## Elementary Linear Algebra 7th Edition

$x_{1} = -t$ $x_{2} = 0$ $x_{3} = t$ $t$ is any real number.
From Equation 2, we can conclude that $x_2 = 0$. Substituting this value into Equation 1 results in the following. $x_{1} + 0x_{2} + x_{3} = 0$ $0x_{1} + x_2 = 0$. We can see that there are 2 pivots and that $x_3$ is a free variable. Therefore, $x_{3} = t$ where $t$ is any real number. We can then substitute this value into Equation 1 and then solve for $x_1$. $x_1 = -t$ $x_2 = 0$ $x_3 = t$