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

$\begin{cases}x_1=-9 \\ x_2=4 \\ x_3 \text{ is free} \end{cases}$
To find the general solution, we use row operations on the augmented matrix $\left[ \begin{array}{cccc} 1 & 4 & 0 & 7\\ 2 & 7 & 0 & 10 \end{array} \right]$ to obtain an equivalent matrix in reduced row echelon form. Then we solve the corresponding system of equations. First, we want to eliminate the 2 in the second row since it is below the leading term of the first row. To do this, we use row-replacement. So we replace row 2 with -2*(row 1)+(row 2), which gives $$\left[ \begin{array}{cccc} 1 & 4 & 0 & 7\\ 0 & -1 & 0 & -4 \end{array} \right].$$ Next, we need to eliminate the 4 above the leading one of the second row. We will again use row-replacement. We replace row 1 with 4*(row 2) + (row 1) to get $$\left[ \begin{array}{cccc} 1 & 0 & 0 & -9\\ 0 & -1 & 0 & -4 \end{array} \right].$$ Thirdly, we multiply row 2 by -1 to obtain the equivalent matrix $$\left[ \begin{array}{cccc} 1 & 0 & 0 & -9\\ 0 & 1 & 0 & 4 \end{array} \right].$$ Now we solve the corresponding system of equations, which is $$x_1=-9\\ x_2=4.$$ From the reduced row echelon form of the matrix, we see that $x_1$ and $x_2$ are the leading variables (since they correspond to the pivot positions of the matrix) and $x_3$ is free. Our system is already solved for the leading variables, and we know $x_3$ is free. So our general solution is $$\begin{cases}x_1=-9 \\ x_2=4 \\ x_3 \text{ is free} \end{cases}.$$