Chapter 10 - 10.2 - Operations with Matrices - 10.2 Exercises - Page 726: 69

Part A: $\begin{bmatrix} 1 & -2 & 3\\ -1 & 3 & -1\\ 2 & -5 & 5\\ \end{bmatrix}$$\begin{bmatrix} x_{1}\\ x_{2}\\ x_{3}\\ \end{bmatrix}$ = $\begin{bmatrix} 9\\ -6\\ 17\\ \end{bmatrix}$ Part B: $\begin{bmatrix} x_{1}\\ x_{2}\\ x_{3}\\ \end{bmatrix}$ = $\begin{bmatrix} 1\\ -1\\ 2\\ \end{bmatrix}$

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