Question:
Use elementary column operation $\mathrm{C}_{2} \rightarrow \mathrm{C}_{2}+2 \mathrm{C}_{1}$ in the following matrix equation :
$\left(\begin{array}{ll}2 & 1 \\ 2 & 0\end{array}\right)=\left(\begin{array}{ll}3 & 1 \\ 2 & 0\end{array}\right)\left(\begin{array}{rr}1 & 0 \\ -1 & 1\end{array}\right)$
Solution:
$\left(\begin{array}{ll}2 & 1 \\ 2 & 0\end{array}\right)=\left(\begin{array}{ll}3 & 1 \\ 2 & 0\end{array}\right)\left(\begin{array}{rr}1 & 0 \\ -1 & 1\end{array}\right)$
Applying $\mathrm{C}_{2} \rightarrow \mathrm{C}_{2}+2 \mathrm{C}_{1}$, we get
$\left(\begin{array}{ll}2 & 5 \\ 2 & 4\end{array}\right)=\left(\begin{array}{ll}3 & 1 \\ 2 & 0\end{array}\right)\left(\begin{array}{rc}1 & 2 \\ -1 & -1\end{array}\right)$