1. Tardigrade
  2. Question
  3. Mathematics
  4. Let A=[5 5 α α 0 α 5 α 0 0 5]. If |A2|=25, then |α| equals to

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Q. Let $A=\begin{bmatrix}5 & 5 \alpha & \alpha \\ 0 & \alpha & 5 \alpha \\ 0 & 0 & 5\end{bmatrix}$. If $\left|A^2\right|=25$, then $|\alpha|$ equals to

Determinants

Solution:

Since, $A=\begin{vmatrix}5 & 5 \alpha & \alpha \\ 0 & \alpha & 5 \alpha \\ 0 & 0 & 5\end{vmatrix}$
$\therefore A^2=\begin{vmatrix}5 & 5 \alpha & \alpha \\ 0 & \alpha & 5 \alpha \\ 0 & 0 & 5\end{vmatrix}\begin{vmatrix}5 & 5 \alpha & \alpha \\ 0 & \alpha & 5 \alpha \\ 0 & 0 & 5\end{vmatrix}$
$=\begin{vmatrix}25 & 25 \alpha+5 \alpha^2 & 10 \alpha+25 \alpha^2 \\ 0 & \alpha^2 & 5 \alpha^2+25 \alpha \\ 0 & 0 & 25\end{vmatrix}$
$\Rightarrow \left|A^2\right|=\begin{vmatrix}25 & 25 \alpha+5 \alpha^2 & 10 \alpha+25 \alpha^2 \\ 0 & \alpha^2 & 5 \alpha^2+25 \alpha \\ 0 & 0 & 25\end{vmatrix}$
$=25\begin{vmatrix}\alpha^2 & 5 \alpha^2+25 \alpha \\ 0 & 25\end{vmatrix}=625 \alpha^2$
$\therefore 625 \alpha^2=25$ (given)
$\Rightarrow |\alpha|=1 / 5$