Question

Consider the system of linear equations
x + y + z = 5,
x + 2y + λ²z = 9,
x + 3y + λz = μ,
where λ, μ ∈ ℝ.Then, which of the following statement is NOT correct?

• A. System has infinite number of solutions if λ = 1 and μ = 13
• B. System is inconsistent if λ = 1 and μ ≠ 13
• C. System is consistent if λ ≠ 1 and μ = 13
• D. System has a unique solution if λ ≠ 1 and μ ≠ 13

Answer 1

The Correct Option is D

To solve this problem, we need to analyze the given system of equations and determine when it has a unique solution, infinite solutions, or is inconsistent. The system of equations is:

• \( x + y + z = 5 \)
• \( x + 2y + \lambda^2z = 9 \)
• \( x + 3y + \lambda z = \mu \)

We'll examine each condition provided in the options by understanding the different scenarios arising from \(\lambda\) and \(\mu\).

Step 1: Analyze the system.

The system can be represented in matrix form as \( A \mathbf{x} = \mathbf{b} \), where:

\(A = \begin{bmatrix} 1 & 1 & 1 \\ 1 & 2 & \lambda^2 \\ 1 & 3 & \lambda \end{bmatrix}, \mathbf{x} = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, \mathbf{b} = \begin{bmatrix} 5 \\ 9 \\ \mu \end{bmatrix}\)

Step 2: Check for infinite solutions.

Infinite solutions occur when the system has dependent equations, meaning the rank of matrix \( A \) is less than the number of variables (here, 3). An easy approach is to check if any of the equations are multiples or linear combinations of each other.

Let's substitute \(\lambda = 1\) and \(\mu = 13\) into the system:

• \( x + y + z = 5 \)
• \( x + 2y + 1^2z = 9 \rightarrow x + 2y + z = 9 \)
• \( x + 3y + 1z = 13 \rightarrow x + 3y + z = 13 \)

The system simplifies to a form where equations become linearly dependent. They do not provide unique values for each variable, leading to infinite solutions.

Step 3: Check for inconsistency.

An inconsistent system does not have any solutions. This can happen if, after row reduction, an equation like \( 0 = c \) (where \( c \neq 0 \)) appears.

For \(\lambda = 1\) and \(\mu \neq 13\), the system would reduce to an inconsistency since modifying the constant term \(\mu\) results in a contradiction after solving.

Step 4: Check for consistency when \(\lambda \neq 1\).

For \(\lambda \neq 1\), the coefficient matrix does not lead to linear dependency, as \( \lambda^2 \neq \lambda \). For \( \mu = 13 \), the matrix is potentially consistent depending on \(\lambda\).

Step 5: Check for uniqueness when \(\lambda \neq 1\) and \(\mu \neq 13\).

If neither value causes the system to degenerate or contradict, altering both \(\lambda\) and \(\mu\) implies differentiability and a unique solution set, as all equations can be differentiated to provide specific values for \( x, y, \) and \( z \).

Conclusion:

The correct answer is: System has a unique solution if \(\lambda \neq 1\) and \(\mu \neq 13\). This statement is NOT correct based on our analysis.

Answer 2

Convert the system to matrix form and perform row reduction:

\[ \left(\begin{array}{ccc|c} 1 & 1 & 1 & 5 \\ 1 & 2 & 2 & 9 \\ 1 & 3 & \lambda & \mu \end{array}\right) \rightarrow \left(\begin{array}{ccc|c} 1 & 1 & 1 & 5 \\ 0 & 1 & 1 & 4 \\ 0 & 2 & \lambda - 1 & \mu - 5 \end{array}\right) \rightarrow \left(\begin{array}{ccc|c} 1 & 1 & 1 & 5 \\ 0 & 1 & 1 & 4 \\ 0 & 0 & \lambda - 3 & \mu - 13 \end{array}\right) \]

For consistency: Unique solution if \(\lambda \neq 3\).

Infinite solutions if \(\lambda = 3\) and \(\mu = 13\).

No solution if \(\lambda = 3\) and \(\mu \neq 13\).

Therefore, the incorrect statement is:

(4) System has unique solution if \(\lambda = 1\) and \(\mu \neq 13\).