91Ó°ÊÓ

Given the system: $$ \begin{aligned} & 2x+3y=5 \\ & 3x+2y=5 \end{aligned} $$ First, we will write the augmented matrix for this system. The augmented matrix is a matrix that consists of the coefficients of the variables and the constants, separated by a vertical line. The matrix for the given system is: $$ \left[\begin{array}{cc|c} 2 & 3 & 5 \\ 3 & 2 & 5 \end{array}\right] $$

Now, we will perform elementary row operations to simplify the matrix to its reduced row-echelon form (RREF). In RREF, the main diagonal elements (the elements from the top-left corner to the bottom-right corner) should be 1, and all elements below the main diagonal should be 0. First, swap row 1 and row 2 to have a leading 1 in the first row: $$ \left[\begin{array}{cc|c} 3 & 2 & 5 \\ 2 & 3 & 5 \end{array}\right] $$ Divide row 1 by 3: $$ \left[\begin{array}{cc|c} 1 & \frac{2}{3} & \frac{5}{3} \\ 2 & 3 & 5 \end{array}\right] $$ Next, replace the second row with the second row minus twice the first row: \((2,3,5) - 2\cdot(1,\frac{2}{3},\frac{5}{3}) \rightarrow (0,2,0)\) $$ \left[\begin{array}{cc|c} 1 & \frac{2}{3} & \frac{5}{3} \\ 0 & 2 & 0 \end{array}\right] $$ Now, divide row 2 by 2: $$ \left[\begin{array}{cc|c} 1 & \frac{2}{3} & \frac{5}{3} \\ 0 & 1 & 0 \end{array}\right] $$ Replace the first row with the first row minus \(\frac{2}{3}\) times the second row: \((1,\frac{2}{3},\frac{5}{3}) - \frac{2}{3}\cdot(0,1,0) \rightarrow (1,0,2)\) $$ \left[\begin{array}{cc|c} 1 & 0 & 2 \\ 0 & 1 & 0 \end{array}\right] $$

The matrix is now in RREF. We can read the solution directly from the matrix: $$ \begin{aligned} x &= 2 \\ y &= 0 \end{aligned} $$ So the solution to the given system of equations is \(x=2\) and \(y=0\).