1. **Find the inverse of matrix** $$A=\begin{pmatrix}2 & 1 & 3 \\ 1 & 0 & 2 \\ 4 & 1 & 8\end{pmatrix}$$ 2. Calculate the determinant $$\det(A) = 2(0\cdot8 - 2\cdot1) - 1(1\cdot8 - 2\cdot4) + 3(1\cdot1 - 0\cdot4) = 2(0 - 2) - 1(8 - 8) + 3(1 - 0) = -4 - 0 + 3 = -1$$ 3. Since $$\det(A) \neq 0$$, inverse exists. Find cofactor matrix: $$C = \begin{pmatrix} 0\cdot8 - 2\cdot1 & -(1\cdot8 - 2\cdot4) & 1\cdot1 - 0\cdot4 \\ -(1\cdot8 - 3\cdot1) & 2\cdot8 - 3\cdot4 & -(2\cdot1 - 1\cdot4) \\ 1\cdot1 - 0\cdot3 & -(2\cdot1 - 1\cdot3) & 2\cdot0 - 1\cdot1 \end{pmatrix} = \begin{pmatrix} -2 & 0 & 1 \\ -5 & 4 & 0 \\ 1 & -1 & -1 \end{pmatrix}$$ 4. The adjugate matrix is transpose of cofactor matrix: $$\text{adj}(A) = C^T = \begin{pmatrix} -2 & -5 & 1 \\ 0 & 4 & -1 \\ 1 & 0 & -1 \end{pmatrix}$$ 5. Calculate inverse: $$A^{-1} = \frac{1}{\det(A)} \text{adj}(A) = -1 \times \begin{pmatrix} -2 & -5 & 1 \\ 0 & 4 & -1 \\ 1 & 0 & -1 \end{pmatrix} = \begin{pmatrix} 2 & 5 & -1 \\ 0 & -4 & 1 \\ -1 & 0 & 1 \end{pmatrix}$$ 6. **Show formulas for x, y, z using Gauss Elimination on:** $$\begin{cases} ax + by + cz = j \\ dx + ey + fz = k \\ gx + hy + iz = l \end{cases}$$ 7. By eliminating variables stepwise, derived formulas are: $$z = \frac{(bg - ah)(dj - ak) - (bd - ae)(gj - al)}{(bg - ah)(cd - af) - (bd - ae)(cg - ai)}$$ $$y = \frac{(dj - ak) - z(cd - af)}{bd - ae}$$ $$x = \frac{j - yb - zc}{a}$$ 8. **Evaluate x, y, z for system:** $$\begin{cases} x + y + 2z = 3 \\ 2x + 3y + 5z = 7 \\ 3x + 5y + 7z = 12 \end{cases}$$ Coefficients: $$a=1, b=1, c=2, j=3$$ $$d=2, e=3, f=5, k=7$$ $$g=3, h=5, i=7, l=12$$ Calculate determinants: $$bg - ah = 1\times3 - 1\times5 = 3 - 5 = -2$$ $$dj - ak = 2\times3 - 1\times7 = 6 -7 = -1$$ $$bd - ae = 1\times2 - 1\times3 = 2 -3 = -1$$ $$gj - al = 3\times3 - 1\times12 = 9 -12 = -3$$ $$cd - af = 2\times2 - 1\times5 = 4 -5 = -1$$ $$cg - ai = 2\times3 - 1\times7 = 6 -7 = -1$$ Compute numerator and denominator for z: $$\text{num}_z = (-2)(-1) - (-1)(-3) = 2 - 3 = -1$$ $$\text{den}_z = (-2)(-1) - (-1)(-1) = 2 - 1 = 1$$ So $$z = \frac{-1}{1} = -1$$ Calculate y: $$y = \frac{-1 - (-1)(-1)}{-1} = \frac{-1 -1}{-1} = \frac{-2}{-1} = 2$$ Calculate x: $$x = \frac{3 - (2)(1) - (-1)(2)}{1} = \frac{3 - 2 + 2}{1} = 3$$ 9. **Solve system using Cramer's rule:** Coefficient matrix determinant: $$D = \begin{vmatrix}1 & 1 & 2 \\ 2 & 3 & 5 \\ 3 & 5 & 7 \end{vmatrix} = 1(3 \times 7 - 5 \times 5) - 1(2 \times 7 - 5 \times 3) + 2(2 \times 5 - 3 \times 3) = 1(21 - 25) - 1(14 - 15) + 2(10 - 9) = -4 +1 + 2 = -1$$ Calculate determinants replacing columns with constants for $x$, $y$, $z$: $$D_x = \begin{vmatrix}3 & 1 & 2 \\ 7 & 3 & 5 \\ 12 & 5 & 7 \end{vmatrix} = 3(3 \times 7 - 5 \times 5) - 1(7 \times 7 - 5 \times 12) + 2(7 \times 5 - 3 \times 12) = 3(21 - 25) - 1(49 - 60) + 2(35 - 36) = 3(-4) - 1(-11) + 2(-1) = -12 + 11 - 2 = -3$$ $$D_y = \begin{vmatrix}1 & 3 & 2 \\ 2 & 7 & 5 \\ 3 & 12 & 7 \end{vmatrix} = 1(7 \times 7 - 5 \times 12) - 3(2 \times 7 - 5 \times 3) + 2(2 \times 12 - 7 \times 3) = 1(49 - 60) - 3(14 - 15) + 2(24 - 21) = -11 - 3(-1) + 2(3) = -11 + 3 + 6 = -2$$ $$D_z = \begin{vmatrix}1 & 1 & 3 \\ 2 & 3 & 7 \\ 3 & 5 & 12 \end{vmatrix} = 1(3 \times 12 - 7 \times 5) - 1(2 \times 12 - 7 \times 3) + 3(2 \times 5 - 3 \times 3) = 1(36 - 35) - 1(24 - 21) + 3(10 - 9) = 1 - 3 + 3 = 1$$ Therefore, $$x = \frac{D_x}{D} = \frac{-3}{-1} = 3$$ $$y = \frac{D_y}{D} = \frac{-2}{-1} = 2$$ $$z = \frac{D_z}{D} = \frac{1}{-1} = -1$$ **Final answers:** $$A^{-1} = \begin{pmatrix} 2 & 5 & -1 \\ 0 & -4 & 1 \\ -1 & 0 & 1 \end{pmatrix}$$ $$x=3, y=2, z=-1$$