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

Answer 1

Answer: (D)

$\left[\begin{array}{cc}36& -32\\ 0& 4\end{array}\right]$

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

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

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

$\Rightarrow A=\left[\begin{array}{cc}k& 0\\ 0& k\end{array}\right]{\left[\begin{array}{cc}1& 2\\ 0& 3\end{array}\right]}^{-1}$ ..... $[\because AB=C\Rightarrow A=C{B}^{-1}]$

Let $B=\left[\begin{array}{cc}1& 2\\ 0& 3\end{array}\right]$

Now, $|B|=3$

Then, $B^{-1}=\frac{1}{|B|}\text{Co-factor matrix of B}$

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

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

$\Rightarrow A=\left[\begin{array}{cc}k& -\frac{2}{3}k\\ 0& \frac{k}{3}\end{array}\right]$ ......... $\left(i\right)$

$|3A|=108$ ........ [Given]

$\Rightarrow 108=|3A|=3|A|=\left|\begin{array}{cc}3k& -2k\\ 0& k\end{array}\right|$

$\Rightarrow 3k^2=108$

$\Rightarrow k^2=36$

$\Rightarrow k=\pm 6$

Take $k=6$

$\Rightarrow A=\left[\begin{array}{cc}6& -4\\ 0& 2\end{array}\right]$ ...... From $\left(i\right)$

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

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

For $k=-6$

$\Rightarrow A=\left[\begin{array}{cc}-6& 4\\ 0& -2\end{array}\right]$ ...... From $\left(i\right)$

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

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

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