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The matrix $\mathrm{A}=\left[\begin{array}{ccc}1 & 3 & 2 \\ 1 & x-1 & 1 \\ 2 & 7 & x-3\end{array}\right]$ will have inverse for every real number $x$ except for
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Verified Answer
The correct answer is:
$x=\frac{11 \pm \sqrt{5}}{2}$
$A=\left[\begin{array}{ccc}1 & 3 & 2 \\ 1 & x-1 & 1 \\ 2 & 7 & x-3\end{array}\right]$
$|\mathrm{A}|=1[(\mathrm{x}-1)(\mathrm{x}-3)-7]-3[(\mathrm{x}-3)-2]+2[7-2(\mathrm{x}-1)]$
$=x^{2}-11 x+29$
If inverse will not exist then $|\mathrm{A}|=0$ $x^{2}-11 x+29=0$
$\mathrm{x}=\frac{11 \pm \sqrt{5}}{2}$
$|\mathrm{A}|=1[(\mathrm{x}-1)(\mathrm{x}-3)-7]-3[(\mathrm{x}-3)-2]+2[7-2(\mathrm{x}-1)]$
$=x^{2}-11 x+29$
If inverse will not exist then $|\mathrm{A}|=0$ $x^{2}-11 x+29=0$
$\mathrm{x}=\frac{11 \pm \sqrt{5}}{2}$
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