Chapter 8 - Section 8.2 - Inconsistent and Dependent Systems and Their Applications - Exercise Set - Page 903: 21

$\left\{ \left( 1,\ -1-z,\ 2,\ z \right) \right\}$

Now we need to convert the given system of equations into matrix form. $\left[ \left. \begin{align} & \begin{matrix} 1 & 1 & -1 & 1 \\ \end{matrix} \\ & \begin{matrix} 2 & -1 & 2 & -1 \\ \end{matrix} \\ & \begin{matrix} -1 & 2 & 1 & 2 \\ \end{matrix} \\ \end{align} \right|\begin{matrix} -2 \\ 7 \\ -1 \\ \end{matrix} \right]$ Now we will solve this matrix as below. $\left[ \left. \begin{align} & \begin{matrix} 1 & 1 & -1 & \ \ 1 \\ \end{matrix} \\ & \begin{matrix} 2 & -1 & 2 & -1 \\ \end{matrix} \\ & \begin{matrix} 0 & 3 & \ \ 0 & \ \ 3 \\ \end{matrix} \\ \end{align} \right|\begin{matrix} -2 \\ 7 \\ -3 \\ \end{matrix} \right]$ $ By,\ {{R}_{1}}+{{R}_{3}}\to {{R}_{3}}$ $\left[ \left. \begin{align} & \begin{matrix} 1 & 1 & -1 & \ \ 1 \\ \end{matrix} \\ & \begin{matrix} 0 & -3 & 4 & -3 \\ \end{matrix} \\ & \begin{matrix} 0 & 3 & \ \ 0 & \ \ 3 \\ \end{matrix} \\ \end{align} \right|\begin{matrix} -2 \\ 11 \\ -3 \\ \end{matrix} \right]$ $ By,\ {{R}_{2}}-2{{R}_{1}}\to {{R}_{2}}$ $\left[ \left. \begin{align} & \begin{matrix} 1 & 1 & -1 & \ \ 1 \\ \end{matrix} \\ & \begin{matrix} 0 & -3 & 4 & -3 \\ \end{matrix} \\ & \begin{matrix} 0 & 0 & \ \ 4 & \ \ 0 \\ \end{matrix} \\ \end{align} \right|\begin{matrix} -2 \\ 11 \\ 8 \\ \end{matrix} \right]$ $ By,\ {{R}_{2}}+{{R}_{3}}\to {{R}_{3}}$ Now, we convert this matrix system into linear equations as below. $\begin{align} w+x-y+z=-2 & \\ -3x+4y-3z=11 & \\ 4y=8 & \\ \end{align}$ Now writing the values of $ w,x $ and $ y $ in terms of $ z $ and solving the equation we get, $\begin{align} & w=1 \\ & x=-1-z \\ & y=2 \\ & z=z \\ \end{align}$ Hence, the solution is: $\left\{ \left( 1,\ -1-z,\ 2,\ z \right) \right\}$