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Model Paper 4 — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSModel Paper 41 Mark
Question
If $\left[\begin{array}{lll}x & -5 & -1\end{array}\right]\left[\begin{array}{lll}1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3\end{array}\right]\left[\begin{array}{l}x \\ 4 \\ 1\end{array}\right]=0$, then the value of $x$ is
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Answer

$(c) \pm 4 \sqrt{3}$
Explanation: 
Given, $\left[\begin{array}{lll}x & -5 & -1\end{array}\right]\left[\begin{array}{lll}1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3\end{array}\right]\left[\begin{array}{l}x \\ 4 \\ 1\end{array}\right]=0$
$\Rightarrow x \times 1+(-5) \times 0+(-1) \times 2 x \times 0+(-5) \times 2+(-1) \times 0$
$x \times 2+(-5) \times 1+(-1) \times 3]\left[\begin{array}{l}x \\ 4 \\ 1\end{array}\right]=0$
$\Rightarrow\left[\begin{array}{lll}x-2 -10 2 x-8\end{array}\right]\left[\begin{array}{l}x \\ 4 \\ 1\end{array}\right]=0$
$\Rightarrow[( x -2) \times x +(-10) \times 4+(2 x -8) \times 1]=0$
$\Rightarrow x ^2-2 x -40+2 x -8=0$
$\Rightarrow x ^2=48$
$\Rightarrow x= \pm \sqrt{48}= \pm 4 \sqrt{3}$

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