91Ó°ÊÓ

First, we will write the given system of equations as an augmented matrix: \[ \left[ \begin{array}{ccc|c} 2 & 3 & -6 & -11 \\ 1 & -2 & 3 & 9 \\ 3 & 1 & 0 & 7 \end{array} \right] \]

Our goal is to create a diagonal matrix through applying row operations. We'll start in the first column, working down the rows. In the end, we'll have a matrix in reduced row-echelon form, from which we can read the solution. First, we will swap rows 1 and 2 to get a 1 in the first row and first column: \[ \left[ \begin{array}{ccc|c} 1 & -2 & 3 & 9 \\ 2 & 3 & -6 & -11 \\ 3 & 1 & 0 & 7 \end{array} \right] \] Now, let's eliminate the first column in rows 2 and 3. We can do this by subtracting 2 times the first row from the second row and subtracting 3 times the first row from the third row: \[ \left[ \begin{array}{ccc|c} 1 & -2 & 3 & 9 \\ 0 & 7 & -12 & -29 \\ 0 & 5 & -9 & -20 \end{array} \right] \] Next, divide the second row by 7 to get a 1 in the second row and second column: \[ \left[ \begin{array}{ccc|c} 1 & -2 & 3 & 9 \\ 0 & 1 & \(-\frac{12}{7}\) & \(-\frac{29}{7}\) \\ 0 & 5 & -9 & -20 \end{array} \right] \] Now, let's eliminate the second column in row 3. We can do this by subtracting 5 times the second row from the third row: \[ \left[ \begin{array}{ccc|c} 1 & -2 & 3 & 9 \\ 0 & 1 & \(-\frac{12}{7}\) & \(-\frac{29}{7}\) \\ 0 & 0 & \(\frac{3}{7}\) & \(\frac{15}{7}\) \end{array} \right] \] Finally, divide the third row by \(\frac{3}{7}\) to get a 1 in the third row and third column: \[ \left[ \begin{array}{ccc|c} 1 & -2 & 3 & 9 \\ 0 & 1 & \(-\frac{12}{7}\) & \(-\frac{29}{7}\) \\ 0 & 0 & 1 & 5 \end{array} \right] \]

Now that our matrix is in reduced row-echelon form, we can read the solution as (x, y, z) from the last column: From the third row: \(z = 5\) From the second row: \(y - \frac{12}{7}z = -\frac{29}{7}\) ⟹ \(y = 3\) From the first row: \(x - 2y + 3z = 9\) ⟹ \(x = 4\) Thus, the solution to the system of equations is \((x, y, z) = (4, 3, 5)\).