## Precalculus (6th Edition) Blitzer

The solution is, ${{x}_{1}}=2,{{x}_{2}}=-2,{{x}_{3}}=3,{{x}_{4}}=4$
Consider the given system of equation \begin{align} & 3{{x}_{1}}+5{{x}_{2}}-8{{x}_{3}}+5{{x}_{4}}=-8 \\ & {{x}_{1}}+2{{x}_{2}}-3{{x}_{3}}+{{x}_{4}}=-7 \\ & 2{{x}_{1}}+3{{x}_{2}}-7{{x}_{3}}+3{{x}_{4}}=-11 \\ & 4{{x}_{1}}+8{{x}_{2}}-10{{x}_{3}}+7{{x}_{4}}=-10 \end{align} Therefore, the system of equations can be written in matrix form as below: $AX=b$ Where $A=\left[ \begin{matrix} 3 & 5 & -8 & 5 \\ 1 & 2 & -3 & 1 \\ 2 & 3 & -7 & 3 \\ 4 & 8 & -10 & 7 \\ \end{matrix} \right];b=\left[ \begin{array}{*{35}{r}} -8 \\ -7 \\ -11 \\ -10 \\ \end{array} \right];X=\left[ \begin{matrix} {{x}_{1}} \\ {{x}_{2}} \\ {{x}_{3}} \\ {{x}_{4}} \\ \end{matrix} \right]$ Consider the augmented matrix $\left[ A|b \right]=\left[ \begin{matrix} 3 & 5 & -8 & 5 & -8 \\ 1 & 2 & -3 & 1 & -7 \\ 2 & 3 & -7 & 3 & -11 \\ 4 & 8 & -10 & 7 & -10 \\ \end{matrix} \right]$ By applying the elementary row operation on $A$ we will convert it to its equivalent upper triangular matrix form. Step 1: Apply the operation ${{{R}'}_{2}}={{R}_{2}}-\frac{1}{3}{{R}_{1}},{{{R}'}_{3}}={{R}_{3}}-\frac{2}{3}{{R}_{1}},{{{R}'}_{4}}={{R}_{4}}-\frac{4}{3}{{R}_{1}}$ $\left[ A|b \right]=\left[ \begin{array}{*{35}{r}} 3 & 5 & -8 & 5 & -8 \\ 0 & \frac{1}{3} & -\frac{1}{3} & -\frac{2}{3} & -\frac{13}{3} \\ 0 & -\frac{1}{3} & -\frac{5}{3} & -\frac{1}{3} & -\frac{17}{3} \\ 0 & \frac{4}{3} & \frac{2}{3} & \frac{1}{3} & -\frac{2}{3} \\ \end{array} \right]$ Step 2: Apply the operation ${{{R}'}_{3}}={{R}_{3}}+{{R}_{2}},{{{R}'}_{4}}={{R}_{4}}-4{{R}_{2}}$ $\left[ A|b \right]=\left[ \begin{array}{*{35}{r}} 3 & 5 & -8 & 5 & -8 \\ 0 & \frac{1}{3} & -\frac{1}{3} & -\frac{2}{3} & -\frac{13}{3} \\ 0 & 0 & -2 & -1 & -10 \\ 0 & 0 & 2 & 3 & 18 \\ \end{array} \right]$ Step 3: Apply the operation ${{{R}'}_{4}}={{R}_{4}}+{{R}_{3}}$ $\left[ A|b \right]=\left[ \begin{array}{*{35}{r}} 3 & 5 & -8 & 5 & -8 \\ 0 & \frac{1}{3} & -\frac{1}{3} & -\frac{2}{3} & -\frac{13}{3} \\ 0 & 0 & -2 & -1 & -10 \\ 0 & 0 & 0 & 2 & 8 \\ \end{array} \right]$ Therefore, the system of equations reduces to \begin{align} & 3{{x}_{1}}+5{{x}_{2}}-8{{x}_{3}}+5{{x}_{4}}=-8 \\ & \frac{1}{3}{{x}_{2}}-\frac{1}{3}{{x}_{3}}-\frac{2}{3}{{x}_{4}}=-\frac{13}{3} \\ & -2{{x}_{3}}-{{x}_{4}}=-10 \\ & 2{{x}_{4}}=8 \\ \end{align} From (IV) we get, ${{x}_{4}}=4$ From (III) we get, \begin{align} & {{x}_{3}}=-\frac{1}{2}\left( {{x}_{4}}-10 \right) \\ & =-\frac{1}{2}\left( 4-10 \right) \\ & =3 \end{align} From (II) we get, \begin{align} & {{x}_{2}}={{x}_{3}}+2{{x}_{4}}-13 \\ & =3+8-13 \\ & =-2 \end{align} From (I) we get, \begin{align} & {{x}_{1}}=\frac{1}{3}\left( -5{{x}_{2}}+8{{x}_{3}}-5{{x}_{4}}-8 \right) \\ & =\frac{1}{3}\left( 10+24-20-8 \right) \\ & =2 \end{align} The solution of the system is ${{x}_{1}}=2,{{x}_{2}}=-2,{{x}_{3}}=3,{{x}_{4}}=4$.