## Precalculus (10th Edition)

Consistent Solution set: $\{(x_1,x_2,x_3,x_4)|x_1=-2-x_4,x_2=2-2x_4,x_3=x_4,x_4\text{ any real number}\}$
We are given the reduced row echelon form of a system of linear equations: $\begin{bmatrix}1&0&0&1&|&-2\\0&1&0&2&|&2\\0&0&1&-1&|&0\\0&0&0&0&|&0\end{bmatrix}$ Write the system of equations corresponding to the given matrix: $\begin{cases} 1x_1+0x_2+0x_3+1x_4=-2\\ 0x_1+1x_2+0x_3+2x_4=2\\ 0x_1+0x_2+1x_3-1x_4=0\\ 0x_1+0x_2+0x_3+0x_4=0 \end{cases}$ $\begin{cases} x_1+x_4=-2\\ x_2+2x_4=2\\ x_3-x_4=0\\ 0=0 \end{cases}$ Because the reduced row echelon form has a row with only zeros, the system is consistent, having infinitely many solutions. Express $x_1,x_2,x_3$ in terms of $x_4$: $x_3-x_4=0\Rightarrow x_3=x_4$ $x_2+2x_4=2\Rightarrow x_2=2-2x_4$ $x_1+x_4=-2\Rightarrow x_1=-2-x_4$ The solution set is: $\{(x_1,x_2,x_3,x_4)|x_1=-2-x_4,x_2=2-2x_4,x_3=x_4,x_4\text{ any real number}\}$