Let `x + y + z = 6, 4x + lambday - lambdaz = 0,3x + 2y - 4z = -5.` The value of `lambda` for which given system of equations does not have a unique solution is

Options

A. 0
B. 1
C. 2
D. 3

Answer

To find the value of \( \lambda \) for which the given system of equations does not have a unique solution, we need to set up the equations in a matrix form and find the determinant. The system of equations is:

1. \( x + y + z = 6 \)
2. \( 4x + \lambda y - \lambda z = 0 \)
3. \( 3x + 2y - 4z = -5 \)

We can express this system in the form of a matrix equation \( AX = B \), where:

\[ A = \begin{bmatrix} 1 & 1 & 1 \\ 4 & \lambda & -\lambda \\ 3 & 2 & -4 \end{bmatrix}, \quad X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, \quad B = \begin{bmatrix} 6 \\ 0 \\ -5 \end{bmatrix} \]

To find the values of \( \lambda \) for which the system does not have a unique solution, we need to calculate the determinant of matrix \( A \) and set it equal to zero:

\[ \text{det}(A) = \begin{vmatrix} 1 & 1 & 1 \\ 4 & \lambda & -\lambda \\ 3 & 2 & -4 \end{vmatrix} \]

Calculating the determinant using the formula for a 3x3 matrix:

\[ \text{det}(A) = 1 \cdot \begin{vmatrix} \lambda & -\lambda \\ 2 & -4 \end{vmatrix} - 1 \cdot \begin{vmatrix} 4 & -\lambda \\ 3 & -4 \end{vmatrix} + 1 \cdot \begin{vmatrix} 4 & \lambda \\ 3 & 2 \end{vmatrix} \]

Calculating each of the 2x2 determinants:

1. \( \begin{vmatrix} \lambda & -\lambda \\ 2 & -4 \end{vmatrix} = \lambda \cdot (-4) - (-\lambda) \cdot 2 = -4\lambda + 2\lambda = -2\lambda \)

2. \( \begin{vmatrix} 4 & -\lambda \\ 3 & -4 \end{vmatrix} = 4 \cdot (-4) - (-\lambda) \cdot 3 = -16 + 3\lambda = 3\lambda - 16 \)

3. \( \begin{vmatrix} 4 & \lambda \\ 3 & 2 \end{vmatrix} = 4 \cdot 2 - \lambda \cdot 3 = 8 - 3\lambda \)

Putting it all together:

\[ \text{det}(A) = 1 \cdot (-2\lambda) - 1 \cdot (3\lambda - 16) + 1 \cdot (8 - 3\lambda) \]

Simplifying:

\[ \text{det}(A) = -2\lambda - 3\lambda + 16 + 8 - 3\lambda \]
\[ = -8\lambda + 24 \]

Setting the determinant equal to zero for non-unique solutions:

\[ -8\lambda + 24 = 0 \]

Solving for \( \lambda \):

\[ -8\lambda = -24 \quad \Rightarrow \quad \lambda = \frac{24}{8} = 3 \]

Thus, the value of \( \lambda \) for which the given system of equations does not have a unique solution is:

\[ \lambda = 3 \]