91Ó°ÊÓ

Write the given system of linear equations as an augmented matrix. The augmented matrix is a 3x3 matrix, where the first two columns represent the coefficients of x and y, and the third column represents the constant terms from the equations. \[ \left[\begin{array}{cc|c} 3 & -2 & -3\\ 2 & 1 & 3\\ 1 & -2 & -5 \end{array} \right] \]

Next, perform the row operations to transform the matrix into its reduced row echelon form. This includes using row swaps, row multiplication, and row addition operations. We will focus on three steps: making a leading 1 or pivot in each row, eliminating all non-zero elements below each pivot, and then eliminating non-zero elements above each pivot. 1. Swap row 1 and row 3 so that the first row has a leading 1: \[ \left[\begin{array}{cc|c} 1 & -2 & -5\\ 2 & 1 & 3\\ 3 & -2 & -3 \end{array} \right] \] 2. Eliminate all non-zero elements below the first pivot: - Subtract 2 times row 1 from row 2: \[ \left[\begin{array}{cc|c} 1 & -2 & -5\\ 0 & 5 & 13\\ 3 & -2 & -3 \end{array} \right] \] - Subtract 3 times row 1 from row 3: \[ \left[\begin{array}{cc|c} 1 & -2 & -5\\ 0 & 5 & 13\\ 0 & 4 & 12 \end{array} \right] \] 3. Create a leading 1 (pivot) in row 2 by dividing it by 5: \[ \left[\begin{array}{cc|c} 1 & -2 & -5\\ 0 & 1 & \frac{13}{5}\\ 0 & 4 & 12 \end{array} \right] \] 4. Eliminate all non-zero elements below the second pivot: - Subtract 4 times row 2 from row 3: \[ \left[\begin{array}{cc|c} 1 & -2 & -5\\ 0 & 1 & \frac{13}{5}\\ 0 & 0 & -\frac{4}{5} \end{array} \right] \] 5. Create a leading 1 (pivot) in row 3 by multiplying it by -5/4: \[ \left[\begin{array}{cc|c} 1 & -2 & -5\\ 0 & 1 & \frac{13}{5}\\ 0 & 0 & 1 \end{array} \right] \] 6. Eliminate all non-zero elements above the third pivot: - Add 5 times row 3 to row 1: \[ \left[\begin{array}{cc|c} 1 & -2 & 0\\ 0 & 1 & \frac{13}{5}\\ 0 & 0 & 1 \end{array} \right] \] 7. Eliminate all non-zero elements above the second pivot: - Add 2 times row 2 to row 1: \[ \left[\begin{array}{cc|c} 1 & 0 & \frac{26}{5}\\ 0 & 1 & \frac{13}{5}\\ 0 & 0 & 1 \end{array} \right] \] Now, the matrix is in reduced row echelon form.

Finally, convert the reduced row echelon form matrix back into a system of linear equations to find the solution: \(\begin{aligned} x &= \frac{26}{5} \\ y &=\frac{13}{5}\\ 1&=1 \end{aligned}\) Since the last equation (1 = 1) is an identity, it means the system has a unique solution: \(\begin{aligned} x &= \frac{26}{5} \\ y &=\frac{13}{5} \end{aligned}\)