# Chapter 8 - Section 8.2 - Inconsistent and Dependent Systems and Their Applications - Exercise Set - Page 903: 15

The solution of the system is $\left\{ \frac{z}{3}+1,\ \frac{z}{3},z \right\}$.

#### Work Step by Step

In this given linear equations, we have to find out the solution; for that we have to follow the following steps. \begin{align} 2x+y-z=2 & \\ 3x+3y-2z=3 & \\ \end{align} Now, converting it into an augmented matrix, we get, \left[ \left. \begin{align} & \begin{matrix} 2 & 1 & -1 \\ \end{matrix} \\ & \begin{matrix} 3 & 3 & -2 \\ \end{matrix} \\ \end{align} \right|\begin{matrix} 2 \\ 3 \\ \end{matrix} \right] Use $2{{R}_{2}}-3{{R}_{1}}\to {{R}_{2}}$ Therefore, \left[ \left. \begin{align} & \begin{matrix} 6 & 3 & -3 \\ \end{matrix} \\ & \begin{matrix} 0 & 3 & -1 \\ \end{matrix} \\ \end{align} \right|\begin{matrix} 6 \\ 0 \\ \end{matrix} \right] Now we will express this matrix into equations as below:. \begin{align} 6x+3y-3z=6 & \\ 3y-z=0 & \\ \end{align} Further we will solve the equations for values of $x, y$ and $z$. \begin{align} & x=\frac{z}{3}+1 \\ & y=\frac{z}{3} \\ & z=z \end{align} Hence, the solution is, $\left\{ \frac{z}{3}+1,\ \frac{z}{3},z \right\}$.

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