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If $A=\left(\begin{array}{ccc}5 & 5 x & x \\ 0 & x & 5 x \\ 0 & 0 & 5\end{array}\right)$ and $\left|A^{2}\right|=25,$ then $|x|$ is equal to
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Verified Answer
The correct answer is:
$\frac{1}{5}$
Given
$$
\begin{aligned}
A &=\left[\begin{array}{ccc}
5 & 5 x & x \\
0 & x & 5 x \\
0 & 0 & 5
\end{array}\right] \\
\therefore \quad A^{2}=A \cdot A
&=\left[\begin{array}{ccc}
5 & 5 x & x \\
0 & x & 5 x \\
0 & 0 & 5
\end{array}\right]\left[\begin{array}{ccc}
5 & 5 x & x \\
0 & x & 5 x \\
0 & 0 & 5
\end{array}\right] \\
&\left[\begin{array}{ccc}
25 & 25 x+5 x^{2} & 5 x+25 x^{2}+5 x \\
0 & x^{2} & 5 x^{2}+25 \\
0 & 0 & 25
\end{array}\right] \\
&=25\left(25 x^{2}-0\right) \\
&=25\left(25 x^{2}\right)
\end{aligned}
$$
since, given that $|A|^{2}=25$
Now, $25\left(25 x^{2}\right)=25$
$\begin{array}{lr}\Rightarrow & 25 x^{2}=1 \\ \Rightarrow & x^{2}=\frac{1}{25} \\ \Rightarrow & x=\pm \frac{1}{5}\end{array}$
$\Rightarrow \quad |x|=\frac{1}{5}$
$$
\begin{aligned}
A &=\left[\begin{array}{ccc}
5 & 5 x & x \\
0 & x & 5 x \\
0 & 0 & 5
\end{array}\right] \\
\therefore \quad A^{2}=A \cdot A
&=\left[\begin{array}{ccc}
5 & 5 x & x \\
0 & x & 5 x \\
0 & 0 & 5
\end{array}\right]\left[\begin{array}{ccc}
5 & 5 x & x \\
0 & x & 5 x \\
0 & 0 & 5
\end{array}\right] \\
&\left[\begin{array}{ccc}
25 & 25 x+5 x^{2} & 5 x+25 x^{2}+5 x \\
0 & x^{2} & 5 x^{2}+25 \\
0 & 0 & 25
\end{array}\right] \\
&=25\left(25 x^{2}-0\right) \\
&=25\left(25 x^{2}\right)
\end{aligned}
$$
since, given that $|A|^{2}=25$
Now, $25\left(25 x^{2}\right)=25$
$\begin{array}{lr}\Rightarrow & 25 x^{2}=1 \\ \Rightarrow & x^{2}=\frac{1}{25} \\ \Rightarrow & x=\pm \frac{1}{5}\end{array}$
$\Rightarrow \quad |x|=\frac{1}{5}$
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