## Intermediate Algebra for College Students (7th Edition)

The answer is $\left[\begin{array}{cc|c} 1 & -\frac{2}{5} & \frac{3}{4} \\ 0 &\frac{18}{5} &-4 \\ \end{array}\right]$.
The given matrix is $\left[\begin{array}{cc|c} 1 & -\frac{2}{5} & \frac{3}{4} \\ 4 &2 &-1 \\ \end{array}\right]$ Perform $-4R_1+R_2$. Multiply row 1 by $-4$ and add to the row 2 as shown below. $\left[\begin{array}{cc|c} 1 & -\frac{2}{5} & \frac{3}{4} \\ -4\cdot 1+4 &-4\cdot (-\frac{2}{5})+ 2 &-4\cdot ( \frac{3}{4} )-1 \\ \end{array}\right]$ Simplify. $\left[\begin{array}{cc|c} 1 & -\frac{2}{5} & \frac{3}{4} \\ -4+4 &\frac{8}{5} +2 &-3-1 \\ \end{array}\right]$ $\left[\begin{array}{cc|c} 1 & -\frac{2}{5} & \frac{3}{4} \\ 0 &\frac{8+10}{5} &-4 \\ \end{array}\right]$ $\left[\begin{array}{cc|c} 1 & -\frac{2}{5} & \frac{3}{4} \\ 0 &\frac{18}{5} &-4 \\ \end{array}\right]$.