Consider the system of linear equations `x+y+z=5`, `x+2y+lambda^(2)z=9,x+3y+lambda z=mu` ,where `lambda,mu in R` .Then,which of the following statement is NOT correct?

Options

A. System has unique solution if `lambda!=1` and `mu!=13`
B. System is consistent if `lambda!=1` and `mu=13`
C. System is inconsistent if `lambda=1` and `mu!=13`
D. System has infinite number of solutions if `lambda=1` and `mu=13`

Answer

To solve the given system of linear equations and determine which statement is NOT correct, we will analyze the equations step by step.

### Given Equations:
1. \( x + y + z = 5 \) (Equation 1)
2. \( x + 2y + \lambda^2 z = 9 \) (Equation 2)
3. \( x + 3y + \lambda z = \mu \) (Equation 3)

### Step 1: Form the Augmented Matrix
We can represent the system of equations in augmented matrix form:
\[ \begin{bmatrix} 1 & 1 & 1 & | & 5 \\ 1 & 2 & \lambda^2 & | & 9 \\ 1 & 3 & \lambda & | & \mu \end{bmatrix} \]

### Step 2: Calculate the Determinant
To find the conditions for consistency and the type of solution, we need to compute the determinant of the coefficient matrix:
\[ D = \begin{vmatrix} 1 & 1 & 1 \\ 1 & 2 & \lambda^2 \\ 1 & 3 & \lambda \end{vmatrix} \]

Calculating this determinant:
\[ D = 1 \cdot \begin{vmatrix} 2 & \lambda^2 \\ 3 & \lambda \end{vmatrix} - 1 \cdot \begin{vmatrix} 1 & \lambda^2 \\ 1 & \lambda \end{vmatrix} + 1 \cdot \begin{vmatrix} 1 & 2 \\ 1 & 3 \end{vmatrix} \]

Calculating the minors:
1. \( \begin{vmatrix} 2 & \lambda^2 \\ 3 & \lambda \end{vmatrix} = 2\lambda - 3\lambda^2 \)
2. \( \begin{vmatrix} 1 & \lambda^2 \\ 1 & \lambda \end{vmatrix} = \lambda - \lambda^2 \)
3. \( \begin{vmatrix} 1 & 2 \\ 1 & 3 \end{vmatrix} = 3 - 2 = 1 \)

Putting it all together:
\[ D = (2\lambda - 3\lambda^2) - (\lambda - \lambda^2) + 1 \]
\[ D = 2\lambda - 3\lambda^2 - \lambda + \lambda^2 + 1 \]
\[ D = -2\lambda^2 + \lambda + 1 \]

### Step 3: Set the Determinant to Zero
For the system to have either no solution or infinitely many solutions, we set the determinant to zero:
\[ -2\lambda^2 + \lambda + 1 = 0 \]
Multiplying through by -1:
\[ 2\lambda^2 - \lambda - 1 = 0 \]

### Step 4: Solve the Quadratic Equation
Using the quadratic formula:
\[ \lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{1 \pm \sqrt{(-1)^2 - 4 \cdot 2 \cdot (-1)}}{2 \cdot 2} \]
\[ = \frac{1 \pm \sqrt{1 + 8}}{4} = \frac{1 \pm 3}{4} \]
Thus, we have:
\[ \lambda = 1 \quad \text{or} \quad \lambda = -\frac{1}{2} \]

### Step 5: Analyze the Solutions
- **For unique solutions**: \( \lambda \neq 1 \) and \( \mu \) can be any real number.
- **For infinite solutions**: \( \lambda = 1 \) and \( \mu = 13 \).
- **For no solutions**: \( \lambda = -\frac{1}{2} \) and \( \mu \neq \text{specific value depending on the equations} \).

### Conclusion
Based on the analysis, we can conclude which statements about the system are NOT correct. The statement that contradicts the conditions derived above is the one that we are looking for.