Mathematics - Matrices Question with Solution | TestHub

A. $\left[\begin{array}{ll}1& 2\\ 0& 3\end{array}\right]$ is a scalar matrix and $|3\mathrm{A}|=108$ suppose the scalar matrix is $\left[\begin{array}{ll}k& 0\\ 0& k\end{array}\right]$ $$\begin{array}{rl}& \therefore A\cdot \left[\begin{array}{ll}1& 2\\ 0& 3\end{array}\right]=\left[\begin{array}{ll}k& 0\\ 0& k\end{array}\right]\\ & \Rightarrow A=\left[\begin{array}{ll}k& 0\\ 0& k\end{array}\right]{\left[\begin{array}{ll}1& 2\\ 0& 3\end{array}\right]}^{-1}\\ & \left[\therefore AB=C\Rightarrow AB{B}^{-1}=C{B}^{-1}\Rightarrow A=C{B}^{-1}\right]\\ & \Rightarrow A=\frac{1}{3}\left[\begin{array}{ll}k& 0\\ 0& k\end{array}\right]\left[\begin{array}{cc}3& -2\\ 0& 1\end{array}\right]\\ & \Rightarrow A=\left[\begin{array}{ll}k& 0\\ 0& k\end{array}\right]\left[\begin{array}{cc}1& -\frac{2}{3}\\ 0& \frac{1}{3}\end{array}\right]\end{array}$$ $\Rightarrow A=\left[\begin{array}{cc}k& -\frac{2}{3}k\\ 0& \frac{k}{3}\end{array}\right]$ $\because |3\mathrm{A}|=108$ $\Rightarrow 108=\left|\begin{array}{cc}3k& -2k\\ 0& k\end{array}\right|$ $\Rightarrow 3k^2=108\Rightarrow k^2=36\Rightarrow k=\pm 6$ For $k=6$ $A=\left[\begin{array}{cc}6& -4\\ 0& 2\end{array}\right]$ $\Rightarrow A^2=\left[\begin{array}{cc}36& -32\\ 0& 4\end{array}\right]$ For $k=-6$ $\Rightarrow A=\left[\begin{array}{cc}-6& 4\\ 0& -2\end{array}\right]$ $\Rightarrow A^2=\left[\begin{array}{cc}36& -32\\ 0& 4\end{array}\right]$