91Ó°ÊÓ

To transform the system of linear equations into the matrix \(A X = B\), we first must identify the coefficients of the variables on the left-hand side of the equations. These will make up matrix \(A\). The variables \(x_1\) and \(x_2\) will constitute the column matrix \(X\), and the constants on the right-hand side of the equations will form column matrix \(B\). Thus, we get: \[A=\begin{bmatrix} 2 & 3 \\ 1 & 4 \end{bmatrix}, X=\begin{bmatrix} x_1 \\ x_2 \end{bmatrix}, B=\begin{bmatrix} 5 \\ 10 \end{bmatrix}\] Hence, the matrix equation \(A X= B\) becomes: \[\begin{bmatrix} 2 & 3 \\ 1 & 4 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} 5 \\ 10 \end{bmatrix}\]

We can use Gauss-Jordan elimination to solve this system. This method involves forming an augmented matrix \(A | B\) and then applying row operations to reach a final form where the solution becomes evident. The augmented matrix for our system is: \[\begin{bmatrix} 2 & 3 & 5 \\ 1 & 4 & 10 \end{bmatrix}\] The main goal of Gauss-Jordan is to create a matrix in reduced row echelon form. We'll start by dividing the first row by 2 for simplicity: \[\begin{bmatrix} 1 & 1.5 & 2.5 \\ 1 & 4 & 10 \end{bmatrix}\] and next, we subtract the first row from the second to eliminate \(x_1\) from the second equation: \[\begin{bmatrix} 1 & 1.5 & 2.5 \\ 0 & 2.5 & 7.5 \end{bmatrix}\] Dividing the second row by 2.5, we yield: \[\begin{bmatrix} 1 & 1.5 & 2.5 \\ 0 & 1 & 3 \end{bmatrix}\] Finally, to eliminate \(x_2\) from the first equation, we subtract 1.5 times the second row from the first: \[\begin{bmatrix} 1 & 0 & -2 \\ 0 & 1 & 3 \end{bmatrix}\] From this we can read off the solutions.

From the last matrix, reading off the solutions, we get \(x_1 = -2\) and \(x_2 = 3\) correspondingly, forming the solution set for the original system of equations.