To determine the value of $\lambda$ such that the system of equations has a unique solution, we need to assess the determinant of the coefficient matrix. If the determinant is nonzero, the system has a unique solution. Otherwise, it does not.

System of Equations

1. $2x_1 + 4x_3 = 6$
2. $3x_1 + x_2 + x_3 = -1$
3. $2x_2 - x_3 = -2$
4. $x_1 - x_2 + \lambda x_3 = -5$

Coefficient Matrix

Rewriting the system in matrix form:

2 & 0 & 4 \\ 3 & 1 & 1 \\ 0 & 2 & -1 \\ 1 & -1 & \lambda \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 6 \\ -1 \\ -2 \\ -5 \end{bmatrix}.$$ The coefficient matrix is: $$A = \begin{bmatrix} 2 & 0 & 4 \\ 3 & 1 & 1 \\ 0 & 2 & -1 \\ 1 & -1 & \lambda \end{bmatrix}.$$ ### Step 1: Determinant of $$A$$ The determinant of $$A$$ is: $$\text{det}(A) = \begin{vmatrix} 2 & 0 & 4 \\ 3 & 1 & 1 \\ 0 & 2 & -1 \\ 1 & -1 & \lambda \end{vmatrix}.$$ Expanding along the first row: \[ \text{det}(A) = 2 \begin{vmatrix} 1 & 1 \\ 2 & -1 & \lambda \end{vmpy as sp # Define the matrix elements and determinant variable λ = sp.Symbol('λ') matrix = sp.Matrix([ [2, 0, 4], [3, 1, 1], [0, 2, -1], [1, -1, λ] ]) # Calculate the determinant of the matrix det_A = matrix.det() The determinant calculation failed because the matrix is not square (it is 4x3). To evaluate the uniqueness of the solution, we instead examine the augmented matrix's rank compared to the coefficient matrix's rank coefficient matrix (3x3) coeff_matrix = sp.Matrix([ [2, 0, 4], [3, 1, 1], [0, 2, -1], [1, -1, λ] ]) # Calculate ranks of the coefficient matrix and augmented matrix rank_coeff_matrix =The coefficient matrix is squareLet's reapproach this problem step-by-step to ensure correctness. First, we need to compute the determinant for the $$4 \times 4$$ matrix to analyze the dependency on $$\lambda$$. Let coefficient matrix (4x4) square_matrix = sp.Matrix([ [2, 0, 4, 0], [3, 1, 1, 0], [0, 2, -1, 0], [1, -1, λ, 1] ]) # Calculate the determinant of the square matrix det_square_matrix = square_matrix.det() det_square_matrixThe determinant of the coefficient matrix is $$18$$, which is independent of $$\lambda$$. This indicates that the system of equations will always have a unique solution, regardless of the value of $$\lambda$$. ### Step 2: Solve the System Now, let us find the solution the augmented matrix (including constants on the right-hand side) augmented_matrix = sp.Matrix([ [2, 0, 4, 6], [3, 1, 1, -1], [0, 2, -1, -2], [1, -1, λ, -5] ]) # Solve the system of equations solution = sp.linsolve((square_matrix[:, :-1], augmented_matrixIt seems there