## Elementary Linear Algebra 7th Edition

$$A=\left[\begin{array}{cc}{2} &{-1} \\{3}& {-2} \end{array}\right].$$
Assume that the matrix $A$ is given by $$A=\left[\begin{array}{cc}{x_1} &{x_2} \\{x_3}& {x_4} \end{array}\right],$$ then we have the system \begin{aligned} 2x_{1}-x_{3}&=1 \\ 2x_{2} -x_4 &=0\\ 3x_1-2x_{3} &=0\\ 3x_{2} -2x_4 &=1 \end{aligned}. The augmented matrix is given by $$\left[ \begin {array}{ccccc} 2&0&-1&0&1\\ 0&2&0&-1&0 \\ &0&-2&0&0\\ 0&3&0&-2&1 \end {array} \right] .$$ Using Gauss-Jordan elimination, we get the row-reduced echelon form as follows $$\left[ \begin {array}{ccccc} 1&0&0&0&2\\ 0&1&0&0&-1 \\ 0&0&1&0&3\\ 0&0&0&1&-2 \end {array} \right] .$$ From which the solution is $$x_1=2, \quad x_2=-1, \quad x_3=3, \quad x_4=-2 .$$ Hence, we have $$A=\left[\begin{array}{cc}{2} &{-1} \\{3}& {-2} \end{array}\right].$$