JEE Main 2018 — Mathematics Algebra

A. $\left[\begin{array}{ll}1 & 2 \\ 0 & 3\end{array}\right]$[10​23​] is a scalar matrix and $|3 \mathrm{~A}|=108$∣3 A∣=108 suppose the scalar matrix is $\left[\begin{array}{ll}k & 0 \\ 0 & k\end{array}\right]$[k0​0k​] $$\begin{aligned} &\therefore A \cdot\left[\begin{array}{ll} 1 & 2 \\ 0 & 3 \end{array}\right]=\left[\begin{array}{ll} k & 0 \\ 0 & k \end{array}\right] \\ &\Rightarrow A=\left[\begin{array}{ll} k & 0 \\ 0 & k \end{array}\right]\left[\begin{array}{ll} 1 & 2 \\ 0 & 3 \end{array}\right]^{-1} \\ &{\left[\therefore A B=C \Rightarrow A B B^{-1}=C B^{-1} \Rightarrow A=C B^{-1}\right]} \\ &\Rightarrow A=\frac{1}{3}\left[\begin{array}{ll} k & 0 \\ 0 & k \end{array}\right]\left[\begin{array}{cc} 3 & -2 \\ 0 & 1 \end{array}\right] \\ &\Rightarrow A=\left[\begin{array}{ll} k & 0 \\ 0 & k \end{array}\right]\left[\begin{array}{cc} 1 & -\frac{2}{3} \\ 0 & \frac{1}{3} \end{array}\right] \end{aligned}$$​∴A⋅[10​23​]=[k0​0k​]⇒A=[k0​0k​][10​23​]−1[∴AB=C⇒ABB−1=CB−1⇒A=CB−1]⇒A=31​[k0​0k​][30​−21​]⇒A=[k0​0k​][10​−32​31​​]​ $\Rightarrow A=\left[\begin{array}{cc}k & -\frac{2}{3} k \\ 0 & \frac{k}{3}\end{array}\right]$⇒A=[k0​−32​k3k​​] $\because|3 \mathrm{~A}|=108$∵∣3 A∣=108 $\Rightarrow 108=\left|\begin{array}{cc}3 k & -2 k \\ 0 & k\end{array}\right|$⇒108=​3k0​−2kk​​ $\Rightarrow 3 k^2=108 \Rightarrow k^2=36 \Rightarrow k=\pm 6$⇒3k2=108⇒k2=36⇒k=±6 For $k=6$k=6 $A=\left[\begin{array}{cc}6 & -4 \\ 0 & 2\end{array}\right]$A=[60​−42​] $\Rightarrow A^2=\left[\begin{array}{cc}36 & -32 \\ 0 & 4\end{array}\right]$⇒A2=[360​−324​] For $k=-6$k=−6 $\Rightarrow A=\left[\begin{array}{cc}-6 & 4 \\ 0 & -2\end{array}\right]$⇒A=[−60​4−2​] $\Rightarrow A^2=\left[\begin{array}{cc}36 & -32 \\ 0 & 4\end{array}\right]$⇒A2=[360​−324​]