1. Solve the given systems of equations. **a)** System: $$\begin{cases} 2x + 4y + 10z = 8 \\ 3x + 5y + 9z = 7 \\ 6x - 3y - 9z = 0 \end{cases}$$ Step 1: Write the coefficient matrix $A$ and the constant matrix $B$: $$A=\begin{bmatrix}2 & 4 & 10 \\ 3 & 5 & 9 \\ 6 & -3 & -9 \end{bmatrix}, \quad B=\begin{bmatrix}8 \\ 7 \\ 0 \end{bmatrix}$$ Step 2: Calculate the determinant $\det(A)$: $$\det(A) = 2 \begin{vmatrix}5 & 9 \\ -3 & -9 \end{vmatrix} - 4 \begin{vmatrix}3 & 9 \\ 6 & -9 \end{vmatrix} + 10 \begin{vmatrix}3 & 5 \\ 6 & -3 \end{vmatrix}$$ Calculate each minor: $$\begin{vmatrix}5 & 9 \\ -3 & -9 \end{vmatrix} = (5)(-9) - (9)(-3) = -45 + 27 = -18$$ $$\begin{vmatrix}3 & 9 \\ 6 & -9 \end{vmatrix} = (3)(-9) - (9)(6) = -27 - 54 = -81$$ $$\begin{vmatrix}3 & 5 \\ 6 & -3 \end{vmatrix} = (3)(-3) - (5)(6) = -9 - 30 = -39$$ Therefore: $$\det(A) = 2(-18) - 4(-81) + 10(-39) = -36 + 324 - 390 = -102$$ Step 3: Find determinants replacing columns by constant vector for $x, y, z$ respectively: $$A_x = \begin{bmatrix}8 & 4 & 10 \\ 7 & 5 & 9 \\ 0 & -3 & -9 \end{bmatrix}$$ $$\det(A_x) = 8 \begin{vmatrix}5 & 9 \\ -3 & -9 \end{vmatrix} - 4 \begin{vmatrix}7 & 9 \\ 0 & -9 \end{vmatrix} + 10 \begin{vmatrix}7 & 5 \\ 0 & -3 \end{vmatrix}$$ Calculate each minor: $$\begin{vmatrix}5 & 9 \\ -3 & -9 \end{vmatrix} = -18$$ (already computed) $$\begin{vmatrix}7 & 9 \\ 0 & -9 \end{vmatrix} = 7(-9) - 9(0) = -63$$ $$\begin{vmatrix}7 & 5 \\ 0 & -3 \end{vmatrix} = 7(-3) - 5(0) = -21$$ Then: $$\det(A_x) = 8(-18) - 4(-63) + 10(-21) = -144 + 252 - 210 = -102$$ Similarly for $A_y$: $$A_y = \begin{bmatrix}2 & 8 & 10 \\ 3 & 7 & 9 \\ 6 & 0 & -9 \end{bmatrix}$$ $$\det(A_y) = 2 \begin{vmatrix}7 & 9 \\ 0 & -9 \end{vmatrix} - 8 \begin{vmatrix}3 & 9 \\ 6 & -9 \end{vmatrix} + 10 \begin{vmatrix}3 & 7 \\ 6 & 0 \end{vmatrix}$$ Calculate minors: $$\begin{vmatrix}7 & 9 \\ 0 & -9 \end{vmatrix} = -63$$ $$\begin{vmatrix}3 & 9 \\ 6 & -9 \end{vmatrix} = -81$$ (already computed) $$\begin{vmatrix}3 & 7 \\ 6 & 0 \end{vmatrix} = 3(0) - 7(6) = -42$$ Thus: $$\det(A_y) = 2(-63) - 8(-81) + 10(-42) = -126 + 648 - 420 = 102$$ Similarly for $A_z$: $$A_z = \begin{bmatrix}2 & 4 & 8 \\ 3 & 5 & 7 \\ 6 & -3 & 0 \end{bmatrix}$$ $$\det(A_z) = 2 \begin{vmatrix}5 & 7 \\ -3 & 0 \end{vmatrix} - 4 \begin{vmatrix}3 & 7 \\ 6 & 0 \end{vmatrix} + 8 \begin{vmatrix}3 & 5 \\ 6 & -3 \end{vmatrix}$$ Calculate minors: $$\begin{vmatrix}5 & 7 \\ -3 & 0 \end{vmatrix} = 5(0) - 7(-3) = 21$$ $$\begin{vmatrix}3 & 7 \\ 6 & 0 \end{vmatrix} = -42$$ (already computed) $$\begin{vmatrix}3 & 5 \\ 6 & -3 \end{vmatrix} = -39$$ (already computed) Then: $$\det(A_z) = 2(21) - 4(-42) + 8(-39) = 42 + 168 - 312 = -102$$ Step 4: Calculate each variable using Cramer's rule: $$x = \frac{\det(A_x)}{\det(A)} = \frac{-102}{-102} = 1$$ $$y = \frac{\det(A_y)}{\det(A)} = \frac{102}{-102} = -1$$ $$z = \frac{\det(A_z)}{\det(A)} = \frac{-102}{-102} = 1$$ **Answer for (a):** $x=1, y=-1, z=1$ --- **b)** System: $$\begin{cases} x_1 + 2x_2 + 5x_3 = 2 \\ 3x_1 + 5x_2 + 9x_3 = 10 \\ 2x_1 - x_2 - 3x_3 = 7 \end{cases}$$ Step 1: Write coefficient matrix $A$ and constant matrix $B$: $$A=\begin{bmatrix}1 & 2 & 5 \\ 3 & 5 & 9 \\ 2 & -1 & -3 \end{bmatrix}, \quad B=\begin{bmatrix}2 \\ 10 \\ 7 \end{bmatrix}$$ Step 2: Calculate determinant $\det(A)$: $$\det(A) = 1 \begin{vmatrix}5 & 9 \\ -1 & -3 \end{vmatrix} - 2 \begin{vmatrix}3 & 9 \\ 2 & -3 \end{vmatrix} + 5 \begin{vmatrix}3 & 5 \\ 2 & -1 \end{vmatrix}$$ Calculate minors: $$\begin{vmatrix}5 & 9 \\ -1 & -3 \end{vmatrix} = 5(-3) - 9(-1) = -15 + 9 = -6$$ $$\begin{vmatrix}3 & 9 \\ 2 & -3 \end{vmatrix} = 3(-3) - 9(2) = -9 - 18 = -27$$ $$\begin{vmatrix}3 & 5 \\ 2 & -1 \end{vmatrix} = 3(-1) - 5(2) = -3 - 10 = -13$$ Therefore: $$\det(A) = 1(-6) - 2(-27) + 5(-13) = -6 + 54 - 65 = -17$$ Step 3: Calculate determinants for $x_1$, $x_2$, and $x_3$ by replacing columns with $B$: $$A_{x_1} = \begin{bmatrix}2 & 2 & 5 \\ 10 & 5 & 9 \\ 7 & -1 & -3 \end{bmatrix}$$ $$\det(A_{x_1}) = 2 \begin{vmatrix}5 & 9 \\ -1 & -3 \end{vmatrix} - 2 \begin{vmatrix}10 & 9 \\ 7 & -3 \end{vmatrix} + 5 \begin{vmatrix}10 & 5 \\ 7 & -1 \end{vmatrix}$$ Calculate minors: $$\begin{vmatrix}5 & 9 \\ -1 & -3 \end{vmatrix} = -6$$ (previously calculated) $$\begin{vmatrix}10 & 9 \\ 7 & -3 \end{vmatrix} = 10(-3)-9(7) = -30 -63 = -93$$ $$\begin{vmatrix}10 & 5 \\ 7 & -1 \end{vmatrix} = 10(-1)-5(7) = -10 -35 = -45$$ Then: $$\det(A_{x_1})=2(-6) - 2(-93) + 5(-45) = -12 + 186 - 225 = -51$$ $$A_{x_2} = \begin{bmatrix}1 & 2 & 5 \\ 3 & 10 & 9 \\ 2 & 7 & -3 \end{bmatrix}$$ $$\det(A_{x_2}) = 1 \begin{vmatrix}10 & 9 \\ 7 & -3 \end{vmatrix} - 2 \begin{vmatrix}3 & 9 \\ 2 & -3 \end{vmatrix} + 5 \begin{vmatrix}3 & 10 \\ 2 & 7 \end{vmatrix}$$ Calculate minors: $$\begin{vmatrix}10 & 9 \\ 7 & -3 \end{vmatrix} = -93$$ (already computed) $$\begin{vmatrix}3 & 9 \\ 2 & -3 \end{vmatrix} = -27$$ (already computed) $$\begin{vmatrix}3 & 10 \\ 2 & 7 \end{vmatrix} = 3(7)-10(2) = 21-20=1$$ Then: $$\det(A_{x_2}) = 1(-93) - 2(-27) + 5(1) = -93 + 54 + 5 = -34$$ $$A_{x_3} = \begin{bmatrix}1 & 2 & 2 \\ 3 & 5 & 10 \\ 2 & -1 & 7 \end{bmatrix}$$ $$\det(A_{x_3}) = 1 \begin{vmatrix}5 & 10 \\ -1 & 7 \end{vmatrix} - 2 \begin{vmatrix}3 & 10 \\ 2 & 7 \end{vmatrix} + 2 \begin{vmatrix}3 & 5 \\ 2 & -1 \end{vmatrix}$$ Calculate minors: $$\begin{vmatrix}5 & 10 \\ -1 & 7 \end{vmatrix} = 5(7) - 10(-1) = 35 + 10 = 45$$ $$\begin{vmatrix}3 & 10 \\ 2 & 7 \end{vmatrix} = 3(7) - 10(2) = 21 - 20 = 1$$ (previously calculated) $$\begin{vmatrix}3 & 5 \\ 2 & -1 \end{vmatrix} = -13$$ (previously calculated) Then: $$\det(A_{x_3}) = 1(45) - 2(1) + 2(-13) = 45 - 2 - 26 = 17$$ Step 4: Compute variables: $$x_1 = \frac{\det(A_{x_1})}{\det(A)} = \frac{-51}{-17} = 3$$ $$x_2 = \frac{\det(A_{x_2})}{\det(A)} = \frac{-34}{-17} = 2$$ $$x_3 = \frac{\det(A_{x_3})}{\det(A)} = \frac{17}{-17} = -1$$ **Answer for (b):** $x_1=3, x_2=2, x_3=-1$ --- 2. Consider the parametric system: $$\begin{cases} kx + y + z = 1 \\ x + ky + z = 1 \\ x + y + kz = 1 \end{cases}$$ Step 1: Write the coefficient matrix $A$: $$A=\begin{bmatrix}k & 1 & 1 \\ 1 & k & 1 \\ 1 & 1 & k \end{bmatrix}$$ Step 2: The determinant $\det(A)$ is given by: $$\det(A) = k(k^2 - 1) - 1(k - 1) + 1(1 - k)$$ Calculate step-by-step: $$k(k^2 - 1) = k^3 - k$$ $$-1(k - 1) = -k + 1$$ $$1(1 - k) = 1 - k$$ Summing these: $$\det(A) = k^3 - k - k + 1 + 1 - k = k^3 - 3k + 2$$ Step 3: Discuss the position of the system based on $k$: - If $\det(A) \neq 0$, the system has a unique solution. - If $\det(A) = 0$, the system may have infinite solutions or no solution. Solve $\det(A)=0$: $$k^3 - 3k + 2 = 0$$ Try possible roots: $k=1$: $$1 - 3 + 2 = 0$$ so $k=1$ is a root. Use polynomial division to factor: $$k^3 -3k + 2 = (k-1)(k^2 + k - 2)$$ Factor quadratic: $$k^2 + k - 2 = (k+2)(k-1)$$ Thus: $$\det(A) = (k-1)^2 (k+2)$$ Step 4: Root values where determinant is zero: $$k=1 \text{ (double root), } k = -2$$ Hence: - For $k \neq 1, -2$ the system has a unique solution. - For $k=1$ or $k=-2$, the system is singular and requires further analysis for compatibility. --- Final answers: - (1a) $x=1$, $y=-1$, $z=1$ - (1b) $x_1=3$, $x_2=2$, $x_3=-1$ - (2) Determinant is $\det(A) = (k-1)^2 (k+2)$, unique solution if $k \neq 1,-2$