Question

Let \(A\) be a matrix such that \(A\cdot\begin{bmatrix}1&2\\0&3\end{bmatrix}\) 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. \(\begin{bmatrix}36&-32\\0&4\end{bmatrix}\)

Hint:

If \(AB=kI\) then \(A=kB^{-1}\); use \(|3A|=9|A|=108\) to find \(k\).

Answer 1

The Correct Option is D

Let \(B=\begin{bmatrix}1&2\\0&3\end{bmatrix}\), \(|B|=3\). \(AB=kI\Rightarrow A=kB^{-1}\).
\(|3A|=9|A|=108\Rightarrow|A|=12\). Also \(|A|=\frac{k^2}{|B|}=\frac{k^2}{3}\), so \(k^2=36\), \(k=\pm6\).
\(B^{-1}=\frac13\begin{bmatrix}3&-2\\0&1\end{bmatrix}\), so \(A=\pm2\begin{bmatrix}3&-2\\0&1\end{bmatrix}\).
\(A^2=4\begin{bmatrix}3&-2\\0&1\end{bmatrix}^2=4\begin{bmatrix}9&-8\\0&1\end{bmatrix}=\begin{bmatrix}36&-32\\0&4\end{bmatrix}\) (the sign of k does not matter).
Answer: (4) \(\begin{bmatrix}36&-32\\0&4\end{bmatrix}\)