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.

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

$\therefore 27|A|=27(4)=108……(i)$

${{\text{Now, }}3A = 3\left[ {\begin{array}{*{20}{l}}
  1&0&1 \\ 
  0&1&2 \\ 
  0&0&4 
\end{array}} \right] = \left[ {\begin{array}{*{20}{l}}
  3&0&3 \\ 
  0&3&6 \\ 
  0&0&{12} 
\end{array}} \right]}$

${\therefore \,|3A| = 3\left| {\begin{array}{*{20}{l}}
  3&6 \\ 
  0&{12} 
\end{array}} \right| – 0\left| {\begin{array}{*{20}{l}}
  0&3 \\ 
  0&{12} 
\end{array}} \right| + 0\left| {\begin{array}{*{20}{l}}
  0&3 \\ 
  3&6 
\end{array}} \right|}$

${\begin{array}{*{20}{l}}
  { = 3(36 – 0) = 3(36) = 108} 
\end{array}}……(ii)$

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

$|3 A|=27|A|$

Hence, the given result is proved.