Question

If $A=\left[\begin{array}{ccc}1& 0& 1\\ 0& 1& 2\\ 0& 0& 4\end{array}\right]$ and $|3A|=k|A|,$ then $k$ is

• A. 27.00
• B. 27
• C. 27.0

Answer 1

Answer: (A), (B), (C)

$A=\left[\begin{array}{ccc}1& 0& 1\\ 0& 1& 2\\ 0& 0& 4\end{array}\right]$ $\Rightarrow |A|=4$
$3A=\left[\begin{array}{ccc}3& 0& 3\\ 0& 3& 6\\ 0& 0& 12\end{array}\right]$
Expanding along the first row
$|3A|=3\left|\begin{array}{cc}3& 6\\ 0& 12\end{array}\right|-0\left|\begin{array}{cc}0& 6\\ 0& 12\end{array}\right|+3\left|\begin{array}{cc}0& 3\\ 0& 0\end{array}\right|$
$=3(3\times 12-6\times 0)-0(0\times 12-0\times 6)+(0\times 0-0\times 3)$
$=3(36-0)+0+3(0+0)$
$=3\times 36$
$=108$
$=27\times 4$
$=27|A|$
Hence $k=27$