## Precalculus (6th Edition) Blitzer

The required solution is $x=\left( 7t+18 \right),y=-\left( 7+3t \right),z=t$.
Consider the given system of equations \begin{align} & 2x+3y-5z=15 \\ & x+2y-z=4 \end{align} Therefore in matrix form the system of equations can be written as $AX=b$ Where $A=\left[ \begin{array}{*{35}{r}} 2 & 3 & -5 \\ 1 & 2 & -1 \\ \end{array} \right];b=\left[ \begin{array}{*{35}{r}} 15 \\ 4 \\ \end{array} \right];X=\left[ \begin{matrix} x \\ y \\ z \\ \end{matrix} \right]$ Consider the augmented matrix $\left[ A|b \right]=\left[ \begin{array}{*{35}{r}} 2 & 3 & -5 & 15 \\ 1 & 2 & -1 & 4 \\ \end{array} \right]$ By applying the elementary row operation on $A$ we will convert it to its equivalent upper triangular matrix form. Step 1: Apply the operation ${{{R}'}_{2}}={{R}_{2}}-\frac{1}{2}{{R}_{1}}$ $\left[ A|b \right]=\left[ \begin{array}{*{35}{r}} 2 & 3 & -5 & 15 \\ 0 & \frac{1}{2} & \frac{3}{2} & -\frac{7}{2} \\ \end{array} \right]$ Therefore, the rank of the augmented matrix $\left[ A|b \right]$ is equal to the rank of the coefficient matrix $A$ -- that is, $\text{rank}\left[ A|b \right]=\text{rank}\left[ A \right]=2$