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

The procedure is shown with the matrix notation for simpler understanding. $x_{1} + 3x_{3} = 2$ $x_{2} - 3x_{4} = 3$ $-2x_{2} + 3x_{3} + 2x_{4} = 1$ $3x_{1} + 7x_{4} = -5$ This can be depicted in the augmented matrix notation as follows : $\begin{bmatrix} 1 &amp; 0 &amp; 3 &amp; 0 &amp; 2\\ 0 &amp; 1 &amp; 0 &amp; -3 &amp; 3\\ 0 &amp; -2 &amp; 3 &amp; 2 &amp; 1\\ 3 &amp; 0 &amp; 0 &amp; 7 &amp; -5 \end{bmatrix}$ To get an $x_{1}$ term in the second row, interchange row 2 and row 4: $\begin{bmatrix} 1 &amp; 0 &amp; 3 &amp; 0 &amp; 2\\ 3 &amp; 0 &amp; 0 &amp; 7 &amp; -5\\ 0 &amp; 1 &amp; 0 &amp; -3 &amp; 3\\ 0 &amp; -2 &amp; 3 &amp; 2 &amp; 1 \end{bmatrix}$ To eliminate the $3x_{1}$ term in the second row, add $-3$ times row 1 to row 2: $\begin{bmatrix} 1 &amp; 0 &amp; 3 &amp; 0 &amp; 2\\ 0 &amp; 0 &amp; -9 &amp; 7 &amp; -11\\ 0 &amp; 1 &amp; 0 &amp; -3 &amp; 3\\ 0 &amp; -2 &amp; 3 &amp; 2 &amp; 1 \end{bmatrix}$ To get an $x_{2}$ term in the second row, interchange row 2 and row 3: $\begin{bmatrix} 1 &amp; 0 &amp; 3 &amp; 0 &amp; 2\\ 0 &amp; 1 &amp; 0 &amp; -3 &amp; 3\\ 0 &amp; 0 &amp; -9 &amp; 7 &amp; -11\\ 0 &amp; -2 &amp; 3 &amp; 2 &amp; 1 \end{bmatrix}$ To get an $x_{2}$ term in the third row, interchange row 3 and row 4: $\begin{bmatrix} 1 &amp; 0 &amp; 3 &amp; 0 &amp; 2\\ 0 &amp; 1 &amp; 0 &amp; -3 &amp; 3\\ 0 &amp; -2 &amp; 3 &amp; 2 &amp; 1\\ 0 &amp; 0 &amp; -9 &amp; 7 &amp; -11 \end{bmatrix}$ To eliminate the $-2x_{2}$ term in the third row, add $2$ times row 2 to row 3: $\begin{bmatrix} 1 &amp; 0 &amp; 3 &amp; 0 &amp; 2\\ 0 &amp; 1 &amp; 0 &amp; -3 &amp; 3\\ 0 &amp; 0 &amp; 3 &amp; -4 &amp; 7\\ 0 &amp; 0 &amp; -9 &amp; 7 &amp; -11 \end{bmatrix}$ To eliminate the $-9x_{3}$ term in the fourth row, add $3$ times row 3 to row 4: $\begin{bmatrix} 1 &amp; 0 &amp; 3 &amp; 0 &amp; 2\\ 0 &amp; 1 &amp; 0 &amp; -3 &amp; 3\\ 0 &amp; 0 &amp; 3 &amp; -4 &amp; 7\\ 0 &amp; 0 &amp; 0 &amp; -5 &amp; 10 \end{bmatrix}$ In the equation form, $x_{1} + 3x_{3} = 2$ $x_{2} - 3x_{4} = 3$ $3x_{3} - 4x_{4} = 7$ $-5x_{4} = 10$ Now, we know the value of $x_{4}$. Back substituting this value into the other equations, we can subsequently find the values of $x_{3}$, $x_{2}$ and $x_{1}$. Hence, a solution exists. Thus, the system is consistent.