## Precalculus (6th Edition) Blitzer

Consider the given system of equations: \begin{align} & 2x-3y+z=1 \\ & x-2y+3z=2 \\ & 3x-4y-z=1 \\ \end{align} Therefore, in matrix form the system of equations can be written as below: $AX=b$ Where $A=\left[ \begin{array}{*{35}{r}} 2 & -3 & 1 \\ 1 & -2 & 3 \\ 3 & -4 & -1 \\ \end{array} \right];b=\left[ \begin{array}{*{35}{r}} 1 \\ 2 \\ 1 \\ \end{array} \right];X=\left[ \begin{matrix} x \\ y \\ z \\ \end{matrix} \right]$ Consider the augmented matrix $\left[ A|b \right]=\left[ \begin{array}{*{35}{r}} 2 & -3 & 1 & 1 \\ 1 & -2 & 3 & 2 \\ 3 & -4 & -1 & 1 \\ \end{array} \right]$ Apply elementary row operation on $A$ to convert it to its equivalent upper triangular matrix form. Step 1: Apply the operation ${{{R}'}_{2}}={{R}_{2}}-\frac{1}{2}{{R}_{1}},{{{R}'}_{3}}={{R}_{3}}-\frac{3}{2}{{R}_{1}}$ to get, $\left[ A|b \right]=\left[ \begin{array}{*{35}{r}} 2 & -3 & 1 & 1 \\ 0 & -\frac{1}{2} & \frac{5}{2} & \frac{3}{2} \\ 0 & \frac{1}{2} & -\frac{5}{2} & -\frac{1}{2} \\ \end{array} \right]$ Step 2: Apply the operation ${{{R}'}_{3}}={{R}_{3}}+{{R}_{2}}$ to get, $\left[ A|b \right]=\left[ \begin{array}{*{35}{r}} 2 & -3 & 1 & 1 \\ 0 & -\frac{1}{2} & \frac{5}{2} & \frac{3}{2} \\ 0 & 0 & 0 & 1 \\ \end{array} \right]$ Therefore, the rank of the augmented matrix $\left[ A|b \right]$ is 3 and the rank of the coefficient matrix $A$ is 2 -- that is $\text{rank}\left[ A|b \right]\ne \text{rank}\left[ A \right]$