Question
If $\text{A}=\begin{bmatrix}3&-2\\4&-2\end{bmatrix}$ and $\text{I}=\begin{bmatrix}1&0\\0&1\end{bmatrix},$ then prove that $A^2 - A + 2I = O.$
✓
Answer
Given: $\text{A}=\begin{bmatrix}3&-2\\4&-2\end{bmatrix}$
Now,
$\text{A}^2=\text{AA}$
$\Rightarrow\text{A}^2=\begin{bmatrix}3&-2\\4&-2\end{bmatrix}\begin{bmatrix}3&-2\\4&-2\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}9-8&-6+4\\12-8&-8+4\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}1&-2\\4&-4\end{bmatrix}$
$\text{A}^2-\text{A}+2\text{I}=\begin{bmatrix}1&-2\\4&-4\end{bmatrix}-\begin{bmatrix}3&-2\\4&-2\end{bmatrix}+2\begin{bmatrix}1&0\\0&1\end{bmatrix}$
$\Rightarrow\text{A}^2-\text{A}+2\text{I}=\begin{bmatrix}1-3&-2+2\\4-4&-4+2\end{bmatrix}\begin{bmatrix}2&0\\0&2\end{bmatrix}$
$\Rightarrow\text{A}^2-\text{A}+2\text{I}=\begin{bmatrix}-2&0\\0&-2\end{bmatrix}\begin{bmatrix}2&0\\0&2\end{bmatrix}$
$\Rightarrow\text{A}^2-\text{A}+2\text{I}=\begin{bmatrix}-2+2&0+0\\0+0&-2+2\end{bmatrix}$
$\Rightarrow\text{A}^2-\text{A}+2\text{I}=\begin{bmatrix}0&0\\0&0\end{bmatrix}$
$\Rightarrow\text{A}^2-\text{A}+2\text{I}=0$
Hence proved.
Now,
$\text{A}^2=\text{AA}$
$\Rightarrow\text{A}^2=\begin{bmatrix}3&-2\\4&-2\end{bmatrix}\begin{bmatrix}3&-2\\4&-2\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}9-8&-6+4\\12-8&-8+4\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}1&-2\\4&-4\end{bmatrix}$
$\text{A}^2-\text{A}+2\text{I}=\begin{bmatrix}1&-2\\4&-4\end{bmatrix}-\begin{bmatrix}3&-2\\4&-2\end{bmatrix}+2\begin{bmatrix}1&0\\0&1\end{bmatrix}$
$\Rightarrow\text{A}^2-\text{A}+2\text{I}=\begin{bmatrix}1-3&-2+2\\4-4&-4+2\end{bmatrix}\begin{bmatrix}2&0\\0&2\end{bmatrix}$
$\Rightarrow\text{A}^2-\text{A}+2\text{I}=\begin{bmatrix}-2&0\\0&-2\end{bmatrix}\begin{bmatrix}2&0\\0&2\end{bmatrix}$
$\Rightarrow\text{A}^2-\text{A}+2\text{I}=\begin{bmatrix}-2+2&0+0\\0+0&-2+2\end{bmatrix}$
$\Rightarrow\text{A}^2-\text{A}+2\text{I}=\begin{bmatrix}0&0\\0&0\end{bmatrix}$
$\Rightarrow\text{A}^2-\text{A}+2\text{I}=0$
Hence proved.
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