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જો $[x-5-1]\left[\begin{array}{lll}1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3\end{array}\right]\left[\begin{array}{l}x \\ 4 \\ 1\end{array}\right]=O$ હોય, તો $x$ શોધો.
$\pm 4 \sqrt{3}$
$\pm 4 \sqrt{3}$
$\pm 4 \sqrt{3}$
$\pm 4 \sqrt{3}$
Solution
We have:
$\left[\begin{array}{ccc}x & -5 & -1\end{array}\right]\left[\begin{array}{ccc}1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3\end{array}\right]\left[\begin{array}{l}x \\ 4 \\ 1\end{array}\right]=0$
$\Rightarrow\left[\begin{array}{lll}x+0-2 & 0-10+0 & 2 x-5-3\end{array}\right]\left[\begin{array}{l}x \\ 4 \\ 1\end{array}\right]=0$
$\Rightarrow\left[\begin{array}{lll}x-2 & -10 & 2 x-8\end{array}\right]\left[\begin{array}{l}x \\ 4 \\ 1\end{array}\right]=0$
$\Rightarrow[x(x-2)-40+2 x-8]=0$
$\Rightarrow\left[x^{2}-2 x-40+2 x-8\right]=[0]$
$\Rightarrow\left[x^{2}-48\right]=[0]$
$\therefore x^{2}-48=0$
$\Rightarrow x^{2}=48$
$\Rightarrow x=\pm 4 \sqrt{3}$
