Question

If $A=\left[\begin{array}{ccc}1& 0& 1\\ 0& 1& 2\\ 0& 0& 4\end{array}\right]$, then show that $|3A|=27|A|$.

Answer 1

Given $A=\left[\begin{array}{ccc}1& 0& 1\\ 0& 1& 2\\ 0& 0& 4\end{array}\right]$
It can be observed that in the first column, two entries are zero. Thus, we expand the matrix A along the first column $\left(C_1\right)$ for finding |A|.
$\therefore |A|=1\left|\begin{array}{cc}1& 2\\ 0& 4\end{array}\right|-0\left|\begin{array}{cc}0& 1\\ 0& 4\end{array}\right|+0\left|\begin{array}{cc}0& 1\\ 1& 2\end{array}\right|=1(1\times 4-0\times 2)=4\therefore 27|A|=27\times 4=108\dots \left(i\right)$

Now, $3A=3\left[\begin{array}{ccc}1& 0& 1\\ 0& 1& 2\\ 0& 0& 4\end{array}\right]=\left[\begin{array}{ccc}3& 0& 3\\ 0& 3& 6\\ 0& 0& 12\end{array}\right]\therefore |3A|=\left[\begin{array}{ccc}3& 0& 3\\ 0& 3& 6\\ 0& 0& 12\end{array}\right]$
It can be observed that in the first column, two entries are zero. Thus, we expand the determinant along the first column $\left(C_1\right)$ for easier calculation.
$|3A|=3\left|\begin{array}{cc}3& 6\\ 0& 12\end{array}\right|-0\left|\begin{array}{cc}0& 3\\ 0& 12\end{array}\right|+0\left|\begin{array}{cc}0& 3\\ 3& 6\end{array}\right|=3(3\times 12-0\times 6)\Rightarrow |3A|=3\left(36\right)=108\dots \left(ii\right)$
From Eqs. (i) and (ii), we get

$|3A|=27|A|$

Hence, the given result is proved.