On using elementary column operations ${C_2} \to {C_2} - 2{C_1}$ in the following matrix equation $\left[ {\begin{array}{cccccccccccccccccccc}1&{ - 3}\\2&4\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}1&{ - 1}\\0&1\end{array}} \right]\left[ {\begin{array}{cccccccccccccccccccc}3&1\\2&4\end{array}} \right]$, we have

Given that, $\left[ {\begin{array}{llllllllllllllllllll}1&{ - 3}\\2&4\end{array}} \right]$

$= \left[ {\begin{array}{cccccccccccccccccccc}1&{ - 1}\\0&1\end{array}} \right]\left[ {\begin{array}{llllllllllllllllllll}3&1\\2&4\end{array}} \right]$

On using ${C_2} \to {C_2} - 2{C_1},$ $\left[ {\begin{array}{cccccccccccccccccccc}1&{ - 5}\\2&0\end{array}} \right]$

$= \left[ {\begin{array}{cccccccccccccccccccc}1&{ - 1}\\0&1\end{array}} \right]\left[ {\begin{array}{cccccccccccccccccccc}3&{ - 5}\\2&0\end{array}} \right]$

Since, on using elementary column operation on $X = AB$,

we apply these operations simultaneously on $X$ and on the second matrix $B$ of the product $AB$ on RHS.