1. **Problem statement:** Given matrix $$A = \begin{pmatrix} -1 & -2 & -5 + \alpha \\ 1 & -4 & 1 \\ 1 & -1 & 3 - \alpha \end{pmatrix}$$ and vector $$b = \begin{pmatrix} -3 \\ -3 \\ 0 \end{pmatrix}$$, solve the following: (a) Find $$\alpha \in \mathbb{R}$$ such that $$\sum_{i=1}^3 a_{ii} = 2$$. (b) Find all $$\alpha \in \mathbb{R}$$ such that $$\det A = 0$$. (c) For $$\alpha = 0$$, find all solutions $$x$$ of the system $$Ax = b$$. --- 2. **Step (a): Sum of diagonal elements equals 2** The diagonal elements are: $$a_{11} = -1, \quad a_{22} = -4, \quad a_{33} = 3 - \alpha$$ Sum: $$\sum_{i=1}^3 a_{ii} = -1 + (-4) + (3 - \alpha) = -2 - \alpha$$ Set equal to 2: $$-2 - \alpha = 2$$ Solve for $$\alpha$$: $$\alpha = -2 - 2 = -4$$ --- 3. **Step (b): Find $$\alpha$$ such that $$\det A = 0$$** Calculate determinant: $$\det A = \begin{vmatrix} -1 & -2 & -5 + \alpha \\ 1 & -4 & 1 \\ 1 & -1 & 3 - \alpha \end{vmatrix}$$ Using cofactor expansion along the first row: $$\det A = (-1) \cdot \begin{vmatrix} -4 & 1 \\ -1 & 3 - \alpha \end{vmatrix} - (-2) \cdot \begin{vmatrix} 1 & 1 \\ 1 & 3 - \alpha \end{vmatrix} + (-5 + \alpha) \cdot \begin{vmatrix} 1 & -4 \\ 1 & -1 \end{vmatrix}$$ Calculate minors: $$M_1 = (-4)(3 - \alpha) - (1)(-1) = -4(3 - \alpha) + 1 = -12 + 4\alpha + 1 = -11 + 4\alpha$$ $$M_2 = 1(3 - \alpha) - 1(1) = 3 - \alpha - 1 = 2 - \alpha$$ $$M_3 = 1(-1) - 1(-4) = -1 + 4 = 3$$ Substitute back: $$\det A = (-1)(-11 + 4\alpha) + 2(2 - \alpha) + (-5 + \alpha)(3)$$ Simplify: $$= 11 - 4\alpha + 4 - 2\alpha - 15 + 3\alpha = (11 + 4 - 15) + (-4\alpha - 2\alpha + 3\alpha) = 0 - 3\alpha = -3\alpha$$ Set determinant to zero: $$-3\alpha = 0 \implies \alpha = 0$$ --- 4. **Step (c): Solve $$Ax = b$$ for $$\alpha = 0$$** Substitute $$\alpha = 0$$ into $$A$$: $$A = \begin{pmatrix} -1 & -2 & -5 \\ 1 & -4 & 1 \\ 1 & -1 & 3 \end{pmatrix}$$ System: $$\begin{cases} -1 x_1 - 2 x_2 - 5 x_3 = -3 \\ 1 x_1 - 4 x_2 + 1 x_3 = -3 \\ 1 x_1 - 1 x_2 + 3 x_3 = 0 \end{cases}$$ From equation 2: $$x_1 = -3 + 4 x_2 - x_3$$ Substitute into equation 3: $$(-3 + 4 x_2 - x_3) - x_2 + 3 x_3 = 0$$ Simplify: $$-3 + 3 x_2 + 2 x_3 = 0 \implies 3 x_2 + 2 x_3 = 3$$ From equation 1: $$-1(-3 + 4 x_2 - x_3) - 2 x_2 - 5 x_3 = -3$$ Simplify: $$3 - 4 x_2 + x_3 - 2 x_2 - 5 x_3 = -3$$ $$3 - 6 x_2 - 4 x_3 = -3$$ $$-6 x_2 - 4 x_3 = -6$$ Divide both sides by -2: $$\cancel{-6} x_2 + \cancel{-4} x_3 = \cancel{-6}$$ $$3 x_2 + 2 x_3 = 3$$ This matches the previous equation from step 3, so the system is consistent. Solve for $$x_2$$: $$3 x_2 = 3 - 2 x_3 \implies x_2 = 1 - \frac{2}{3} x_3$$ Recall $$x_1$$: $$x_1 = -3 + 4 x_2 - x_3 = -3 + 4 \left(1 - \frac{2}{3} x_3\right) - x_3 = -3 + 4 - \frac{8}{3} x_3 - x_3 = 1 - \frac{11}{3} x_3$$ Let $$x_3 = t \in \mathbb{R}$$ be a free parameter. **Final solution:** $$x = \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = \begin{pmatrix} 1 - \frac{11}{3} t \\ 1 - \frac{2}{3} t \\ t \end{pmatrix}, \quad t \in \mathbb{R}$$