1. **State the problem:** Solve the system of linear equations: $$x+(b-1)y-z=4$$ $$2x+by+4z=4b$$ $$(b+1)x+7y+2z=20$$ 2. **Write the system in matrix form:** $$\begin{pmatrix}1 & b-1 & -1 \\ 2 & b & 4 \\ b+1 & 7 & 2\end{pmatrix} \begin{pmatrix}x \\ y \\ z\end{pmatrix} = \begin{pmatrix}4 \\ 4b \\ 20\end{pmatrix}$$ 3. **Use Cramer's rule or matrix methods to solve for $x$, $y$, and $z$.** 4. **Calculate the determinant $D$ of the coefficient matrix:** $$D = \begin{vmatrix}1 & b-1 & -1 \\ 2 & b & 4 \\ b+1 & 7 & 2\end{vmatrix}$$ Expanding along the first row: $$D = 1 \cdot \begin{vmatrix}b & 4 \\ 7 & 2\end{vmatrix} - (b-1) \cdot \begin{vmatrix}2 & 4 \\ b+1 & 2\end{vmatrix} + (-1) \cdot \begin{vmatrix}2 & b \\ b+1 & 7\end{vmatrix}$$ Calculate each minor: $$\begin{vmatrix}b & 4 \\ 7 & 2\end{vmatrix} = b \cdot 2 - 4 \cdot 7 = 2b - 28$$ $$\begin{vmatrix}2 & 4 \\ b+1 & 2\end{vmatrix} = 2 \cdot 2 - 4(b+1) = 4 - 4b - 4 = -4b$$ $$\begin{vmatrix}2 & b \\ b+1 & 7\end{vmatrix} = 2 \cdot 7 - b(b+1) = 14 - b^2 - b$$ Substitute back: $$D = 1(2b - 28) - (b-1)(-4b) - 1(14 - b^2 - b)$$ $$= 2b - 28 + 4b(b-1) - 14 + b^2 + b$$ $$= 2b - 28 + 4b^2 - 4b - 14 + b^2 + b$$ $$= (4b^2 + b^2) + (2b - 4b + b) + (-28 - 14)$$ $$= 5b^2 - b - 42$$ 5. **Check if $D \neq 0$ to ensure a unique solution.** 6. **Find determinants $D_x$, $D_y$, and $D_z$ by replacing the respective columns with the constants vector:** $$D_x = \begin{vmatrix}4 & b-1 & -1 \\ 4b & b & 4 \\ 20 & 7 & 2\end{vmatrix}$$ Calculate $D_x$: $$D_x = 4 \cdot \begin{vmatrix}b & 4 \\ 7 & 2\end{vmatrix} - (b-1) \cdot \begin{vmatrix}4b & 4 \\ 20 & 2\end{vmatrix} + (-1) \cdot \begin{vmatrix}4b & b \\ 20 & 7\end{vmatrix}$$ Calculate minors: $$\begin{vmatrix}b & 4 \\ 7 & 2\end{vmatrix} = 2b - 28$$ $$\begin{vmatrix}4b & 4 \\ 20 & 2\end{vmatrix} = 4b \cdot 2 - 4 \cdot 20 = 8b - 80$$ $$\begin{vmatrix}4b & b \\ 20 & 7\end{vmatrix} = 4b \cdot 7 - b \cdot 20 = 28b - 20b = 8b$$ Substitute: $$D_x = 4(2b - 28) - (b-1)(8b - 80) - 1(8b)$$ $$= 8b - 112 - (b-1)(8b - 80) - 8b$$ $$= 8b - 112 - (8b^2 - 80b - 8b + 80) - 8b$$ $$= 8b - 112 - (8b^2 - 88b + 80) - 8b$$ $$= 8b - 112 - 8b^2 + 88b - 80 - 8b$$ $$= -8b^2 + (8b + 88b - 8b) - (112 + 80)$$ $$= -8b^2 + 88b - 192$$ 7. **Calculate $D_y$ by replacing the second column:** $$D_y = \begin{vmatrix}1 & 4 & -1 \\ 2 & 4b & 4 \\ b+1 & 20 & 2\end{vmatrix}$$ Calculate $D_y$: $$D_y = 1 \cdot \begin{vmatrix}4b & 4 \\ 20 & 2\end{vmatrix} - 4 \cdot \begin{vmatrix}2 & 4 \\ b+1 & 2\end{vmatrix} + (-1) \cdot \begin{vmatrix}2 & 4b \\ b+1 & 20\end{vmatrix}$$ Calculate minors: $$\begin{vmatrix}4b & 4 \\ 20 & 2\end{vmatrix} = 8b - 80$$ $$\begin{vmatrix}2 & 4 \\ b+1 & 2\end{vmatrix} = 4 - 4b - 4 = -4b$$ $$\begin{vmatrix}2 & 4b \\ b+1 & 20\end{vmatrix} = 2 \cdot 20 - 4b(b+1) = 40 - 4b^2 - 4b$$ Substitute: $$D_y = 1(8b - 80) - 4(-4b) - 1(40 - 4b^2 - 4b)$$ $$= 8b - 80 + 16b - 40 + 4b^2 + 4b$$ $$= 4b^2 + (8b + 16b + 4b) - (80 + 40)$$ $$= 4b^2 + 28b - 120$$ 8. **Calculate $D_z$ by replacing the third column:** $$D_z = \begin{vmatrix}1 & b-1 & 4 \\ 2 & b & 4b \\ b+1 & 7 & 20\end{vmatrix}$$ Calculate $D_z$: $$D_z = 1 \cdot \begin{vmatrix}b & 4b \\ 7 & 20\end{vmatrix} - (b-1) \cdot \begin{vmatrix}2 & 4b \\ b+1 & 20\end{vmatrix} + 4 \cdot \begin{vmatrix}2 & b \\ b+1 & 7\end{vmatrix}$$ Calculate minors: $$\begin{vmatrix}b & 4b \\ 7 & 20\end{vmatrix} = b \cdot 20 - 4b \cdot 7 = 20b - 28b = -8b$$ $$\begin{vmatrix}2 & 4b \\ b+1 & 20\end{vmatrix} = 40 - 4b^2 - 4b$$ $$\begin{vmatrix}2 & b \\ b+1 & 7\end{vmatrix} = 14 - b^2 - b$$ Substitute: $$D_z = 1(-8b) - (b-1)(40 - 4b^2 - 4b) + 4(14 - b^2 - b)$$ $$= -8b - (b-1)(40 - 4b^2 - 4b) + 56 - 4b^2 - 4b$$ Expand $(b-1)(40 - 4b^2 - 4b)$: $$= b(40 - 4b^2 - 4b) - 1(40 - 4b^2 - 4b) = 40b - 4b^3 - 4b^2 - 40 + 4b^2 + 4b = 40b - 4b^3 - 40 + 4b$$ $$= -4b^3 + 44b - 40$$ Substitute back: $$D_z = -8b - (-4b^3 + 44b - 40) + 56 - 4b^2 - 4b$$ $$= -8b + 4b^3 - 44b + 40 + 56 - 4b^2 - 4b$$ $$= 4b^3 - 4b^2 + (-8b - 44b - 4b) + (40 + 56)$$ $$= 4b^3 - 4b^2 - 56b + 96$$ 9. **Find the solutions:** $$x = \frac{D_x}{D} = \frac{-8b^2 + 88b - 192}{5b^2 - b - 42}$$ $$y = \frac{D_y}{D} = \frac{4b^2 + 28b - 120}{5b^2 - b - 42}$$ $$z = \frac{D_z}{D} = \frac{4b^3 - 4b^2 - 56b + 96}{5b^2 - b - 42}$$ **Final answer:** $$\boxed{x = \frac{-8b^2 + 88b - 192}{5b^2 - b - 42}, \quad y = \frac{4b^2 + 28b - 120}{5b^2 - b - 42}, \quad z = \frac{4b^3 - 4b^2 - 56b + 96}{5b^2 - b - 42}}$$ Note: The solution exists and is unique if $5b^2 - b - 42 \neq 0$.