# Chapter 8 - Section 8.1 - Matrix Solutions to Linear Systems - Exercise Set - Page 893: 13

The new matrix is, $\left[ \begin{matrix} 1 & -3 & 2 & 5 \\ 1 & 5 & -5 & 0 \\ 3 & 0 & 4 & 7 \\ \end{matrix} \right]$

#### Work Step by Step

Consider the given matrix, $\left[ \begin{matrix} 2 & -6 & 4 & 10 \\ 1 & 5 & -5 & 0 \\ 3 & 0 & 4 & 7 \\ \end{matrix} \right]$ The operation $\frac{1}{2}{{R}_{1}}$ implies that each element of row one will be multiplied by $\frac{1}{2}$. The new matrix is obtained after performing row operation ${{R}_{1}}\to \frac{1}{2}{{R}_{1}}$. So, $\left[ \begin{matrix} \frac{1}{2}\times 2 & \frac{1}{2}\times (-6) & \frac{1}{2}\times 4 & \frac{1}{2}\times 10 \\ 1 & 5 & -5 & 0 \\ 3 & 0 & 4 & 7 \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & -3 & 2 & 5 \\ 1 & 5 & -5 & 0 \\ 3 & 0 & 4 & 7 \\ \end{matrix} \right]$ Therefore, the new matrix is, $\left[ \begin{matrix} 1 & -3 & 2 & 5 \\ 1 & 5 & -5 & 0 \\ 3 & 0 & 4 & 7 \\ \end{matrix} \right]$

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