1. **Problem a:** Solve the simultaneous equations using matrix method: $$\begin{cases} 2x + 5y = 13 \\ 2x - 3y = 5 \end{cases}$$ 2. Write the system in matrix form $AX = B$ where $$A = \begin{bmatrix} 2 & 5 \\ 2 & -3 \end{bmatrix}, \quad X = \begin{bmatrix} x \\ y \end{bmatrix}, \quad B = \begin{bmatrix} 13 \\ 5 \end{bmatrix}$$ 3. Find the determinant of $A$: $$|A| = (2)(-3) - (2)(5) = -6 - 10 = -16$$ Since $|A| \neq 0$, $A$ is invertible. 4. Find the inverse of $A$: $$A^{-1} = \frac{1}{|A|} \begin{bmatrix} -3 & -5 \\ -2 & 2 \end{bmatrix} = \frac{1}{-16} \begin{bmatrix} -3 & -5 \\ -2 & 2 \end{bmatrix} = \begin{bmatrix} \frac{3}{16} & \frac{5}{16} \\ \frac{1}{8} & -\frac{1}{8} \end{bmatrix}$$ 5. Multiply $A^{-1}$ by $B$ to find $X$: $$X = A^{-1}B = \begin{bmatrix} \frac{3}{16} & \frac{5}{16} \\ \frac{1}{8} & -\frac{1}{8} \end{bmatrix} \begin{bmatrix} 13 \\ 5 \end{bmatrix} = \begin{bmatrix} \frac{3}{16} \times 13 + \frac{5}{16} \times 5 \\ \frac{1}{8} \times 13 - \frac{1}{8} \times 5 \end{bmatrix} = \begin{bmatrix} \frac{39}{16} + \frac{25}{16} \\ \frac{13}{8} - \frac{5}{8} \end{bmatrix} = \begin{bmatrix} \frac{64}{16} \\ \frac{8}{8} \end{bmatrix} = \begin{bmatrix} 4 \\ 1 \end{bmatrix}$$ 6. **Answer for a:** $x=4$, $y=1$ --- 7. **Problem b:** Find the determinant of matrix $$M = \begin{bmatrix} 2 & 3 & -6 \\ 0 & 5 & -1 \\ 1 & -2 & 6 \end{bmatrix}$$ 8. Use cofactor expansion along the first row: $$|M| = 2 \times \begin{vmatrix} 5 & -1 \\ -2 & 6 \end{vmatrix} - 3 \times \begin{vmatrix} 0 & -1 \\ 1 & 6 \end{vmatrix} + (-6) \times \begin{vmatrix} 0 & 5 \\ 1 & -2 \end{vmatrix}$$ 9. Calculate each minor: $$\begin{vmatrix} 5 & -1 \\ -2 & 6 \end{vmatrix} = (5)(6) - (-1)(-2) = 30 - 2 = 28$$ $$\begin{vmatrix} 0 & -1 \\ 1 & 6 \end{vmatrix} = (0)(6) - (-1)(1) = 0 + 1 = 1$$ $$\begin{vmatrix} 0 & 5 \\ 1 & -2 \end{vmatrix} = (0)(-2) - (5)(1) = 0 - 5 = -5$$ 10. Substitute back: $$|M| = 2 \times 28 - 3 \times 1 + (-6) \times (-5) = 56 - 3 + 30 = 83$$ 11. **Answer for b:** $|M| = 83$ --- 12. **Problem c:** Given $$A = \begin{bmatrix} x & 4 \\ 3x & 2x \end{bmatrix}, \quad |A| = -16$$ 13. Calculate determinant: $$|A| = x \times 2x - 4 \times 3x = 2x^2 - 12x$$ 14. Set equal to $-16$: $$2x^2 - 12x = -16$$ 15. Rearrange: $$2x^2 - 12x + 16 = 0$$ 16. Divide entire equation by 2: $$\cancel{2}x^2 - \cancel{2} \times 6x + \cancel{2} \times 8 = 0 \Rightarrow x^2 - 6x + 8 = 0$$ 17. Factor quadratic: $$x^2 - 6x + 8 = (x - 2)(x - 4) = 0$$ 18. Solve for $x$: $$x - 2 = 0 \Rightarrow x = 2$$ $$x - 4 = 0 \Rightarrow x = 4$$ 19. **Answer for c:** $x = 2$ or $x = 4$