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Algebra of Matrices — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsAlgebra of Matrices5 Marks
Question
If $\text{A}=\begin{bmatrix}3&-5\\-4&2\end{bmatrix},$ find $A^2 - 5A - 14.$
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Answer

Given: $\text{A}=\begin{bmatrix}3&-5\\-4&2\end{bmatrix}$
Now,
$\text{A}^2=\text{AA}$
$\Rightarrow\text{A}^2=\begin{bmatrix}3&-5\\-4&2\end{bmatrix}\begin{bmatrix}3&-5\\-4&2\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}9+20&-15-10\\-12-8&20+4\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}29&-25\\-20&24\end{bmatrix}$
$\text{A}^2-5\text{A}-14\text{I}$
$\Rightarrow\text{A}^2-5\text{A}-14\text{I}=\begin{bmatrix}29&-25\\-20&24\end{bmatrix}-5\begin{bmatrix}3&-5\\-4&2\end{bmatrix}-14\begin{bmatrix}1&0\\0&1\end{bmatrix}$
$\Rightarrow\text{A}^2-5\text{A}-14\text{I}=\begin{bmatrix}29&-25\\-20&24\end{bmatrix}-\begin{bmatrix}15&-25\\-20&10\end{bmatrix}-\begin{bmatrix}14&0\\0&14\end{bmatrix}$
$\Rightarrow\text{A}^2-5\text{A}-14\text{I}=\begin{bmatrix}29-15-14&-25+25+0\\-20+20+0&24-10-14\end{bmatrix}$
$ \Rightarrow\text{A}^2-5\text{A}-14\text{I}=\begin{bmatrix}0&0\\0&0\end{bmatrix}$

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