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 R` .Which one of the following statements is "NOT" correct ?

Options

A. The system is consistent if `lambda!=(1)/(2)`
B. The system has unique solution if `lambda!=(1)/(2)` and `mune1, 15`
C. The system has infinite number of solutions if `lambda=(1)/(2)` and `mu=15`
D. The system is inconsistent if `lambda=(1)/(2)` and `mu!=1`

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. \( x + y + z = 4\mu \) (Equation 1)
2. \( x + 2y + 2\lambda z = 10\mu \) (Equation 2)
3. \( x + 3y + 4\lambda^2 z = \mu^2 + 15 \) (Equation 3)

### Step 1: Form the Augmented Matrix
We can represent the system of equations in an augmented matrix form:

\[ \begin{bmatrix} 1 & 1 & 1 & | & 4\mu \\ 1 & 2 & 2\lambda & | & 10\mu \\ 1 & 3 & 4\lambda^2 & | & \mu^2 + 15 \end{bmatrix} \]

### Step 2: Row Reduction
We will perform row operations to simplify the matrix.

- Subtract Row 1 from Row 2 and Row 3:

\[ R_2 \rightarrow R_2 - R_1 \\ R_3 \rightarrow R_3 - R_1 \]

This gives us:

\[ \begin{bmatrix} 1 & 1 & 1 & | & 4\mu \\ 0 & 1 & (2\lambda - 1) & | & (10\mu - 4\mu) \\ 0 & 2 & (4\lambda^2 - 1) & | & (\mu^2 + 15 - 4\mu) \end{bmatrix} \]

This simplifies to:

\[ \begin{bmatrix} 1 & 1 & 1 & | & 4\mu \\ 0 & 1 & (2\lambda - 1) & | & 6\mu \\ 0 & 2 & (4\lambda^2 - 1) & | & (\mu^2 - 4\mu + 15) \end{bmatrix} \]

### Step 3: Further Row Reduction
Next, we can simplify Row 3 by subtracting 2 times Row 2 from Row 3:

\[ R_3 \rightarrow R_3 - 2R_2 \]

This gives us:

\[ \begin{bmatrix} 1 & 1 & 1 & | & 4\mu \\ 0 & 1 & (2\lambda - 1) & | & 6\mu \\ 0 & 0 & (4\lambda^2 - 1 - 2(2\lambda - 1)) & | & (\mu^2 - 4\mu + 15 - 12\mu) \end{bmatrix} \]

The third row simplifies to:

\[ 0 = (4\lambda^2 - 4\lambda + 1) \quad \text{and} \quad \mu^2 - 16\mu + 15 \]

### Step 4: Finding Conditions for Solutions
1. For the system to have a unique solution, the determinant must be non-zero, which occurs when \( \lambda \neq \frac{1}{2} \).
2. For infinite solutions, we need both conditions \( 4\lambda^2 - 4\lambda + 1 = 0 \) and \( \mu^2 - 16\mu + 15 = 0 \).

### Step 5: Solving for \( \lambda \) and \( \mu \)
- The quadratic \( 4\lambda^2 - 4\lambda + 1 = 0 \) gives \( \lambda = \frac{1}{2} \).
- The quadratic \( \mu^2 - 16\mu + 15 = 0 \) gives \( \mu = 1 \) or \( \mu = 15 \).

### Conclusion
Now, we can analyze the statements provided:

1. **Statement A**: There exists a consistent solution for all \( \lambda, \mu \in \mathbb{R} \) - **NOT correct** (as it depends on specific values).
2. **Statement B**: There exists a unique solution when \( \lambda \neq \frac{1}{2} \) - **correct**.
3. **Statement C**: There are infinite solutions when \( \lambda = \frac{1}{2} \) and \( \mu = 1 \) or \( \mu = 15 \) - **correct**.
4. **Statement D**: The system is inconsistent for \( \lambda = \frac{1}{2} \) and \( \mu \) not equal to 1 or 15 - **correct**.

Thus, the statement that is "NOT" correct is **Statement A**.