## Algebra and Trigonometry 10th Edition

NOTE: Gauss-Jordan elimination will reduce until the matrix is in reduced row-echelon form. $\begin{bmatrix} 2 & -1 & 3 & |24\\ 0 & 2 & -1 & |14\\ 7 & -5 & 0 & |6\\ \end{bmatrix}$ ~ $\begin{bmatrix} 2 & -1 & 3 & |24\\ 0 & 2 & -1 & |14\\ 0 & -3 & -21 & |-156\\ \end{bmatrix}$ ~ $\begin{bmatrix} 2 & -1 & 3 & |24\\ 0 & 2 & -1 & |14\\ 0 & 1 & 7 & |52\\ \end{bmatrix}$ ~ $\begin{bmatrix} 2 & -1 & 3 & |24\\ 0 & 2 & -1 & |14\\ 0 & 0 & -15 & |-90\\ \end{bmatrix}$ ~ $\begin{bmatrix} 2 & -1 & 3 & |24\\ 0 & 2 & -1 & |14\\ 0 & 0 & 1 & |6\\ \end{bmatrix}$ ~ $\begin{bmatrix} 2 & -1 & 0 & |6\\ 0 & 2 & 0 & |20\\ 0 & 0 & 1 & |6\\ \end{bmatrix}$ ~ $\begin{bmatrix} 2 & -1 & 0 & |6\\ 0 & 1 & 0 & |10\\ 0 & 0 & 1 & |6\\ \end{bmatrix}$ ~ $\begin{bmatrix} 2 & 0 & 0 & |16\\ 0 & 1 & 0 & |10\\ 0 & 0 & 1 & |6\\ \end{bmatrix}$ ~ $\begin{bmatrix} 1 & 0 & 0 & |8\\ 0 & 1 & 0 & |10\\ 0 & 0 & 1 & |6\\ \end{bmatrix}$ From reduced row-echelon form the solution is simple: x = 8 y = 10 z = 6