Chapter 10 - 10.2 - Operations with Matrices - 10.2 Exercises - Page 726: 72

Part A: $\begin{bmatrix} 1 & -1 & 4\\ 1 & 3 & 0\\ 0 & -6 & 5\\ \end{bmatrix}$$\begin{bmatrix} x_{1}\\ x_{2}\\ x_{3}\\ \end{bmatrix}$ = $\begin{bmatrix} 17\\ -11\\ 40\\ \end{bmatrix}$ Part B: $\begin{bmatrix} x_{1}\\ x_{2}\\ x_{3}\\ \end{bmatrix}$ = $\begin{bmatrix} 4\\ -5\\ 2\\ \end{bmatrix}$

Part A: $\begin{bmatrix} 1 & -1 & 4\\ 1 & 3 & 0\\ 0 & -6 & 5\\ \end{bmatrix}$$\begin{bmatrix} x_{1}\\ x_{2}\\ x_{3}\\ \end{bmatrix}$ = $\begin{bmatrix} 17\\ -11\\ 40\\ \end{bmatrix}$ Part B: Gauss Jordan Elimination: $\begin{bmatrix} 1 & -1 & 4 & |17\\ 1 & 3 & 0 & |-11\\ 0 & -6 & 5 & |40\\ \end{bmatrix}$ ~ $\begin{bmatrix} 1 & -1 & 4 & |17\\ 0 & -4 & 4 & |28\\ 0 & -6 & 5 & |40\\ \end{bmatrix}$ ~ $\begin{bmatrix} 1 & -1 & 4 & |17\\ 0 & 1 & -1 & |-7\\ 0 & -6 & 5 & |40\\ \end{bmatrix}$ ~ $\begin{bmatrix} 1 & -1 & 4 & |17\\ 0 & 1 & -1 & |-7\\ 0 & 0 & -1 & |-2\\ \end{bmatrix}$ ~ $\begin{bmatrix} 1 & -1 & 0 & |9\\ 0 & 1 & 0 & |-5\\ 0 & 0 & 1 & |2\\ \end{bmatrix}$ ~ $\begin{bmatrix} 1 & 0 & 0 & |4\\ 0 & 1 & 0 & |-5\\ 0 & 0 & 1 & |2\\ \end{bmatrix}$ Therefore the answer is: $\begin{bmatrix} x_{1}\\ x_{2}\\ x_{3}\\ \end{bmatrix}$ = $\begin{bmatrix} 4\\ -5\\ 2\\ \end{bmatrix}$