If $A = \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4 \end{bmatrix}$,then show that $|3A| = 27|A|$.

(N/A) The given matrix is $A = \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4 \end{bmatrix}$.
First,we calculate the determinant of $A$ by expanding along the first column:
$|A| = 1 \begin{vmatrix} 1 & 2 \\ 0 & 4 \end{vmatrix} - 0 \begin{vmatrix} 0 & 1 \\ 0 & 4 \end{vmatrix} + 0 \begin{vmatrix} 0 & 1 \\ 1 & 2 \end{vmatrix} = 1(4 - 0) = 4$.
Therefore,$27|A| = 27 \times 4 = 108$ $(i)$.
Now,we find $3A$:
$3A = 3 \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4 \end{bmatrix} = \begin{bmatrix} 3 & 0 & 3 \\ 0 & 3 & 6 \\ 0 & 0 & 12 \end{bmatrix}$.
Calculating the determinant of $3A$ by expanding along the first column:
$|3A| = 3 \begin{vmatrix} 3 & 6 \\ 0 & 12 \end{vmatrix} - 0 + 0 = 3(36 - 0) = 3 \times 36 = 108$ $(ii)$.
From equations $(i)$ and $(ii)$,we have $|3A| = 27|A|$.