Answer
$$\left[\begin{array}{cc}{a} &{b} \\{c}& {d} \end{array}\right]=\left[\begin{array}{cc}{48} &{-31} \\{-7}& {6} \end{array}\right].$$
Work Step by Step
We have the system $$ \begin{aligned} 2a+3b&=3 \\ a+b &=17\\ 2c+3d &=4\\ c+d &=-1 \end{aligned}. $$ The augmented matrix is given by $$ \left[ \begin {array}{ccccc} 2&3&0&0&3\\ 1&1&0&0&17 \\ 0&0&2&3&4\\ 0&0&1&1&-1 \end {array} \right] . $$ Using Gauss-Jordan elimination, we get the row-reduced echelon form as follows $$\left[ \begin {array}{ccccc} 1&0&0&0&48\\ 0&1&0&0&- 31\\ 0&0&1&0&-7\\ 0&0&0&1&6 \end {array} \right] .$$ From which the solution is $$a=48, \quad b=-31 \quad c=-7, \quad d=6 .$$ Hence, we have $$\left[\begin{array}{cc}{a} &{b} \\{c}& {d} \end{array}\right]=\left[\begin{array}{cc}{48} &{-31} \\{-7}& {6} \end{array}\right].$$
