1. **State the problem:** We have matrix $$A = \begin{bmatrix} 2 & 6 & 23 \\ -5 & -12 & -45 \\ 5 & 15 & 60 \end{bmatrix}$$ and the system: $$\begin{cases} 2x_1 + 6x_2 + 23x_3 = -4 \\ -5x_1 - 12x_2 - 45x_3 = 2 \\ 5x_1 + 15x_2 + 60x_3 = -3 \end{cases}$$ We need to (a) compute $$\det(A)$$ and (b) solve for $$x_1, x_2, x_3$$ using Cramer's rule. 2. **Compute $$\det(A)$$:** Using cofactor expansion along the first row: $$\det(A) = 2 \cdot \det\begin{bmatrix} -12 & -45 \\ 15 & 60 \end{bmatrix} - 6 \cdot \det\begin{bmatrix} -5 & -45 \\ 5 & 60 \end{bmatrix} + 23 \cdot \det\begin{bmatrix} -5 & -12 \\ 5 & 15 \end{bmatrix}$$ Calculate each minor: $$\det\begin{bmatrix} -12 & -45 \\ 15 & 60 \end{bmatrix} = (-12)(60) - (-45)(15) = -720 + 675 = -45$$ $$\det\begin{bmatrix} -5 & -45 \\ 5 & 60 \end{bmatrix} = (-5)(60) - (-45)(5) = -300 + 225 = -75$$ $$\det\begin{bmatrix} -5 & -12 \\ 5 & 15 \end{bmatrix} = (-5)(15) - (-12)(5) = -75 + 60 = -15$$ Substitute back: $$\det(A) = 2(-45) - 6(-75) + 23(-15) = -90 + 450 - 345 = 15$$ 3. **Apply Cramer's rule:** Cramer's rule states: $$x_i = \frac{\det(A_i)}{\det(A)}$$ where $$A_i$$ is matrix $$A$$ with the $$i$$-th column replaced by the constants vector $$b = \begin{bmatrix} -4 \\ 2 \\ -3 \end{bmatrix}$$. - For $$x_1$$, replace first column: $$A_1 = \begin{bmatrix} -4 & 6 & 23 \\ 2 & -12 & -45 \\ -3 & 15 & 60 \end{bmatrix}$$ Calculate $$\det(A_1)$$: $$= -4 \cdot \det\begin{bmatrix} -12 & -45 \\ 15 & 60 \end{bmatrix} - 6 \cdot \det\begin{bmatrix} 2 & -45 \\ -3 & 60 \end{bmatrix} + 23 \cdot \det\begin{bmatrix} 2 & -12 \\ -3 & 15 \end{bmatrix}$$ Calculate minors: $$\det\begin{bmatrix} -12 & -45 \\ 15 & 60 \end{bmatrix} = -45$$ (from above) $$\det\begin{bmatrix} 2 & -45 \\ -3 & 60 \end{bmatrix} = 2 \cdot 60 - (-45)(-3) = 120 - 135 = -15$$ $$\det\begin{bmatrix} 2 & -12 \\ -3 & 15 \end{bmatrix} = 2 \cdot 15 - (-12)(-3) = 30 - 36 = -6$$ Substitute: $$\det(A_1) = -4(-45) - 6(-15) + 23(-6) = 180 + 90 - 138 = 132$$ - For $$x_2$$, replace second column: $$A_2 = \begin{bmatrix} 2 & -4 & 23 \\ -5 & 2 & -45 \\ 5 & -3 & 60 \end{bmatrix}$$ Calculate $$\det(A_2)$$: $$= 2 \cdot \det\begin{bmatrix} 2 & -45 \\ -3 & 60 \end{bmatrix} - (-4) \cdot \det\begin{bmatrix} -5 & -45 \\ 5 & 60 \end{bmatrix} + 23 \cdot \det\begin{bmatrix} -5 & 2 \\ 5 & -3 \end{bmatrix}$$ Calculate minors: $$\det\begin{bmatrix} 2 & -45 \\ -3 & 60 \end{bmatrix} = -15$$ (from above) $$\det\begin{bmatrix} -5 & -45 \\ 5 & 60 \end{bmatrix} = -75$$ (from above) $$\det\begin{bmatrix} -5 & 2 \\ 5 & -3 \end{bmatrix} = (-5)(-3) - 2(5) = 15 - 10 = 5$$ Substitute: $$\det(A_2) = 2(-15) - (-4)(-75) + 23(5) = -30 - 300 + 115 = -215$$ - For $$x_3$$, replace third column: $$A_3 = \begin{bmatrix} 2 & 6 & -4 \\ -5 & -12 & 2 \\ 5 & 15 & -3 \end{bmatrix}$$ Calculate $$\det(A_3)$$: $$= 2 \cdot \det\begin{bmatrix} -12 & 2 \\ 15 & -3 \end{bmatrix} - 6 \cdot \det\begin{bmatrix} -5 & 2 \\ 5 & -3 \end{bmatrix} + (-4) \cdot \det\begin{bmatrix} -5 & -12 \\ 5 & 15 \end{bmatrix}$$ Calculate minors: $$\det\begin{bmatrix} -12 & 2 \\ 15 & -3 \end{bmatrix} = (-12)(-3) - 2(15) = 36 - 30 = 6$$ $$\det\begin{bmatrix} -5 & 2 \\ 5 & -3 \end{bmatrix} = 5$$ (from above) $$\det\begin{bmatrix} -5 & -12 \\ 5 & 15 \end{bmatrix} = -15$$ (from above) Substitute: $$\det(A_3) = 2(6) - 6(5) + (-4)(-15) = 12 - 30 + 60 = 42$$ 4. **Calculate solutions:** $$x_1 = \frac{\det(A_1)}{\det(A)} = \frac{132}{15} = \frac{44}{5} = 8.8$$ $$x_2 = \frac{\det(A_2)}{\det(A)} = \frac{-215}{15} = -\frac{43}{3} \approx -14.3333$$ $$x_3 = \frac{\det(A_3)}{\det(A)} = \frac{42}{15} = \frac{14}{5} = 2.8$$ **Final answers:** $$\det(A) = 15$$ $$x_1 = \frac{44}{5}, \quad x_2 = -\frac{43}{3}, \quad x_3 = \frac{14}{5}$$