## Linear Algebra and Its Applications (5th Edition)

$x_{1} = 5$ $x_{2} = 3$ $x_{3} = -1$
The procedure is shown with the matrix notation for simpler understanding. $x_{1} - 3x_{3} = 8$ $2x_{1} + 2x_{2} + 9x_{3} = 7$ $x_{2} + 5x_{3} = -2$ This can be depicted in the augmented matrix notation as follows : $\begin{bmatrix} 1 & 0 & -3 & 8 \\ 2 & 2 & 9 & 7 \\ 0 & 1 & 5 & -2 \end{bmatrix}$ To eliminate the $2x_{1}$ term in the second equation, add $-2$ times row 1 to row 2: $\begin{bmatrix} 1 & 0 & -3 & 8 \\ 0 & 2 & 15 & -9 \\ 0 & 1 & 5 & -2 \end{bmatrix}$ To get an $x_{2}$ term in the second row, interchange row 2 and row 3: $\begin{bmatrix} 1 & 0 & -3 & 8 \\ 0 & 1 & 5 & -2\\ 0 & 2 & 15 & -9 \end{bmatrix}$ Next we use the $x_{2}$ term in the second equation to eliminate the $2x_{2}$ term from the third equation. Add $-2$ times row 2 to row 3: $\begin{bmatrix} 1 & 0 & -3 & 8 \\ 0 & 1 & 5 & -2\\ 0 & 0 & 5 & -5 \end{bmatrix}$ Now the augmented matrix is in a triangular form. For interpreting it, we go back to the equation notation: $x_{1} - 3x_{3} = 8$ $x_{2} + 5x_{3} = -2$ $5x_{3} = -5$ From $5x_{3} = -5$, we get $x_{3} = -1$ Using this value in $x_{2} + 5x_{3} = -2$, we get: $x_{2} + 5(-1) = -2$ or, $x_{2} = 3$ Using the value of $x_{3}$ in the equation $x_{1} - 3x_{3} = 8$, $x_{1} - 3(-1) = 8$ or, $x_{1} = 5$ Hence, $x_{1} = 5$ ; $x_{2} = 3$ ; $x_{3} = -1$ .