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MATRICES — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsMATRICES5 Marks
Question
$ \text{Find}\ \frac{1}{2}(\text{A}+\text{A}')\ \text{and}\frac{1}{2}(\text{A}-\text{A}'),\ \text{when}\ \text{A}=\begin{bmatrix}0&\text{a}&\text{b}\\-\text{a}&0&\text{c}\\-\text{b}&-\text{c}&0\end{bmatrix}$
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Answer

The given matrix is $\text{A}=\begin{bmatrix}0&\text{a}&\text{b}\\-\text{a}&0&\text{c}\\-\text{b}&-\text{c}&0\end{bmatrix} $
$\therefore \text{A}'=\begin{bmatrix}0&-\text{a}&-\text{b}\\\text{a}&0&-\text{c}\\\text{b}&\text{c}&0\end{bmatrix}$
$\text{A + A}'=\begin{bmatrix}0&\text{a}&\text{b}\\-\text{a}&0&\text{c}\\-\text{b}&-\text{c}&0\end{bmatrix}+\begin{bmatrix}0&-\text{a}&-\text{b}\\\text{a}&0&-\text{c}\\\text{b}&\text{c}&0\end{bmatrix}=\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}$
$\therefore\ \frac{1}{2}\text{(A + A}')=\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}$
Now, $\text{A} - \text{A}'=\begin{bmatrix}0&\text{a}&\text{b}\\-\text{a}&0&\text{c}\\-\text{b}&-\text{c}&0\end{bmatrix}-\begin{bmatrix}0&-\text{a}&-\text{b}\\\text{a}&0&-\text{c}\\\text{b}&\text{c}&0\end{bmatrix}$$=\begin{bmatrix}0&2\text{a}&2\text{b}\\-2\text{a}&0&2\text{c}\\-2\text{b}&-2\text{c}&0\end{bmatrix}$
$\therefore\ \frac{1}{2}\text{(A} -\text{A}')=\begin{bmatrix}0&\text{a}&\text{b}\\-\text{a}&0&\text{c}\\-\text{b}&-\text{c}&0\end{bmatrix}$

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