Let $A$ be a matrix such that $A$. $\left[\begin{array}{ll}1 & 2 \ 0 & 3\end{array} ight]$ is a scalar matrix and $|3 A|=108$. Then $A^2$ equals

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Let $A$ be a matrix such that $A$. $\left[\begin{array}{ll}1 & 2 \ 0 & 3\end{array}ight]$ is a scalar matrix and $|3 A|=108$. Then $A^2$ equals

Aicharya · Accepted Answer

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

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

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