Mathematics - Matrices Question with Solution | TestHub

Let $A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$

Now, $A.\left[\begin{array}{cc}1& 2\\ 0& 3\end{array}\right]=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]\left[\begin{array}{cc}1& 2\\ 0& 3\end{array}\right]=\left[\begin{array}{cc}a& 2a+3b\\ c& 2c+3d\end{array}\right]$ is a scalar.

$\therefore c =0, 2a+3b=0, a=2c+3d\Rightarrow a=3d$

Given, $\left|3A\right|=108$

$\Rightarrow \left|A\right|=\frac{108}{3^2}=12\Rightarrow ad-bc=ad=12$

$\Rightarrow \left(3d\right)d=12\Rightarrow d^2=4 \left[\because a=3d\right]$

$\therefore a^2={\left(3d\right)}^2=9d^2==9\times 4=36$ and

$\because 2a+3b=0\Rightarrow b=\frac{-2a}{3}=\frac{-2\times 6}{3}\Rightarrow b=-4$

Now, $A^2=\left[\begin{array}{cc}a& b\\ 0& d\end{array}\right]\left[\begin{array}{cc}a& b\\ 0& d\end{array}\right]=\left[\begin{array}{cc}{a}^2& ab+bd\\ 0& d^2\end{array}\right]=\left[\begin{array}{cc}36& -32\\ 0& 4\end{array}\right]$