Matrices A and B-1 are given by A = [1, 1, -1; x, 2, 1; x-2, x+1, 0] and B−1 = 1/8 * [1, -1, 1; 2, 1, 0; 0, 3, 1]. Calculate the value of x given that det(A) = 13. Hence find A-1 and (AB)-1.

$$A = \begin{pmatrix} 1 & 2 & 1 \\ x & 1 & x+1 \\ 2 & x-2 & 0 \end{pmatrix}$$

The determinant of A is given by:

$$\det(A) = 1 \begin{vmatrix} 1 & x+1 \\ x-2 & 0 \end{vmatrix} - 2 \begin{vmatrix} x & x+1 \\ 2 & 0 \end{vmatrix} + 1 \begin{vmatrix} x & 1 \\ 2 & x-2 \end{vmatrix}$$

$$\det(A) = 1(1 \cdot 0 - (x+1)(x-2)) - 2(x \cdot 0 - (x+1) \cdot 2) + 1(x(x-2) - 1 \cdot 2)$$ $$\det(A) = -(x^2 - 2x + x - 2) - 2(-2x - 2) + (x^2 - 2x - 2)$$ $$\det(A) = -(x^2 - x - 2) + 4x + 4 + x^2 - 2x - 2$$ $$\det(A) = -x^2 + x + 2 + 4x + 4 + x^2 - 2x - 2$$ $$\det(A) = (-x^2 + x^2) + (x + 4x - 2x) + (2 + 4 - 2)$$ $$\det(A) = 3x + 4$$

Given $\det(A) = 13$:

$$3x + 4 = 13$$ $$3x = 13 - 4$$ $$3x = 9$$ $$x = \boxed{\mathbf{3}}$$

This satisfies the condition $x > 0$.

$$A = \begin{pmatrix} 1 & 2 & 1 \\ 3 & 1 & 4 \\ 2 & 1 & 0 \end{pmatrix}$$

The cofactor matrix $C$ is:

$$C_{11} = \begin{vmatrix} 1 & 4 \\ 1 & 0 \end{vmatrix} = -4 \quad C_{12} = -\begin{vmatrix} 3 & 4 \\ 2 & 0 \end{vmatrix} = 8 \quad C_{13} = \begin{vmatrix} 3 & 1 \\ 2 & 1 \end{vmatrix} = 1$$ $$C_{21} = -\begin{vmatrix} 2 & 1 \\ 1 & 0 \end{vmatrix} = 1 \quad C_{22} = \begin{vmatrix} 1 & 1 \\ 2 & 0 \end{vmatrix} = -2 \quad C_{23} = -\begin{vmatrix} 1 & 2 \\ 2 & 1 \end{vmatrix} = 3$$ $$C_{31} = \begin{vmatrix} 2 & 1 \\ 1 & 4 \end{vmatrix} = 7 \quad C_{32} = -\begin{vmatrix} 1 & 1 \\ 3 & 4 \end{vmatrix} = -1 \quad C_{33} = \begin{vmatrix} 1 & 2 \\ 3 & 1 \end{vmatrix} = -5$$ $$C = \begin{pmatrix} -4 & 8 & 1 \\ 1 & -2 & 3 \\ 7 & -1 & -5 \end{pmatrix}$$

The adjugate matrix is the transpose of the cofactor matrix:

$$\text{adj}(A) = C^T = \begin{pmatrix} -4 & 1 & 7 \\ 8 & -2 & -1 \\ 1 & 3 & -5 \end{pmatrix}$$

Since $\det(A) = 13$, the inverse matrix $A^{-1}$ is:

$$A^{-1} = \frac{1}{\det(A)} \text{adj}(A) = \boxed{\mathbf{\frac{1}{13} \begin{pmatrix} -4 & 1 & 7 \\ 8 & -2 & -1 \\ 1 & 3 & -5 \end{pmatrix}}}$$

Given $B^{-1}$:

$$B^{-1} = \frac{1}{8} \begin{pmatrix} 1 & 2 & -1 \\ 2 & 1 & 0 \\ 0 & 3 & 1 \end{pmatrix}$$ $$(AB)^{-1} = \frac{1}{8} \begin{pmatrix} 1 & 2 & -1 \\ 2 & 1 & 0 \\ 0 & 3 & 1 \end{pmatrix} \cdot \frac{1}{13} \begin{pmatrix} -4 & 1 & 7 \\ 8 & -2 & -1 \\ 1 & 3 & -5 \end{pmatrix}$$ $$(AB)^{-1} = \frac{1}{8 \times 13} \begin{pmatrix} (1)(-4)+(2)(8)+(-1)(1) & (1)(1)+(2)(-2)+(-1)(3) & (1)(7)+(2)(-1)+(-1)(-5) \\ (2)(-4)+(1)(8)+(0)(1) & (2)(1)+(1)(-2)+(0)(3) & (2)(7)+(1)(-1)+(0)(-5) \\ (0)(-4)+(3)(8)+(1)(1) & (0)(1)+(3)(-2)+(1)(3) & (0)(7)+(3)(-1)+(1)(-5) \end{pmatrix}$$ $$(AB)^{-1} = \frac{1}{104} \begin{pmatrix} -4+16-1 & 1-4-3 & 7-2+5 \\ -8+8+0 & 2-2+0 & 14-1+0 \\ 0+24+1 & 0-6+3 & 0-3-5 \end{pmatrix}$$ $$(AB)^{-1} = \boxed{\mathbf{\frac{1}{104} \begin{pmatrix} 11 & -6 & 10 \\ 0 & 0 & 13 \\ 25 & -3 & -8 \end{pmatrix}}}$$