Question

Consider the system of linear equations \( x + y + z = 4\mu \), \( x + 2y + 2\lambda z = 10\mu \), \( x + 3y + 4\lambda^2 z = \mu^2 + 15 \), where \( \lambda, \mu \in \mathbb{R} \). Which one of the following statements is NOT correct?

• A. The system has a unique solution if \( \lambda \neq \frac{1}{2} \) and \( \mu \neq 1, 15 \).
• B. The system has an infinite number of solutions if \( \lambda = \frac{1}{2} \) and \( \mu = 15 \).
• C. The system is inconsistent if \( \lambda = \frac{1}{2} \) and \( \mu \neq 1 \).
• D. The system is consistent if \( \lambda \neq \frac{1}{2} \).

Answer 1

The Correct Option is C

To determine which statement is NOT correct, let's analyze the given system of linear equations:

• \( x + y + z = 4\mu \)
• \( x + 2y + 2\lambda z = 10\mu \)
• \( x + 3y + 4\lambda^2 z = \mu^2 + 15 \)

We need to check the conditions under which the system has a unique solution, infinite solutions, or is inconsistent. We will consider each case as per the given options:

1. The system has a unique solution if \( \lambda \neq \frac{1}{2} \) and \( \mu \neq 1, 15 \):

For the system to have a unique solution, the determinant of the coefficient matrix must be non-zero. The coefficient matrix is:

\( \begin{vmatrix} 1 & 1 & 1 \\ 1 & 2 & 2\lambda \\ 1 & 3 & 4\lambda^2 \end{vmatrix} \)

Calculating the determinant, we have:

\( \begin{vmatrix} 1 & 1 & 1 \\ 1 & 2 & 2\lambda \\ 1 & 3 & 4\lambda^2 \end{vmatrix} = 1(2 \cdot 4\lambda^2 - 2\lambda \cdot 3) - 1(1 \cdot 4\lambda^2 - 2\lambda \cdot 1) + 1(1 \cdot 3 - 2 \cdot 1) \)

Continuing,

\( = 8\lambda^2 - 6\lambda - 4\lambda^2 + 2\lambda + 3 - 2 \)

\( = 4\lambda^2 - 4\lambda + 1 \)

The determinant simplifies to

\( (2\lambda - 1)^2 \).

The determinant is zero when \( \lambda = \frac{1}{2} \), leading to no unique solutions. Hence, if \( \lambda \neq \frac{1}{2} \), the determinant is non-zero, and the system may have a unique solution, given that \( \mu \) does not affect determinant independence. Thus, this statement may be correct.

1. The system has an infinite number of solutions if \( \lambda = \frac{1}{2} \) and \( \mu = 15 \):

If \( \lambda = \frac{1}{2} \), the determinant is zero (\( (2 \times \frac{1}{2} - 1)^2 = 0 \)), indicating possible infinite solutions. To check if these exist, substitute \(\lambda = \frac{1}{2}\) and \(\mu = 15\) in the equations and verify consistency:

• Substitute \(\lambda = \frac{1}{2}\) in the equations:
• \( x + y + z = 60 \)
• \( x + 2y + z = 150 \)
• \( x + 3y + z = 240 \)

Checking compatibility, for \(\mu = 15\), the intervention proves consistent.

1. The system is inconsistent if \( \lambda = \frac{1}{2} \) and \( \mu \neq 1 \):

For \(\lambda = \frac{1}{2}\), the determinant is zero. To check inconsistency, see if multiplying equations doesn't provide a valid propagation:

• Given equation manipulation for \(\mu\neq 15\) shows disagreement; hence inconsistency holds true.

1. The system is consistent if \( \lambda \neq \frac{1}{2} \):

If \(\lambda \neq \frac{1}{2}\), the determinant is non-zero, and hence, the system is logically consistent.

After evaluating all statements, it's verified that option 3 (The system is inconsistent if \(\lambda = \frac{1}{2}\) and \(\mu \neq 1\)) is NOT correct.

Answer 2

Write the system of equations in matrix form:

\[ \begin{bmatrix} 1 & 1 & 1 \\ 1 & 2 & 2 \\ 1 & 3 & 4\lambda \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 4\mu \\ 10\mu \\ \mu^2 + 15 \end{bmatrix} \]

Let the coefficient matrix be \( A \):

\[ A = \begin{bmatrix} 1 & 1 & 1 \\ 1 & 2 & 2 \\ 1 & 3 & 4\lambda \end{bmatrix} \]

Calculate the determinant of \( A \):

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

For unique solutions, \( \text{det}(A) \neq 0 \) or \( \lambda \neq \frac{1}{2} \).

For infinite solutions, \( \lambda = \frac{1}{2} \), and consistency depends on the rank of the augmented matrix with specific values of \( \mu \).

The system is inconsistent if \( \lambda = \frac{1}{2} \) and \( \mu \neq 1 \)