Intermediate Algebra for College Students (7th Edition)

The answer is $\left[\begin{array}{ccc|c} 1& -1 & 5&-6 \\ 0 &6 &-16&28 \\ 1&3&2&5\\ \end{array}\right]$.
The given matrix is $\left[\begin{array}{ccc|c} 1& -1 & 5&-6 \\ 3 &3 &-1& 10 \\ 1&3&2&5\\ \end{array}\right]$ Perform $-3R_1+R_2$. Multiply row one by $-3$ and add to the row two as shown below. $\left[\begin{array}{ccc|c} 1& -1 & 5&-6 \\ -3\cdot 1 +3 &-3\cdot (-1)+3 &-3\cdot(5)-1&-3\cdot (-6)+ 10 \\ 1&3&2&5\\ \end{array}\right]$ Simplify. $\left[\begin{array}{ccc|c} 1& -1 & 5&-6 \\ -3 +3 &3+3 &-15-1&18+ 10 \\ 1&3&2&5\\ \end{array}\right]$ $\left[\begin{array}{ccc|c} 1& -1 & 5&-6 \\ 0 &6 &-16&28 \\ 1&3&2&5\\ \end{array}\right]$.