# Chapter 8 - Section 8.2 - Inconsistent and Dependent Systems and Their Applications - Exercise Set - Page 902: 1

The system of equations has no solution

#### Work Step by Step

The provided system of equations is: \begin{align} 5x+12y+z=10 & \\ 2x+5y+2z=-1 & \\ x+2y-3z=5 & \\ \end{align} The given system of equations can be written in matrix form as below: $\left[ \left. \begin{matrix} 5 & 12 & 1 \\ 2 & 5 & 2 \\ 1 & 2 & -3 \\ \end{matrix} \right|\begin{matrix} 10 \\ -1 \\ 5 \\ \end{matrix} \right]$ Now solve this matrix. $\left[ \left. \begin{matrix} 5 & 12 & 1 \\ 0 & 1 & 8 \\ 1 & 2 & -3 \\ \end{matrix} \right|\begin{matrix} 10 \\ -11 \\ 5 \\ \end{matrix} \right]$ $By,\ {{R}_{2}}-2{{R}_{3}}\to {{R}_{2}}$ $\left[ \left. \begin{matrix} 5 & 12 & 1 \\ 0 & 1 & 8 \\ 0 & 2 & 16 \\ \end{matrix} \right|\begin{matrix} 10 \\ -11 \\ -15 \\ \end{matrix} \right]$ $By,\ {{R}_{1}}-5{{R}_{3}}\to {{R}_{3}}$ $\left[ \left. \begin{matrix} 5 & 12 & 1 \\ 0 & 1 & 8 \\ 0 & 0 & 0 \\ \end{matrix} \right|\begin{matrix} 10 \\ -11 \\ -7 \\ \end{matrix} \right]$ $By,\ 2{{R}_{2}}-{{R}_{3}}\to {{R}_{3}}$ Convert the last row in equation form: \begin{align} 0x+0y+0z=-7 & \\ 0=-7 & \\ \end{align} Which is not possible.

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