Question

Let $A$ be a matrix such that $A\cdot \left[\begin{array}{cc}1& 2\\ 0& 3\end{array}\right]$ is a scalar matrix and $|3A|=108$. Then $A^2$ equals :

• A. $\left[\begin{array}{cc}4& -32\\ 0& 36\end{array}\right]$
• B. $\left[\begin{array}{cc}36& 0\\ -32& 4\end{array}\right]$
• C. $\left[\begin{array}{cc}4& 0\\ -32& 36\end{array}\right]$
• D. $\left[\begin{array}{cc}36& -32\\ 0& 4\end{array}\right]$

Answer 1

The correct option is D $\left[\begin{array}{cc}36& -32\\ 0& 4\end{array}\right]$

Given, $A\cdot \left[\begin{array}{cc}1& 2\\ 0& 3\end{array}\right]$

$\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]\cdot \left[\begin{array}{cc}1& 2\\ 0& 3\end{array}\right]=\left[\begin{array}{cc}x& 0\\ 0& x\end{array}\right]\Rightarrow \left[\begin{array}{cc}a& 2a+3b\\ c& 2c+3d\end{array}\right]=\left[\begin{array}{cc}x& 0\\ 0& x\end{array}\right]\Rightarrow a=x,b=-\frac{2x}{3},c=0,d=\frac{x}{3}$
We know that,
$|3A|=108\Rightarrow |A|=12\Rightarrow \left|\begin{array}{cc}x& -\frac{2x}{3}\\ 0& \frac{x}{3}\end{array}\right|=12\Rightarrow x^2=36$
Now,
$A^2=\left[\begin{array}{cc}{x}^2& -\frac{8x^2}{9}\\ 0& \frac{x^2}{9}\end{array}\right]\Rightarrow A^2=\left[\begin{array}{cc}36& -32\\ 0& 4\end{array}\right]$