On using row operation `R_(1)rArrR_(1)-3R_(2)` in the following matrix equation `[{:(4,2),(3,3):}]=[{:(1,2),(0,3):}][{:(2,0),(1,1):}]` we have

Options

A. `[{:(-5,-7),(3,3):}]=[{:(1,-7),(0,3):}][{:(2,0),(1,1):}]`
B. `[{:(-5,-7),(3,3):}]=[{:(1,2),(0,3):}][{:(-1,-3),(1,1):}]`
C. `[{:(-1,-7),(3,3):}]=[{:(1,2),(1,-7):}][{:(2,0),(1,1):}]`
D. `[{:(4,2),(-5,-7):}]=[{:(1,2),(-3,-3):}]=[{:(1,2),(-3,-3):}][{:(2,0),(1,1):}]`

Explanation

To solve the problem, we need to perform the row operation \( R_1 \rightarrow R_1 - 3R_2 \) on the left-hand side (LHS) of the matrix equation. The original matrix equation is:

\[
\begin{pmatrix}
4 & 2 \\
3 & 3
\end{pmatrix}
=
...