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

$A = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&0&1\\0&1&2\\0&0&4\end{array}} \right]$

$\Rightarrow$ $3A = 3\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&0&1\\0&1&2\\0&0&4\end{array}} \right] = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}3&0&3\\0&3&6\\0&0&{12}\end{array}} \right]$

L.H.S. $= |3A| = \left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}3&0&3\\0&3&6\\0&0&{12}\end{array}} \right|$

$= 3\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}3&6\\0&{12}\end{array}} \right| - 0\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}0&6\\0&{12}\end{array}} \right| + 3\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}0&3\\0&0\end{array}} \right|$

$= 3 \times 36 - 0 + 3\left( 0 \right) = 108$

R.H.S.$=$ $274|A| = 27\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&0&1\\0&1&2\\0&0&4\end{array}} \right|$

$= 24\left[ {1\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&2\\0&4\end{array}} \right| - 0\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}0&2\\0&4\end{array}} \right| + 1\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}0&1\\0&0\end{array}} \right|} \right]$

$= 24[1(4) - 0 + 0] = 108$

Hence, $|3A| = 27|A|$