## Precalculus (6th Edition) Blitzer

In this matrix, we will convert all the rows into a linear equation and solve them. Translating every row of the matrix into equation form we get, \begin{align} x+y+2z=19 & \\ y+2z=13 & \\ z=5 & \\ \end{align} Substitute, $z=5$ in $y+2z=13$ to get, \begin{align} & y+2\times 5=13 \\ & y+10=13 \\ & y=13-10 \\ & y=3 \end{align} Substitute, $z=5$, $y=3$ in $x+y+2z=19$ to get, \begin{align} & x+3+2\times 5=19 \\ & x+3+10=19 \\ & x=19-13 \\ & x=6 \end{align} So, the values are: \begin{align} & x=6 \\ & y=3 \\ & z=5 \end{align} Hence, the system has one solution.