1. **Solve system a using Cramer's rule:** Given: $$\begin{cases} 2x + 4y + 10z = 8 \\ 3x + 5y + 9z = 7 \\ 6x - 3y - 9z = 0 \end{cases}$$ Step 1: Compute the determinant of the coefficient matrix $D$: $$D = \begin{vmatrix} 2 & 4 & 10 \\ 3 & 5 & 9 \\ 6 & -3 & -9 \end{vmatrix}$$ Calculate $D$: $$D = 2\begin{vmatrix}5 & 9 \\ -3 & -9\end{vmatrix} - 4\begin{vmatrix}3 & 9 \\ 6 & -9\end{vmatrix} + 10\begin{vmatrix}3 & 5 \\ 6 & -3\end{vmatrix}$$ $$= 2(5 \times -9 - (-3) \times 9) - 4(3 \times -9 - 6 \times 9) + 10(3 \times -3 - 6 \times 5)$$ $$= 2(-45 + 27) - 4(-27 - 54) + 10(-9 - 30)$$ $$= 2(-18) - 4(-81) + 10(-39) = -36 + 324 - 390 = -102$$ Step 2: Compute $D_x$, replacing first column with constants: $$D_x = \begin{vmatrix}8 & 4 & 10 \\ 7 & 5 & 9 \\ 0 & -3 & -9 \end{vmatrix}$$ $$= 8\begin{vmatrix}5 & 9 \\ -3 & -9\end{vmatrix} - 4\begin{vmatrix}7 & 9 \\ 0 & -9\end{vmatrix} + 10\begin{vmatrix}7 & 5 \\ 0 & -3\end{vmatrix}$$ $$= 8(-45 + 27) - 4( -63 - 0) + 10(-21 - 0)$$ $$= 8(-18) - 4(-63) + 10(-21) = -144 + 252 - 210 = -102$$ Step 3: Compute $D_y$, replacing second column: $$D_y = \begin{vmatrix}2 & 8 & 10 \\ 3 & 7 & 9 \\ 6 & 0 & -9 \end{vmatrix}$$ $$= 2\begin{vmatrix}7 & 9 \\ 0 & -9\end{vmatrix} - 8\begin{vmatrix}3 & 9 \\ 6 & -9\end{vmatrix} + 10\begin{vmatrix}3 & 7 \\ 6 & 0\end{vmatrix}$$ $$= 2(-63) - 8( -27 - 54) + 10(0 - 42)$$ $$= -126 - 8(-81) + 10(-42) = -126 + 648 - 420 = 102$$ Step 4: Compute $D_z$, replacing third column: $$D_z = \begin{vmatrix}2 & 4 & 8 \\ 3 & 5 & 7 \\ 6 & -3 & 0 \end{vmatrix}$$ $$= 2\begin{vmatrix}5 & 7 \\ -3 & 0\end{vmatrix} - 4\begin{vmatrix}3 & 7 \\ 6 & 0\end{vmatrix} + 8\begin{vmatrix}3 & 5 \\ 6 & -3\end{vmatrix}$$ $$= 2(0 + 21) - 4(0 - 42) + 8(-9 - 30)$$ $$= 42 + 168 - 312 = -102$$ Step 5: Solve for variables: $$x = \frac{D_x}{D} = \frac{-102}{-102} = 1$$ $$y = \frac{D_y}{D} = \frac{102}{-102} = -1$$ $$z = \frac{D_z}{D} = \frac{-102}{-102} = 1$$ 2. **Solve system b using Cramer's rule:** Given: $$\begin{cases} x_1 + 2x_2 + 5x_3 = 2 \\ 3x_1 + 5x_2 + 9x_3 = 10 \\ 2x_1 - x_2 - 3x_3 = 7 \end{cases}$$ Step 1: Determinant $D$: $$D = \begin{vmatrix}1 & 2 & 5 \\ 3 & 5 & 9 \\ 2 & -1 & -3 \end{vmatrix}$$ Calculate: $$D = 1\begin{vmatrix}5 & 9 \\ -1 & -3\end{vmatrix} - 2\begin{vmatrix}3 & 9 \\ 2 & -3\end{vmatrix} + 5\begin{vmatrix}3 & 5 \\ 2 & -1\end{vmatrix}$$ $$= 1(5 \times -3 - (-1) \times 9) - 2(3 \times -3 - 2 \times 9) + 5(3 \times -1 - 2 \times 5)$$ $$= 1(-15 + 9) - 2(-9 - 18) + 5(-3 - 10)$$ $$= -6 + 54 - 65 = -17$$ Step 2: Compute $D_{x_1}$ replacing first column: $$D_{x_1} = \begin{vmatrix}2 & 2 & 5 \\ 10 & 5 & 9 \\ 7 & -1 & -3 \end{vmatrix}$$ $$= 2\begin{vmatrix}5 & 9 \\ -1 & -3\end{vmatrix} - 2\begin{vmatrix}10 & 9 \\ 7 & -3\end{vmatrix} + 5\begin{vmatrix}10 & 5 \\ 7 & -1\end{vmatrix}$$ $$= 2(-15 + 9) - 2(-30 - 63) + 5(-10 - 35)$$ $$= 2(-6) - 2(-93) + 5(-45) = -12 + 186 - 225 = -51$$ Step 3: Compute $D_{x_2}$ replacing second column: $$D_{x_2} = \begin{vmatrix}1 & 2 & 5 \\ 3 & 10 & 9 \\ 2 & 7 & -3 \end{vmatrix}$$ $$= 1\begin{vmatrix}10 & 9 \\ 7 & -3\end{vmatrix} - 2\begin{vmatrix}3 & 9 \\ 2 & -3\end{vmatrix} + 5\begin{vmatrix}3 & 10 \\ 2 & 7\end{vmatrix}$$ $$= 1(10 \times -3 - 7 \times 9) - 2(3 \times -3 - 2 \times 9) + 5(3 \times 7 - 2 \times 10)$$ $$= 1(-30 - 63) - 2(-9 - 18) + 5(21 - 20)$$ $$= -93 + 54 + 5 = -34$$ Step 4: Compute $D_{x_3}$ replacing third column: $$D_{x_3} = \begin{vmatrix}1 & 2 & 2 \\ 3 & 5 & 10 \\ 2 & -1 & 7 \end{vmatrix}$$ $$= 1\begin{vmatrix}5 & 10 \\ -1 & 7\end{vmatrix} - 2\begin{vmatrix}3 & 10 \\ 2 & 7\end{vmatrix} + 2\begin{vmatrix}3 & 5 \\ 2 & -1\end{vmatrix}$$ $$= 1(5 \times 7 - (-1) \times 10) - 2(3 \times 7 - 2 \times 10) + 2(3 \times -1 - 2 \times 5)$$ $$= 1(35 + 10) - 2(21 - 20) + 2(-3 - 10)$$ $$= 45 - 2(1) + 2(-13) = 45 - 2 - 26 = 17$$ Step 5: Solve: $$x_1 = \frac{D_{x_1}}{D} = \frac{-51}{-17} = 3$$ $$x_2 = \frac{D_{x_2}}{D} = \frac{-34}{-17} = 2$$ $$x_3 = \frac{D_{x_3}}{D} = \frac{17}{-17} = -1$$ 3. **Solve the parameterized system using Cramer's rule:** Given: $$\begin{cases} kx + y + z = 1 \\ x + ky + z = 1 \\ x + y + kz = 1 \end{cases}$$ Step 1: Determinant $D$ of coefficient matrix: $$D = \begin{vmatrix} k & 1 & 1 \\ 1 & k & 1 \\ 1 & 1 & k \end{vmatrix}$$ Calculate $D$: $$D = k(k \times k - 1 \times 1) - 1(1 \times k - 1 \times 1) + 1(1 \times 1 - k \times 1)$$ $$= k(k^2 - 1) - (k - 1) + (1 - k)$$ $$= k^3 - k - k + 1 + 1 - k = k^3 - 3k + 2$$ Step 2: Compute $D_x$, replacing first column with constants: $$D_x = \begin{vmatrix} 1 & 1 & 1 \\ 1 & k & 1 \\ 1 & 1 & k \end{vmatrix} = 1(k \times k - 1 \times 1) - 1(1 \times k - 1 \times 1) + 1(1 \times 1 - k \times 1)$$ $$= 1(k^2 - 1) - (k - 1) + (1 - k) = k^2 - 1 - k + 1 + 1 - k = k^2 - 2k + 1$$ $$= (k - 1)^2$$ Step 3: Similarly, $D_y$ replacing second column: $$D_y = \begin{vmatrix} k & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & k \end{vmatrix}$$ Calculate: $$= k(1 \times k - 1 \times 1) - 1(1 \times k - 1 \times 1) + 1(1 \times 1 - 1 \times 1)$$ $$= k(k - 1) - (k - 1) + 0 = k(k - 1) - (k - 1) = (k - 1)(k - 1) = (k - 1)^2$$ Step 4: Compute $D_z$, replacing third column: $$D_z = \begin{vmatrix} k & 1 & 1 \\ 1 & k & 1 \\ 1 & 1 & 1 \end{vmatrix}$$ Calculate: $$= k(k \times 1 - 1 \times 1) - 1(1 \times 1 - 1 \times 1) + 1(1 \times 1 - k \times 1)$$ $$= k(k - 1) - 0 + (1 - k) = k^2 - k + 1 - k = k^2 - 2k + 1 = (k - 1)^2$$ Step 5: Solve for variables: $$x = \frac{D_x}{D} = \frac{(k - 1)^2}{k^3 - 3k + 2}$$ $$y = \frac{D_y}{D} = \frac{(k - 1)^2}{k^3 - 3k + 2}$$ $$z = \frac{D_z}{D} = \frac{(k - 1)^2}{k^3 - 3k + 2}$$ **Note:** When $D = k^3 - 3k + 2 = 0$, the system does not have a unique solution.