Question

Let A be a matrix such that $A\ast \left[\begin{array}{cc}1& 2\\ 0& 3\end{array}\right]$ 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 $\left[\begin{array}{cc}36& -32\\ 0& 4\end{array}\right]$

$A\cdot \left[\begin{array}{cc}1& 2\\ 0& 3\end{array}\right]$ is a scalar matrix and $|3A|=108$

Let the scalar matrix be $\left[\begin{array}{cc}k& 0\\ 0& k\end{array}\right]$

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

$A=\left[\begin{array}{cc}k& 0\\ 0& k\end{array}\right]$${\left[\begin{array}{cc}1& 2\\ 0& 3\end{array}\right]}^{-1}$

$⟹$$A=\left[\begin{array}{cc}k& 0\\ 0& k\end{array}\right]$$\left[\begin{array}{cc}1& -2/3\\ 0& 1/3\end{array}\right]$

$⟹A=$$\left[\begin{array}{cc}k& -2k/3\\ 0& k/3\end{array}\right]$

$|A|=\frac{k^2}{3}$

$|3A|=108$

As we know $kA|=k^n|A|$

$|3A|=3^2\frac{k^2}{3}$

$⟹k^2=36⟹k=\pm 6$

So, putting the value of $k$ in the matrix $A$

$A=$$\left[\begin{array}{cc}6& (-2/3)\left(6\right)\\ 0& 6/3\end{array}\right]$

$A=\left[\begin{array}{cc}6& -4\\ 0& 2\end{array}\right]$

So, $A^2=\left[\begin{array}{cc}6& -4\\ 0& 2\end{array}\right]$$\left[\begin{array}{cc}6& -4\\ 0& 2\end{array}\right]$

$⟹A^2=$$\left[\begin{array}{cc}36& -32\\ 0& 4\end{array}\right]$

Hence, this is the answer.