91Ó°ÊÓ

First, we express the system of linear equations as an augmented matrix. \( \left[ \begin{array}{cccc|c} 3 & -2 & 1 & 0 & 4\\ 1 & 3 & -4 & 0 & -3\\ 2 & -3 & 5 & 0 & 7\\ 1 & -8 & 9 & 0 & 10 \end{array} \right] \)

To simplify the matrix, we must perform row operations that will eliminate variables in the columns below the diagonal. We will interchange rows, multiply rows by non-zero constants, and add/subtract rows. Step 1: Swap rows 1 and 2 to place '1' in the top left corner. \( \left[ \begin{array}{cccc|c} 1 & 3 & -4 & 0 & -3\\ 3 & -2 & 1 & 0 & 4\\ 2 & -3 & 5 & 0 & 7\\ 1 & -8 & 9 & 0 & 10 \end{array} \right] \) Step 2: Subtract 3 times the first row from the second row. Subtract 2 times the first row from the third row. Subtract the first row from the fourth row. \( \left[ \begin{array}{cccc|c} 1 & 3 & -4 & 0 & -3\\ 0 & -11 & 13 & 0 & 13\\ 0 & -9 & 13 & 0 & 13\\ 0 & -11 & 13 & 0 & 13 \end{array} \right] \) Step 3: Multiply the second row by -1/11. \( \left[ \begin{array}{cccc|c} 1 & 3 & -4 & 0 & -3\\ 0 & 1 & -13/11 & 0 & -13/11\\ 0 & -9 & 13 & 0 & 13\\ 0 & -11 & 13 & 0 & 13 \end{array} \right] \) Step 4: Add 3 times the second row to the first row. Add 9 times the second row to the third row. Add 11 times the second row to the fourth row. \( \left[ \begin{array}{cccc|c} 1 & 0 & 1 & 0 & 0\\ 0 & 1 & -13/11 & 0 & -13/11\\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 \end{array} \right] \)

Now that the matrix is in reduced row-echelon form, we can see that there are infinitely many solutions to the system of linear equations. This is because there is only one non-zero row (corresponding to x and y), leaving the last two variables (z and the one missing from the original system) as free variables. We can rewrite the RREF matrix back as a system of linear equations: \( x + z = 0 \\ y - \frac{13}{11}z = -\frac{13}{11} \\ \) The solution is, with \(t\) as a free variable: \( x = -t \\ y = \frac{13}{11}t - \frac{13}{11} \\ z = t \)