Question
If $\text{A}=\begin{bmatrix}1&2&0\\3&-4&5\\0&-1&3\end{bmatrix},$ compute $A^2 - 4A + 3I_3.$
✓
Answer
Given: $\text{A}=\begin{bmatrix}1&2&0\\3&-4&5\\0&-1&3\end{bmatrix}$
Now,
$\text{A}^2=\text{AA}$
$\Rightarrow\text{A}^2=\begin{bmatrix}1&2&0\\3&-4&5\\0&-1&3\end{bmatrix}\begin{bmatrix}1&2&0\\3&-4&5\\0&-1&3\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}1+6+0&2-8-0&0+10+0\\3-12+0&6+16-5&0-20+15\\0-3+0&0+4-3&0-5+9\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}7&-6&10\\-9&17&-5\\-3&1&4\end{bmatrix}$
$\text{A}^2-4\text{A}+3\text{I}_3$
$\Rightarrow\text{A}^2-4\text{A}+3\text{I}_3=\begin{bmatrix}7&-6&10\\-9&17&-5\\-3&1&4\end{bmatrix}-4\begin{bmatrix}1&2&0\\3&-4&5\\0&-1&3\end{bmatrix}+3\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$
$\Rightarrow\text{A}^2-4\text{A}+3\text{I}_3=\begin{bmatrix}7&-6&10\\-9&17&-5\\-3&1&4\end{bmatrix}-\begin{bmatrix}4&8&0\\12&-16&20\\0&-4&12\end{bmatrix}+\begin{bmatrix}3&0&0\\0&3&0\\0&0&3\end{bmatrix}$
$\Rightarrow\text{A}^2-4\text{A}+3\text{I}_3=\begin{bmatrix}7-4+3&-6-8+0&10-0+0\\-9-12+0&17+16+3&-5-20+0\\-3-0+0&1+4+0&4-12+3\end{bmatrix}$
$\Rightarrow\text{A}^2-4\text{A}+3\text{I}_3=\begin{bmatrix}6&-14&10\\-21&36&-25\\-3&5&-5\end{bmatrix}$
Now,
$\text{A}^2=\text{AA}$
$\Rightarrow\text{A}^2=\begin{bmatrix}1&2&0\\3&-4&5\\0&-1&3\end{bmatrix}\begin{bmatrix}1&2&0\\3&-4&5\\0&-1&3\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}1+6+0&2-8-0&0+10+0\\3-12+0&6+16-5&0-20+15\\0-3+0&0+4-3&0-5+9\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}7&-6&10\\-9&17&-5\\-3&1&4\end{bmatrix}$
$\text{A}^2-4\text{A}+3\text{I}_3$
$\Rightarrow\text{A}^2-4\text{A}+3\text{I}_3=\begin{bmatrix}7&-6&10\\-9&17&-5\\-3&1&4\end{bmatrix}-4\begin{bmatrix}1&2&0\\3&-4&5\\0&-1&3\end{bmatrix}+3\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$
$\Rightarrow\text{A}^2-4\text{A}+3\text{I}_3=\begin{bmatrix}7&-6&10\\-9&17&-5\\-3&1&4\end{bmatrix}-\begin{bmatrix}4&8&0\\12&-16&20\\0&-4&12\end{bmatrix}+\begin{bmatrix}3&0&0\\0&3&0\\0&0&3\end{bmatrix}$
$\Rightarrow\text{A}^2-4\text{A}+3\text{I}_3=\begin{bmatrix}7-4+3&-6-8+0&10-0+0\\-9-12+0&17+16+3&-5-20+0\\-3-0+0&1+4+0&4-12+3\end{bmatrix}$
$\Rightarrow\text{A}^2-4\text{A}+3\text{I}_3=\begin{bmatrix}6&-14&10\\-21&36&-25\\-3&5&-5\end{bmatrix}$
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