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

The procedure is shown with the matrix notation for simpler understanding. $x_{1} - 2x_{4} = -3$ $2x_{2} + 2x_{3} = 0$ $x_{3} + 3x_{4} = 1$ $-2x_{1} + 3x_{2} + 2x_{3} + x_{4} = 5$ This can be depicted in the augmented matrix notation as follows : $\begin{bmatrix} 1 &amp; 0 &amp; 0 &amp; -2 &amp; -3\\ 0 &amp; 2 &amp; 2 &amp; 0 &amp; 0\\ 0 &amp; 0 &amp; 1&amp; 3 &amp; 1\\ -2 &amp; 3 &amp; 2 &amp; 1 &amp; 5 \end{bmatrix}$ To eliminate the $-2x_{1}$ term in the fourth row, add $2$ times row 1 to row 4: $\begin{bmatrix} 1 &amp; 0 &amp; 0 &amp; -2 &amp; -3\\ 0 &amp; 2 &amp; 2 &amp; 0 &amp; 0\\ 0 &amp; 0 &amp; 1&amp; 3 &amp; 1\\ 0 &amp; 3 &amp; 2 &amp; -3 &amp; -1 \end{bmatrix}$ To eliminate the $3x_{2}$ term in the fourth row, add $-3/2$ times row 2 to row 4: $\begin{bmatrix} 1 &amp; 0 &amp; 0 &amp; -2 &amp; -3\\ 0 &amp; 2 &amp; 2 &amp; 0 &amp; 0\\ 0 &amp; 0 &amp; 1&amp; 3 &amp; 1\\ 0 &amp; 0 &amp; -1 &amp; -3 &amp; -1 \end{bmatrix}$ To eliminate the $-x_{3}$ term in the fourth row, add $1$ times row 3 to row 4: $\begin{bmatrix} 1 &amp; 0 &amp; 0 &amp; -2 &amp; -3\\ 0 &amp; 2 &amp; 2 &amp; 0 &amp; 0\\ 0 &amp; 0 &amp; 1&amp; 3 &amp; 1\\ 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 \end{bmatrix}$ $x_{1} - 2x_{4} = -3$ $2x_{2} + 2x_{3} = 0$ $x_{3} + 3x_{4} = 1$ $0x_{4} = 0$ Now, we know that the last equation is true for infinite values of $x_{4}$. Back substituting these values into the other equations, we can find that there will be infinite subsequent values of $x_{3}$, $x_{2}$ and $x_{1}$ for each value of $x_{4}$. Hence, infinitely many solutions exist. Thus, the system is consistent.