If f(x) = x^2 - 2x, find f(A), where A= 0&1&2 4&5&0 0&2&3 - Vidyadip

Question

If $f(x) = x^2 - 2x$, find f(A), where $\text{A}=\begin{bmatrix}0&1&2\\4&5&0\\0&2&3\end{bmatrix}$

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Answer

Given: $f(x) = x^2 - 2x$
$f(A) = A^2 - 2A$
Now,
$\text{A}^2=\text{AA}$
$ \Rightarrow\text{A}^2=\begin{bmatrix}0&1&2\\4&5&0\\0&2&3\end{bmatrix}\begin{bmatrix}0&1&2\\4&5&0\\0&2&3\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}0+4+0&0+5+4&0+0+6\\0+20+0&4+25+0&8+0+0\\0+8+0&0+10+6&0+0+9\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}4&9&6\\20&29&8\\8&16&9\end{bmatrix}$
$\text{f(A)}=\text{A}^2-2\text{A}$
$\Rightarrow\text{f(A)}=\begin{bmatrix}4&9&6\\20&29&8\\8&16&9\end{bmatrix}-2\begin{bmatrix}0&1&2\\4&5&0\\0&2&3\end{bmatrix}$
$ \Rightarrow\text{f(A)}=\begin{bmatrix}4&9&6\\20&29&8\\8&16&9\end{bmatrix}-\begin{bmatrix}0&2&4\\8&10&0\\0&4&6\end{bmatrix}$
$ \Rightarrow\text{f(A)}=\begin{bmatrix}4-0&9-2&6-4\\20-8&29-10&8-0\\8-0&16-4&9-6\end{bmatrix}$
$\Rightarrow\text{f(A)}=\begin{bmatrix}4&7&2\\12&19&8\\8&12&3\end{bmatrix}$

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