## Precalculus (6th Edition) Blitzer

The new matrix is, $\left[ \begin{matrix} 1 & -5 & 2 & -2 & 4 \\ 0 & 1 & -3 & -1 & 0 \\ 0 & 15 & -4 & 5 & -6 \\ 0 & -19 & 12 & -6 & 13 \\ \end{matrix} \right]$
Consider the given matrix, $\left[ \begin{matrix} 1 & -5 & 2 & -2 & 4 \\ 0 & 1 & -3 & -1 & 0 \\ 3 & 0 & 2 & -1 & 6 \\ -4 & 1 & 4 & 2 & -3 \\ \end{matrix} \right]$ The operation $-3{{R}_{1}}+{{R}_{3}}$ implies that elements of the first row will be multiplied by $-3$ and then added with the corresponding elements of the third row. The operation $4{{R}_{1}}+{{R}_{4}}$ implies that elements of the first row will be multiplied by $4$ and then added with the corresponding elements of the fourth row. The new matrix is obtained after performing the row operation ${{R}_{3}}\to -3{{R}_{1}}+{{R}_{3}},{{R}_{4}}\to 4{{R}_{1}}+{{R}_{4}}$ , $\left[ \begin{matrix} 1 & -5 & 2 & -2 & 4 \\ 0 & 1 & -3 & -1 & 0 \\ -3+3 & 15+0 & -6+2 & 6-1 & -12+6 \\ 4-4 & -20+1 & 8+4 & -8+2 & 16-3 \\ \end{matrix} \right]$ Therefore, the new matrix is $\left[ \begin{matrix} 1 & -5 & 2 & -2 & 4 \\ 0 & 1 & -3 & -1 & 0 \\ 0 & 15 & -4 & 5 & -6 \\ 0 & -19 & 12 & -6 & 13 \\ \end{matrix} \right]$