If, then show that

The given matrix is $\mathrm{A}=\left[\begin{array}{lll}1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4\end{array}\right]$.

It can be observed that in the first column, two entries are zero. Thus, we expand along the first column (C1) for easier calculation.

$|\mathrm{A}|=1\left|\begin{array}{ll}1 & 2 \\ 0 & 4\end{array}\right|-0\left|\begin{array}{ll}0 & 1 \\ 0 & 4\end{array}\right|+0\left|\begin{array}{ll}0 & 1 \\ 1 & 2\end{array}\right|=1(4-0)-0+0=4$

$\therefore 27|\mathrm{~A}|=27(4)=108$   ....(i)

Now, $3 \mathrm{~A}=3\left[\begin{array}{lll}1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4\end{array}\right]=\left[\begin{array}{ccc}3 & 0 & 3 \\ 0 & 3 & 6 \\ 0 & 0 & 12\end{array}\right]$

$\therefore|3 \mathrm{~A}|=3\left|\begin{array}{cc}3 & 6 \\ 0 & 12\end{array}\right|-0\left|\begin{array}{cc}0 & 3 \\ 0 & 12\end{array}\right|+0\left|\begin{array}{ll}0 & 3 \\ 3 & 6\end{array}\right|$

$=3(36-0)=3(36)=108$    ...(ii)

From equations (i) and (ii), we have:

$|3 A|=27|A|$

Hence, the given result is proved.