Question

Let A be a matrix such that $A .\begin{bmatrix}1&2\\ 0&3\end{bmatrix}$ is a scalar matrix and $| 3A | =108$. Then $A^2$ equals :

• A. $\begin{bmatrix}4&-32\\ 0&36\end{bmatrix}$
• B. $\begin{bmatrix} 36 & 0 \\ -32 & 4 \end{bmatrix}$
• C. $\begin{bmatrix}4& 0 \\ -32 & 36 \end{bmatrix}$
• D. $\begin{bmatrix} 36 & -32 \\ 0 & 4 \end{bmatrix}$

Answer 1

Answer: (D)

$A = \begin{bmatrix}a&b\\ c&d\end{bmatrix}$
Now A. $\begin{bmatrix}1&2\\ 0&3\end{bmatrix} = \begin{bmatrix}a&b\\ c&d\end{bmatrix} \begin{bmatrix}1&2\\ 0&3\end{bmatrix}$
$= \begin{bmatrix}a&2a&+&3b\\ c&2c&+&3d\end{bmatrix}$ is scalar
$\therefore c = 0, \quad2a + 3b = 0, \quad a = 2c + 3d \quad a = 3d \,\therefore \quad a^{2} = 9d^{2} = 36$
$\left|3A\right| = 108$
$\therefore \left|A\right| = 12 = ad-bc = ad$
$\therefore d^{2} = 4$
Now $A^{2} = \begin{bmatrix}a&b\\ 0&d\end{bmatrix}\begin{bmatrix}a&b\\ 0&d\end{bmatrix} = \begin{bmatrix}a^{2}&ab+bd\\ 0&d^{2}\end{bmatrix}$