## Precalculus (6th Edition) Blitzer

The solution set of an inconsistent system of equations is $\phi$.
A system in which there is no particular solution set for the variables is called an inconsistent system. So, in the Gaussian elimination method, a system is known to be inconsistent if there are 0's left of the line and a nonzero value to the right. Example: \begin{align} & 3x+y+3z=14 \\ & 7x+5y+8z=32 \\ & x+3y+2z=9\, \end{align} The matrix corresponding to the system of equations is as follows: $\left[ \begin{matrix} 3 & 1 & 3 & 14 \\ 7 & 5 & 3 & 32 \\ 1 & 3 & 2 & 9 \\ \end{matrix} \right]$ To obtain the desired results, using elementary row transformation, we will find the echelon form of the matrix. Apply ${{R}_{1}}\leftrightarrow {{R}_{3}}$: $\left[ \begin{matrix} 1 & 3 & 2 & 9 \\ 7 & 5 & 8 & 32 \\ 3 & 1 & 3 & 9 \\ \end{matrix} \right]$ Apply ${{R}_{3}}\to {{R}_{3}}-3{{R}_{1}}$ and ${{R}_{2}}\to {{R}_{2}}-7{{R}_{1}}$: $\left[ \begin{matrix} 1 & 3 & 2 & 9 \\ 0 & -16 & -6 & -31 \\ 0 & -8 & -3 & -13 \\ \end{matrix} \right]$ Apply ${{R}_{3}}\to {{R}_{3}}-\frac{1}{2}{{R}_{2}}$: $\left[ \begin{matrix} 1 & 3 & 2 & 9 \\ 0 & -16 & -6 & -31 \\ 0 & 0 & 0 & \frac{5}{2} \\ \end{matrix} \right]$ Apply ${{R}_{2}}\to -\frac{1}{16}{{R}_{2}}$: $\left[ \begin{matrix} 1 & 3 & 2 & 9 \\ 0 & 1 & \frac{3}{8} & \frac{31}{16} \\ 0 & 0 & 0 & \frac{5}{2} \\ \end{matrix} \right]$ From the above matrix: $0x+0y+0z=\frac{5}{2}$, which is not possible since for any values of $x,\ y$, and $z$, $0\ne \frac{5}{2}$. Thus, there is no solution to the given problem. Therefore, the solution set of an inconsistent system of equations is $\phi$.