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If the matrix
$A=\left[\begin{array}{ccc}2-x & 1 & 1 \\ 1 & 3-x & 0 \\ -1 & -3 & -x\end{array}\right]$
is singular, then what is the solution set $S ?$
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$A=\left[\begin{array}{ccc}2-x & 1 & 1 \\ 1 & 3-x & 0 \\ -1 & -3 & -x\end{array}\right]$
is singular, then what is the solution set $S ?$
Solution:
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Verified Answer
The correct answer is:
$\mathrm{S}=\{0,2,3\}$
Let $A=\left[\begin{array}{ccc}2-x & 1 & 1 \\ 1 & 3-x & 0 \\ -1 & -3 & -x\end{array}\right]$
Since, this matrix is singular.
$\therefore |A|=0$
$\Rightarrow\left|\begin{array}{ccc}2-x & 1 & 1 \\ 1 & 3-x & 0 \\ -1 & -3 & -x\end{array}\right|=0$
Applying $\mathrm{R}_{2} \rightarrow \mathrm{R}_{2}+\mathrm{R}_{3}$
$\Rightarrow\left|\begin{array}{ccc}2-x & 1 & 1 \\ 0 & -x & -x \\ -1 & -3 & -x\end{array}\right|=0$
$\Rightarrow(2-x)\left(x^{2}-3 x\right)-1(-x)+1(-x)=0$
$\Rightarrow(2-x)(x)(x-3)=0$
$\Rightarrow x=2,0,3$
Hence, solution set $S=\{0,2,3\}$
Since, this matrix is singular.
$\therefore |A|=0$
$\Rightarrow\left|\begin{array}{ccc}2-x & 1 & 1 \\ 1 & 3-x & 0 \\ -1 & -3 & -x\end{array}\right|=0$
Applying $\mathrm{R}_{2} \rightarrow \mathrm{R}_{2}+\mathrm{R}_{3}$
$\Rightarrow\left|\begin{array}{ccc}2-x & 1 & 1 \\ 0 & -x & -x \\ -1 & -3 & -x\end{array}\right|=0$
$\Rightarrow(2-x)\left(x^{2}-3 x\right)-1(-x)+1(-x)=0$
$\Rightarrow(2-x)(x)(x-3)=0$
$\Rightarrow x=2,0,3$
Hence, solution set $S=\{0,2,3\}$
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