1. The problem asks to put a system of equations into matrix form and solve parts a), b), c), and d). Since the exact system is not provided, I will demonstrate the general process of converting a system to matrix form and solving it step-by-step. 2. **Matrix form of a system:** A system of linear equations can be written as $$AX = B$$ where $$A$$ is the coefficient matrix, $$X$$ is the column vector of variables, and $$B$$ is the constants vector. 3. **Example system:** Suppose the system is: $$\begin{cases} 2x + 3y = 5 \\ 4x - y = 1 \end{cases}$$ 4. **Matrix form:** $$A = \begin{bmatrix} 2 & 3 \\ 4 & -1 \end{bmatrix}, \quad X = \begin{bmatrix} x \\ y \end{bmatrix}, \quad B = \begin{bmatrix} 5 \\ 1 \end{bmatrix}$$ 5. **a) Write the matrix equation:** $$AX = B$$ 6. **b) Find the determinant of $$A$$:** $$\det(A) = (2)(-1) - (3)(4) = -2 - 12 = -14$$ 7. **c) Find the inverse of $$A$$ (if $$\det(A) \neq 0$$):** $$A^{-1} = \frac{1}{\det(A)} \begin{bmatrix} -1 & -3 \\ -4 & 2 \end{bmatrix} = \frac{1}{-14} \begin{bmatrix} -1 & -3 \\ -4 & 2 \end{bmatrix} = \begin{bmatrix} \frac{1}{14} & \frac{3}{14} \\ \frac{2}{7} & -\frac{1}{7} \end{bmatrix}$$ 8. **d) Solve for $$X$$:** $$X = A^{-1}B = \begin{bmatrix} \frac{1}{14} & \frac{3}{14} \\ \frac{2}{7} & -\frac{1}{7} \end{bmatrix} \begin{bmatrix} 5 \\ 1 \end{bmatrix} = \begin{bmatrix} \frac{1}{14} \times 5 + \frac{3}{14} \times 1 \\ \frac{2}{7} \times 5 - \frac{1}{7} \times 1 \end{bmatrix} = \begin{bmatrix} \frac{5}{14} + \frac{3}{14} \\ \frac{10}{7} - \frac{1}{7} \end{bmatrix} = \begin{bmatrix} \frac{8}{14} \\ \frac{9}{7} \end{bmatrix} = \begin{bmatrix} \frac{4}{7} \\ \frac{9}{7} \end{bmatrix}$$ 9. **Final answer:** $$x = \frac{4}{7}, \quad y = \frac{9}{7}$$ This completes the matrix form and solution steps for the example system.