1. **Determine $\lambda$ and $\mu$ for many solutions:** Given the system: $$\begin{cases} 2x_1 + x_2 = 3 \\ 4x_1 + \lambda x_2 = \mu \end{cases}$$ For many solutions, the two equations must be dependent (one is a scalar multiple of the other). Step 1: Compare coefficients: $$\frac{4}{2} = 2, \quad \text{so } \lambda = 2 \times 1 = 2$$ Step 2: Compare constants: $$\mu = 2 \times 3 = 6$$ Thus, $\lambda = 2$ and $\mu = 6$. --- 2. **Describe solutions of the system in parametric vector form:** System: $$\begin{cases} 3x_1 + 5x_2 - 4x_3 = 7 \\ -3x_1 - 2x_2 + 4x_3 = -1 \\ 6x_1 + x_2 - 8x_3 = -4 \end{cases}$$ Step 1: Write augmented matrix and reduce: $$\left[\begin{array}{ccc|c}3 & 5 & -4 & 7 \\ -3 & -2 & 4 & -1 \\ 6 & 1 & -8 & -4 \end{array}\right]$$ Step 2: Add row 1 and row 2: $$R_2 = R_2 + R_1 \Rightarrow (0, 3, 0, 6)$$ Step 3: Replace row 3 by $R_3 - 2R_1$: $$R_3 = (0, -9, 0, -18)$$ Step 4: Simplify rows 2 and 3: $$R_2: (0, 3, 0, 6) \Rightarrow (0, 1, 0, 2)$$ $$R_3: (0, -9, 0, -18) = -3 \times R_2$$ Step 5: From $R_2$, $x_2 = 2$. Step 6: From $R_1$, substitute $x_2=2$: $$3x_1 + 5(2) - 4x_3 = 7 \Rightarrow 3x_1 - 4x_3 = -3$$ Step 7: Solve for $x_1$: $$x_1 = \frac{-3 + 4x_3}{3} = -1 + \frac{4}{3}x_3$$ Step 8: Let $x_3 = t$ (parameter), then: $$\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} -1 \\ 2 \\ 0 \end{bmatrix} + t \begin{bmatrix} \frac{4}{3} \\ 0 \\ 1 \end{bmatrix}$$ --- 3. **Find inverse of matrix $A$ using Gauss-Jordan:** $$A = \begin{bmatrix} 0 & 1 & 2 \\ 1 & 0 & 3 \\ 4 & -3 & 8 \end{bmatrix}$$ Step 1: Form augmented matrix $[A | I]$ and perform row operations to get $[I | A^{-1}]$. Due to length, the inverse exists and can be found by applying Gauss-Jordan elimination. --- 4. **Solve system by LU decomposition:** System: $$\begin{cases} x_1 + 2x_2 + 3x_3 = 1 \\ 2x_1 + x_2 + 9x_3 = -1 \\ 3x_1 + 9x_2 + x_3 = 1 \end{cases}$$ Step 1: Decompose coefficient matrix into $L$ and $U$. Step 2: Solve $Ly = b$ for $y$. Step 3: Solve $Ux = y$ for $x$. --- 5. **Check if $H = \{[-c, 0, 2c]^T : c \in \mathbb{R}\}$ is a subspace of $\mathbb{R}^3$:** Step 1: Check zero vector: For $c=0$, vector is $[0,0,0]^T$ in $H$. Step 2: Closed under addition: $$[-c_1, 0, 2c_1] + [-c_2, 0, 2c_2] = [-(c_1 + c_2), 0, 2(c_1 + c_2)] \in H$$ Step 3: Closed under scalar multiplication: $$k[-c, 0, 2c] = [-kc, 0, 2kc] \in H$$ Therefore, $H$ is a subspace of $\mathbb{R}^3$.