Answer
$(1,0,1,-2)$
Work Step by Step
Step 1. Set up the augmented matrix of the system of equations:
$\begin{array} \\A=\\ \end{array}
\begin{bmatrix} 1&1&1&1&0\\1&-1&-4&-1&-1\\1&-2&0&4&-7\\2&2&3&4&-3 \end{bmatrix}
\begin{array} \\ \\-R_2+R_1\to R_2\\-R_3+R_1\to R_3\\R_4-2R_1\to R_4 \\ \end{array}$
Step 2. Do the row operations indicated on the right side of the matrix:
$\begin{array} \\A=\\ \end{array}
\begin{bmatrix} 1&1&1&1&0\\0&2&5&2&1\\0&3&1&-3&7\\0&0&1&2&-3 \end{bmatrix}
\begin{array} \\ \\ \\-2R_3+3R_2\to R_3\\ \\ \end{array}$
Step 3. Do the row operations indicated on the right side of the matrix:
$\begin{array} \\A=\\ \end{array}
\begin{bmatrix} 1&1&1&1&0\\0&2&5&2&1\\0&0&13&12&-11\\0&0&1&2&-3 \end{bmatrix}
\begin{array} \\ \\ \\ \\13R_4-R_3\to R_4 \\ \end{array}$
Step 4. Do the row operations indicated on the right side of the matrix:
$\begin{array} \\A=\\ \end{array}
\begin{bmatrix} 1&1&1&1&0\\0&2&5&2&1\\0&0&13&12&-11\\0&0&0&14&-28 \end{bmatrix}
\begin{array} \\ \\ \\ \\ \\ \end{array}$
Step 5. The last row gives $w=-2$, back-substitute to get other solutions $13z-24=-11$ or $z=1$, $2y+5-4=1$ or $y=0$, and $x+0+1-2=0$ or $x=1$
Step 6. Conclusion: the solution to the system is $(1,0,1,-2)$
