Search any question & find its solution
Question:
Answered & Verified by Expert
If $=\left[\begin{array}{ccc}5 & 5 \alpha & \alpha \\ 0 & \alpha & 5 \alpha \\ 0 & 0 & 5\end{array}\right]$ and $\operatorname{det}\left(A^2\right)=25$, then $|\alpha|=$
Options:
Solution:
2629 Upvotes
Verified Answer
The correct answer is:
$\frac {1}{5}$
$\operatorname{det}\left(A^2\right)=25$
$\begin{aligned} & (\operatorname{det} A)^2=25 \\ & |A|=\left|\begin{array}{ccc}5 & 5 \alpha & \alpha \\ 0 & \alpha & 5 \alpha \\ 0 & 0 & 5\end{array}\right|=5(5 \alpha) \\ & \therefore \quad|A|=25 \alpha \Rightarrow(25 \alpha)^2=25 \Rightarrow \alpha^2=\frac{1}{25} \\ & \therefore \quad|\alpha|=\frac{1}{5} .\end{aligned}$
$\begin{aligned} & (\operatorname{det} A)^2=25 \\ & |A|=\left|\begin{array}{ccc}5 & 5 \alpha & \alpha \\ 0 & \alpha & 5 \alpha \\ 0 & 0 & 5\end{array}\right|=5(5 \alpha) \\ & \therefore \quad|A|=25 \alpha \Rightarrow(25 \alpha)^2=25 \Rightarrow \alpha^2=\frac{1}{25} \\ & \therefore \quad|\alpha|=\frac{1}{5} .\end{aligned}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.