Answer
$(2,1,-1,-1)$
Work Step by Step
Step 1. Set up the augmented matrix of the system of equations:
$\begin{array} \\A=\\ \end{array}
\begin{bmatrix} 1&0&3&0&-1\\0&1&0&-4&5\\0&2&1&1&0\\2&1&5&-4&4 \end{bmatrix}
\begin{array} \\ \\ \\R_3-2R_2\to R_3\\R_4-2R_1\to R_4 \\ \end{array}$
Step 2. Do the row operations indicated on the right side of the matrix:
$\begin{array} \\A=\\ \end{array}
\begin{bmatrix} 1&0&3&0&-1\\0&1&0&-4&5\\0&0&1&9&-10\\0&1&-1&-4&6 \end{bmatrix}
\begin{array} \\ \\ \\ \\-R_4+R_2\to R_4 \\ \end{array}$
Step 3. Do the row operations indicated on the right side of the matrix:
$\begin{array} \\A=\\ \end{array}
\begin{bmatrix} 1&0&3&0&-1\\0&1&0&-4&5\\0&0&1&9&-10\\0&0&1&0&-1 \end{bmatrix}
\begin{array} \\ \\ \\ \\ \\ \end{array}$
Step 4. The last row gives $z=-1$, back substitute into the third row to get $-1+9w=-10$ or $w=-1$, repeat the process for the first two rows to get $y-4(-1)=5$ or $y=1$, and $x+3(-1)=-1$ or $x=2$
Step 5.Conclusion: the solution to the system is $(2,1,-1,-1)$
