If $k>1$ and the determinant of the matrix $A^2$, where $A=\left[\begin{array}{ccc}k & k \alpha & \alpha \\…
If $k>1$ and the determinant of the matrix $A^2$, where $A=\left[\begin{array}{ccc}k & k \alpha & \alpha \\ 0 & \alpha & k \alpha \\ 0 & 0 & k\end{array}\right]$, is $k^2$, then $|\alpha|$ equal to
$\frac{1}{k^2}$
$k$
$k^2$
$\frac{1}{k}$
Solution
$
\begin{aligned}
& \text { Given } A=\left[\begin{array}{rrr}
k & k \alpha & \alpha \\
0 & \alpha & k \alpha \\
0 & 0 & k
\end{array}\right] \\
& \therefore \quad|A|=\alpha k^2 \\
& \text { Now } \quad \begin{aligned}
\left|A^2\right| & =|A|^2 \\
& =\left(\alpha k^2\right)^2 \\
& =\alpha^2 k^4
\end{aligned}
\end{aligned}
$
According to given condition,
$
\begin{array}{rlrl}
& & \left|A^2\right| & =k^2 \\
& \therefore & \alpha^4 k^4 & =k^2 \\
\Rightarrow & \alpha^4 & =\frac{1}{k^2} \\
\Rightarrow & & \alpha \mid & =\frac{1}{k}
\end{array}
$