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Algebra of Matrices — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsAlgebra of Matrices5 Marks
Question
Let $\text{A}=\begin{bmatrix}-1&1&-1\\3&-3&3\\5&5&5\end{bmatrix}$ and $\text{B}=\begin{bmatrix}0&4&3\\1&-3&-3\\-1&4&4\end{bmatrix},$ compute $A^2 - B^2.$
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Answer

Given: $\text{A}=\begin{bmatrix}-1&1&-1\\3&-3&3\\5&5&5\end{bmatrix}$
Now,
$\text{A}^2=\text{AA}$
$\Rightarrow\text{A}^2=\begin{bmatrix}-1&1&-1\\3&-3&3\\5&5&5\end{bmatrix}\begin{bmatrix}-1&1&-1\\3&-3&3\\5&5&5\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}1+3-5&-1-3-5&1+3-5\\-3-9+15&3+9+15&-3-9+15\\-5+15+25&5-15+25&-5+15+25\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}-1&-9&-1\\3&27&3\\35&15&35\end{bmatrix}$
$\text{B}^2=\text{BB}$
$\Rightarrow\text{B}^2=\begin{bmatrix}0&4&3\\1&-3&-3\\-1&4&4\end{bmatrix}\begin{bmatrix}0&4&3\\1&-3&-3\\-1&4&4\end{bmatrix}$
$ \Rightarrow\text{B}^2=\begin{bmatrix}0+4-3&0-12+12&0-12+12\\0-3+3&4+9-12&3+9-12\\0+4-4&-4-12+16&-3-12+16\end{bmatrix}$
$\Rightarrow\text{B}^2=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$
$\text{A}^2-\text{B}^2$
$\Rightarrow\text{A}^2-\text{B}^2=\begin{bmatrix}-1&-9&-1\\3&27&3\\35&15&35\end{bmatrix}-\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$
$\Rightarrow\text{A}^2-\text{B}^2=\begin{bmatrix}-1-1&-9-0&-1-0\\3-0&27-1&3-0\\35-0&15-0&35-1\end{bmatrix}$
$\Rightarrow\text{A}^2-\text{B}^2=\begin{bmatrix}-2&-1&-9\\3&26&3\\35&15&34\end{bmatrix}$

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