College Algebra, Chapter 7, 7.1, Section 7.1, Problem 44

Find the complete solution of the system

$\left\{ \begin{array}{ccccc} 3x & + y & & = & 2 \\ -4x & + 3y & + z & = & 4 \\ 2x & + 5y & + z & = & 0 \end{array} \right.$


We transform the system into row-echelon form

$\left[ \begin{array}{cccc} 3 & 1 & 0 & 2 \\ -4 & 3 & 1 & 4 \\ 2 & 5 & 1 & 0 \end{array} \right]$

$\displaystyle \frac{1}{3} R_1$

$\left[ \begin{array}{cccc} 1 & \displaystyle \frac{1}{3} & 0 & \displaystyle \frac{2}{3} \\ -4 & 3 & 1 & 4 \\ 2 & 5 & 1 & 0 \end{array} \right]$

$\displaystyle R_2 + 4 R_1 \to R_2$

$\left[ \begin{array}{cccc} 1 & \displaystyle \frac{1}{3} & 0 & \displaystyle \frac{2}{3} \\ 0 & \displaystyle \frac{13}{3} & 1 & \displaystyle \frac{20}{3} \\ 2 & 5 & 1 & 0 \end{array} \right]$

$\displaystyle R_3 - 2 R_1 \to R_3$

$\left[ \begin{array}{cccc} 1 & \displaystyle \frac{1}{3} & 0 & \displaystyle \frac{2}{3} \\ 0 & \displaystyle \frac{13}{3} & 1 & \displaystyle \frac{20}{3} \\ 0 & \displaystyle \frac{13}{3} & 1 & \displaystyle \frac{-4}{3} \end{array} \right]$

$\displaystyle \frac{3}{13} R_2$

$\left[ \begin{array}{cccc} 1 & \displaystyle \frac{1}{3} & 0 & \displaystyle \frac{2}{3} \\ 0 & 1 & \displaystyle \frac{3}{13} & \displaystyle \frac{20}{13} \\ 0 & \displaystyle \frac{13}{3} & 1 & \displaystyle \frac{-4}{3} \end{array} \right]$

$\displaystyle R_3 - \frac{13}{3} R_2 \to R_3$

$\left[ \begin{array}{cccc} 1 & \displaystyle \frac{1}{3} & 0 & \displaystyle \frac{2}{3} \\ 0 & 1 & \displaystyle \frac{3}{13} & \displaystyle \frac{20}{13} \\ 0 & 0 & 0 & -8 \end{array} \right]$

$\displaystyle \frac{-1}{8} R_3$

$\left[ \begin{array}{cccc} 1 & \displaystyle \frac{1}{3} & 0 & \displaystyle \frac{2}{3} \\ 0 & 1 & \displaystyle \frac{3}{13} & \displaystyle \frac{20}{13} \\ 0 & 0 & 0 & 1 \end{array} \right]$

This last matrix is in row-echelon form, so we stop the Gaussian Elimination Process. Now if we translate the last row back into equation form, we get $0x + 0y + 0z = 1$ or $0 = 1$, which is false. This means that the system has no solution or it is inconsistent.