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If $A=\left[\begin{array}{lll}x & 2 & 1 \\ 2 & x & 1 \\ 2 & 1 & 0\end{array}\right]$ and $\operatorname{det}\left(A^3\right)=125$, then $x=$
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Verified Answer
The correct answer is:
$1 / 3$
$|A|=\left|\begin{array}{lll}x & 2 & 1 \\ 2 & x & 1 \\ 2 & 1 & 0\end{array}\right|=x(0-1)-2(0-2)+1(2-2 x)$
$=6-3 x$
According to question, $\left|A^3\right|=125$
$\Rightarrow \quad|A|^3=5^3 \quad\left[\because\left|A^3\right|=|A|^3\right]$
$\begin{aligned} \Rightarrow & & |A| & =5 \\ \Rightarrow & & 6-3 x & =5 \\ \Rightarrow & & x & =1 / 3\end{aligned}$
$=6-3 x$
According to question, $\left|A^3\right|=125$
$\Rightarrow \quad|A|^3=5^3 \quad\left[\because\left|A^3\right|=|A|^3\right]$
$\begin{aligned} \Rightarrow & & |A| & =5 \\ \Rightarrow & & 6-3 x & =5 \\ \Rightarrow & & x & =1 / 3\end{aligned}$
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