## Precalculus (6th Edition) Blitzer

The new matrix is, $\left[ \begin{matrix} 1 & -4 & 2 & 3 \\ 1 & -4 & 4 & 0 \\ 2 & 0 & 7 & 4 \\ \end{matrix} \right]$
Consider the given matrix, $\left[ \begin{matrix} 3 & -12 & 6 & 9 \\ 1 & -4 & 4 & 0 \\ 2 & 0 & 7 & 4 \\ \end{matrix} \right]$ The operation $\frac{1}{3}{{R}_{1}}$ implies that each element of row one will be multiplied by $\frac{1}{3}$. The new matrix is obtained after performing the row operation ${{R}_{1}}\to \frac{1}{3}{{R}_{1}}$. $\left[ \begin{matrix} \frac{1}{3}\times 3 & \frac{1}{3}\times (-12) & \frac{1}{3}\times 6 & \frac{1}{3}\times 9 \\ 1 & -4 & 4 & 0 \\ 2 & 0 & 7 & 4 \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & -4 & 2 & 3 \\ 1 & -4 & 4 & 0 \\ 2 & 0 & 7 & 4 \\ \end{matrix} \right]$ Therefore, the new matrix is, $\left[ \begin{matrix} 1 & -4 & 2 & 3 \\ 1 & -4 & 4 & 0 \\ 2 & 0 & 7 & 4 \\ \end{matrix} \right]$