1. Let's state the problem: Solve a system of linear equations using the matrix method. 2. Write the system of equations in matrix form as $AX = B$, where $A$ is the coefficient matrix, $X$ is the variable matrix, and $B$ is the constants matrix. 3. Compute the inverse of matrix $A$, denoted as $A^{-1}$, if it exists. 4. Multiply both sides of the equation by $A^{-1}$ to get $X = A^{-1}B$. 5. Calculate the product $A^{-1}B$ to find the values of the variables in $X$. Example: Solve the system $$\begin{cases} 2x + 3y = 5 \\ 4x - y = 1 \end{cases}$$ Step 1: Write matrices $$A=\begin{bmatrix} 2 & 3 \\ 4 & -1 \end{bmatrix}, X=\begin{bmatrix} x \\ y \end{bmatrix}, B=\begin{bmatrix} 5 \\ 1 \end{bmatrix}$$ Step 2: Calculate determinant of $A$: $$\det(A) = (2)(-1) - (4)(3) = -2 - 12 = -14\neq 0$$ Step 3: Find inverse of $A$: $$A^{-1} = \frac{1}{-14}\begin{bmatrix} -1 & -3 \\ -4 & 2 \end{bmatrix} = \begin{bmatrix} \frac{1}{14} & \frac{3}{14} \\ \frac{2}{7} & -\frac{1}{7} \end{bmatrix}$$ Step 4: Multiply $A^{-1} B$: $$X = \begin{bmatrix} \frac{1}{14} & \frac{3}{14} \\ \frac{2}{7} & -\frac{1}{7} \end{bmatrix} \begin{bmatrix} 5 \\ 1 \end{bmatrix} = \begin{bmatrix} \frac{5}{14} + \frac{3}{14} \\ \frac{10}{7} - \frac{1}{7} \end{bmatrix} = \begin{bmatrix} \frac{8}{14} \\ \frac{9}{7} \end{bmatrix} = \begin{bmatrix} \frac{4}{7} \\ \frac{9}{7} \end{bmatrix}$$ Final answer: $$x=\frac{4}{7},\quad y=\frac{9}{7}.$$