Question

Let A be a matrix such that $A.\left[\begin{array}{cc}1& 2\\ 0& 3\end{array}\right]$ 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

Answer: (D)

$A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$
Now A. $\left[\begin{array}{cc}1& 2\\ 0& 3\end{array}\right]=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]\left[\begin{array}{cc}1& 2\\ 0& 3\end{array}\right]$
$=\left[\begin{array}{cccc}a& 2a& +& 3b\\ c& 2c& +& 3d\end{array}\right]$ is scalar
$\therefore c=0,2a+3b=0,a=2c+3da=3d\therefore a^2=9d^2=36$
$\left|3A\right|=108$
$\therefore \left|A\right|=12=ad-bc=ad$
$\therefore d^2=4$
Now $A^2=\left[\begin{array}{cc}a& b\\ 0& d\end{array}\right]\left[\begin{array}{cc}a& b\\ 0& d\end{array}\right]=\left[\begin{array}{cc}{a}^2& ab+bd\\ 0& d^2\end{array}\right]$