## Precalculus (6th Edition) Blitzer

The solution is, ${{x}_{1}}=2-37t,{{x}_{2}}=16t,{{x}_{3}}=1-7t,{{x}_{4}}=t$.
Consider the given system of equations \begin{align} & {{x}_{1}}+4{{x}_{2}}+3{{x}_{3}}-6{{x}_{4}}=5 \\ & {{x}_{1}}+3{{x}_{2}}+{{x}_{3}}-4{{x}_{4}}=3 \\ & 2{{x}_{1}}+8{{x}_{2}}+7{{x}_{3}}-5{{x}_{4}}=11 \\ & 2{{x}_{1}}+5{{x}_{2}}-6{{x}_{4}}=4 \end{align} Therefore, in matrix form the system of equations can be written as $AX=b$ Where $A=\left[ \begin{array}{*{35}{r}} 1 & 4 & 3 & -6 \\ 1 & 3 & 1 & -4 \\ 2 & 8 & 7 & -5 \\ 2 & 5 & 0 & -6 \\ \end{array} \right];b=\left[ \begin{array}{*{35}{r}} 5 \\ 3 \\ 11 \\ 4 \\ \end{array} \right];X=\left[ \begin{matrix} {{x}_{1}} \\ {{x}_{2}} \\ {{x}_{3}} \\ {{x}_{4}} \\ \end{matrix} \right]$ Consider the augmented matrix $\left[ A|b \right]=\left[ \begin{array}{*{35}{r}} 1 & 4 & 3 & -6 & 5 \\ 1 & 3 & 1 & -4 & 3 \\ 2 & 8 & 7 & -5 & 11 \\ 2 & 5 & 0 & -6 & 4 \\ \end{array} \right]$ By applying elementary row operation on $A$ we will convert it to its equivalent upper triangular matrix form. Step 1: Apply the operation ${{{R}'}_{2}}={{R}_{2}}-{{R}_{1}},{{{R}'}_{3}}={{R}_{3}}-2{{R}_{1}},{{{R}'}_{4}}={{R}_{4}}-2{{R}_{1}}$ $\left[ A|b \right]=\left[ \begin{array}{*{35}{r}} 1 & 4 & 3 & -6 & 5 \\ 0 & -1 & -2 & 2 & -2 \\ 0 & 0 & 1 & 7 & 1 \\ 0 & -3 & -6 & 6 & -6 \\ \end{array} \right]$ Step 2: Apply the operation ${{{R}'}_{4}}={{R}_{4}}-3{{R}_{3}}$ $\left[ A|b \right]=\left[ \begin{array}{*{35}{r}} 1 & 4 & 3 & -6 & 5 \\ 0 & -1 & -2 & 2 & -2 \\ 0 & 0 & 1 & 7 & 1 \\ 0 & 0 & 0 & 0 & 0 \\ \end{array} \right]$ Therefore, the rank of the augmented matrix $\left[ A|b \right]$ is equal to the rank of the coefficient matrix $A$ -- that is, $\text{rank}\left[ A|b \right]=\text{rank}\left[ A \right]=3$