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

(A) The given matrix is $A=\left[\begin{array}{lll}1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4\end{array}\right]$.
It can be observed that in the first column,two entries are zero. Thus,we expand along the first column $(C_1)$ for easier calculation.
$|A|=1\left|\begin{array}{ll}1 & 2 \\ 0 & 4\end{array}\right|-0\left|\begin{array}{ll}0 & 1 \\ 0 & 4\end{array}\right|+0\left|\begin{array}{ll}0 & 1 \\ 1 & 2\end{array}\right|=1(4-0)-0+0=4$.
$\therefore 27|A|=27(4)=108 \dots (i)$.
Now,$3A = 3\left[\begin{array}{lll}1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4\end{array}\right] = \left[\begin{array}{lll}3 & 0 & 3 \\ 0 & 3 & 6 \\ 0 & 0 & 12\end{array}\right]$.
$\therefore |3A| = 3\left|\begin{array}{ll}3 & 6 \\ 0 & 12\end{array}\right| - 0\left|\begin{array}{ll}0 & 6 \\ 0 & 12\end{array}\right| + 0\left|\begin{array}{ll}0 & 3 \\ 0 & 0\end{array}\right| = 3(36 - 0) = 108 \dots (ii)$.
From equations $(i)$ and $(ii)$,we have $|3A| = 27|A|$.
Hence,the given result is proved.