})=0$
$(\text{A}-2\text{I})(\text{A}-3\text{I})$
$=\begin{pmatrix}\begin{bmatrix}4&2\\-1&-1 \end{bmatrix}-2\begin{bmatrix}1&0\\0&1 \end{bmatrix}\end{pmatrix}\begin{pmatrix}\begin{bmatrix}4&2\\-1&-1 \end{bmatrix}-3\begin{bmatrix}1&0\\0&1 \end{bmatrix} \end{pmatrix}$
$=\begin{pmatrix}\begin{bmatrix}4&2\\-1&-1 \end{bmatrix}-\begin{bmatrix}2&0\\0&2 \end{bmatrix} \end{pmatrix}\begin{pmatrix}\begin{bmatrix}4&2\\-1&-1 \end{bmatrix}-\begin{bmatrix}3&0\\0&3 \end{bmatrix} \end{pmatrix}$
$=\begin{pmatrix}\begin{bmatrix}4-2&2-0\\-1-0&1-2 \end{bmatrix} \end{pmatrix}\begin{pmatrix}\begin{bmatrix}4-3&2-0\\-1-0&1-3 \end{bmatrix} \end{pmatrix}$
$=\begin{bmatrix}2&2\\-1&-1 \end{bmatrix}\begin{bmatrix}1&2\\-1&-2 \end{bmatrix}$
$=\begin{bmatrix}2-2&4-4\\-1+1&-2+2 \end{bmatrix}$
$=\begin{bmatrix}0 & 0 \\0 & 0 \end{bmatrix}$
$=0$
Hence,
$(\text{A}-2\text{I})(\text{A}-3\text{I})=0$
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