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Matrices — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsMatrices3 Marks
Question
Find $\frac{1}{2}\left(\mathrm{~A}+\mathrm{A}^{\prime}\right)$ and $\frac{1}{2}\left(\mathrm{~A}-\mathrm{A}^{\prime}\right)$ when $A=\left[\begin{array}{ccc}0 & a & b \\ -a & 0 & c \\ -b & -c & 0\end{array}\right]$.
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Answer

Given: $A = \left[ {\begin{array}{*{20}{c}} 0&a&b \\ { - a}&0&c \\ { - b}&{ - c}&0 \end{array}} \right]$
$\therefore A' = \left[ {\begin{array}{*{20}{c}} 0&{ - a}&{ - b} \\ a&0&{ - c} \\ b&c&0 \end{array}} \right]$
Now, $A + A' = \left[ {\begin{array}{*{20}{c}} 0&a&b \\ { - a}&0&c \\ { - b}&{ - c}&0 \end{array}} \right] + \left[ {\begin{array}{*{20}{c}} 0&{ - a}&{ - b} \\ a&0&{ - c} \\ b&c&0 \end{array}} \right]$
$ = \left[ {\begin{array}{*{20}{c}} {0 + 0}&{a - a}&{b - b} \\ { - a + a}&{0 + 0}&{c - c} \\ { - b + b}&{ - c + c}&{0 + 0} \end{array}} \right]$
$ = \left[ {\begin{array}{*{20}{c}} 0&0&0 \\ 0&0&0 \\ 0&0&0 \end{array}} \right]$
$\therefore \frac{1}{2}(A + A') = \frac{1}{2}\left[ {\begin{array}{*{20}{c}} 0&0&0 \\ 0&0&0 \\ 0&0&0 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 0&0&0 \\ 0&0&0 \\ 0&0&0 \end{array}} \right]$
Now, $A - A' = \left[ {\begin{array}{*{20}{c}} 0&a&b \\ { - a}&0&c \\ { - b}&{ - c}&0 \end{array}} \right] - \left[ {\begin{array}{*{20}{c}} 0&{ - a}&{ - b} \\ a&0&{ - c} \\ b&c&0 \end{array}} \right]$
$ = \left[ {\begin{array}{*{20}{c}} {0 - 0}&{a + a}&{b + b} \\ { - a - a}&{0 - 0}&{c + c} \\ { - b - b}&{ - c - c}&{0 - 0} \end{array}} \right]$
$ = \left[ {\begin{array}{*{20}{c}} 0&{2a}&{2b} \\ { - 2a}&0&{2c} \\ { - 2b}&{ - 2c}&0 \end{array}} \right]$
$\therefore \frac{1}{2}(A - A') = \frac{1}{2}\left[ {\begin{array}{*{20}{c}} 0&{2a}&{2b} \\ { - 2a}&0&{2c} \\ { - 2b}&{ - 2c}&0 \end{array}} \right]$
$ = \left[ {\begin{array}{*{20}{c}} 0&a&b \\ { - a}&0&c \\ { - b}&{ - c}&0 \end{array}} \right]$

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