## Precalculus (6th Edition) Blitzer

The new matrix is, $\left[ \begin{matrix} 1 & -1 & 1 & 1 & 3 \\ 0 & 1 & -2 & -1 & 0 \\ 0 & 2 & 1 & 2 & 5 \\ 0 & 6 & -3 & -1 & -9 \\ \end{matrix} \right]$
Consider the given matrix, $\left[ \begin{matrix} 1 & -1 & 1 & 1 & 3 \\ 0 & 1 & -2 & -1 & 0 \\ 2 & 0 & 3 & 4 & 11 \\ 5 & 1 & 2 & 4 & 6 \\ \end{matrix} \right]$ The operation $-2{{R}_{1}}+{{R}_{3}}$ implies that elements of the first row will be multiplied by $-2$ and then added with corresponding element of the third row. The operation $-5{{R}_{1}}+{{R}_{4}}$ implies that elements of the first row will be multiplied by $-5$ and then added with the corresponding elements of the fourth row. The new matrix is obtained from the row operations ${{R}_{3}}\to -2{{R}_{1}}+{{R}_{3}},{{R}_{4}}\to -5{{R}_{1}}+{{R}_{4}}$. So, $\left[ \begin{matrix} 1 & -1 & 1 & 1 & 3 \\ 0 & 1 & -2 & -1 & 0 \\ -2+2 & 2+0 & -2+3 & -2+4 & -6+11 \\ -5+5 & 5+1 & -5+2 & -5+4 & -15+6 \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & -1 & 1 & 1 & 3 \\ 0 & 1 & -2 & -1 & 0 \\ 0 & 2 & 1 & 2 & 5 \\ 0 & 6 & -3 & -1 & -9 \\ \end{matrix} \right]$ Therefore, the new matrix is $\left[ \begin{matrix} 1 & -1 & 1 & 1 & 3 \\ 0 & 1 & -2 & -1 & 0 \\ 0 & 2 & 1 & 2 & 5 \\ 0 & 6 & -3 & -1 & -9 \\ \end{matrix} \right]$