1. **State the problem:** We are given a system of linear equations: $$2x - 3y = 7$$ $$3x + y = 5$$ and a 3x3 matrix: $$\begin{bmatrix} 2 & -3 & 6 \\ 2 & -1 & 2 \\ -10 & 5 & 2 \end{bmatrix}$$ We need to solve the system and evaluate the determinant of the matrix. 2. **Solve the system of equations:** The system is: $$2x - 3y = 7 \quad (1)$$ $$3x + y = 5 \quad (2)$$ We can use substitution or elimination. Let's use substitution. From equation (2): $$y = 5 - 3x$$ Substitute into equation (1): $$2x - 3(5 - 3x) = 7$$ $$2x - 15 + 9x = 7$$ $$11x - 15 = 7$$ $$11x = 7 + 15$$ $$11x = 22$$ $$x = \frac{22}{11}$$ $$x = 2$$ Substitute $x=2$ back into $y = 5 - 3x$: $$y = 5 - 3(2)$$ $$y = 5 - 6$$ $$y = -1$$ 3. **Evaluate the determinant of the matrix:** The matrix is: $$A = \begin{bmatrix} 2 & -3 & 6 \\ 2 & -1 & 2 \\ -10 & 5 & 2 \end{bmatrix}$$ The determinant of a 3x3 matrix $A = \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix}$ is: $$\det(A) = a(ei - fh) - b(di - fg) + c(dh - eg)$$ Calculate each term: $$a = 2, b = -3, c = 6$$ $$d = 2, e = -1, f = 2$$ $$g = -10, h = 5, i = 2$$ Calculate: $$ei - fh = (-1)(2) - (2)(5) = -2 - 10 = -12$$ $$di - fg = (2)(2) - (2)(-10) = 4 + 20 = 24$$ $$dh - eg = (2)(5) - (-1)(-10) = 10 - 10 = 0$$ Now: $$\det(A) = 2(-12) - (-3)(24) + 6(0) = -24 + 72 + 0 = 48$$ **Final answers:** - Solution to the system: $x=2$, $y=-1$ - Determinant of the matrix: $48$