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Matrices — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsMatrices3 Marks
Question
Find $x$, if $\left[\begin{array}{lll}x & -5 & -1\end{array}\right]\left[\begin{array}{lll}1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3\end{array}\right]\left[\begin{array}{l}x \\ 4 \\ 1\end{array}\right]=0$
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Answer

Given: $\left[ {\begin{array}{*{20}{c}} x&{ - 5}&{ - 1} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 1&0&2 \\ 0&2&1 \\ 2&0&3 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} x \\ 4 \\ 1 \end{array}} \right] = 0$
$\Rightarrow \left[ {\begin{array}{*{20}{c}} {x - 0 - 2}&{0 - 10 - 0}&{2x - 5 - 3} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} x \\ 4 \\ 1 \end{array}} \right] = 0$
$\Rightarrow \left[ {\begin{array}{*{20}{c}} {x - 2}&{ - 10}&{2x - 8} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} x \\ 4 \\ 1 \end{array}} \right] = 0$
$\Rightarrow [(x - 2)x - 10(4) + (2x - 8)1] = 0$
$\Rightarrow [x^2 - 2x - 40 + 2x - 8] = 0$
$\Rightarrow {\left[ {{x^2} - 48} \right]_{1 \times 1}} = {\left[ 0 \right]_{1 \times 1}}$
Equating corresponding entries, we have
$x^2 - 48 = 0$
$\Rightarrow x^2 = 48$
$\Rightarrow x = \pm 4\sqrt 3$

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