Answer
$$\left[\begin{array}{cc}{a} &{b} \\{c}& {d} \end{array}\right]=\left[\begin{array}{cc}{7} &{-4} \\{-\frac{1}{2}}& {\frac{7}{2}} \end{array}\right].$$
Work Step by Step
We have the system
$$
\begin{aligned}
a+2c&=6 \\
b +2d &=3\\
3a+4c &=19\\
3b +4d &=2
\end{aligned}.
$$
The augmented matrix is given by
$$
\left[ \begin {array}{ccccc} 1&0&2&0&6\\ 0&1&0&2&3
\\ 3&0&4&0&19\\ 0&3&0&4&2
\end {array} \right]
.
$$
Using Gauss-Jordan elimination, we get the row-reduced echelon form as follows
$$\left[ \begin {array}{ccccc} 1&0&0&0&7\\ 0&1&0&0&-4
\\ 0&0&1&0&-\frac{1}{2}\\ 0&0&0&1&\frac{7}{2}
\end {array} \right]
.$$
From which the solution is
$$a=7, \quad b=-4, \quad c=-\frac{1}{2}, \quad d=\frac{7}{2} .$$
Hence, we have
$$\left[\begin{array}{cc}{a} &{b} \\{c}& {d} \end{array}\right]=\left[\begin{array}{cc}{7} &{-4} \\{-\frac{1}{2}}& {\frac{7}{2}} \end{array}\right].$$
