If system of equation `4x-lambday+2z=0 , 2x+2y+z=0 , mux+2y+3z=0` has non trivial solution than

Options

A. `lambda=6 , mu =2`
B. `lambda in R , mu = 6`
C. `lambda = 5 , mu in R`
D. None of these

Answer

To determine the values of \(\lambda\) and \(\mu\) for which the system of equations has a non-trivial solution, we need to analyze the determinant of the coefficient matrix formed by the given equations.

The system of equations is:
1. \(4x - \lambda y + 2z = 0\)
2. \(2x + 2y + z = 0\)
3. \(\mu x + 2y + 3z = 0\)

### Step 1: Form the Coefficient Matrix
The coefficient matrix \(A\) for the system can be written as:
\[ A = \begin{bmatrix} 4 & -\lambda & 2 \\ 2 & 2 & 1 \\ \mu & 2 & 3 \end{bmatrix} \]

### Step 2: Calculate the Determinant
To find the values of \(\lambda\) and \(\mu\) for which the system has a non-trivial solution, we need to set the determinant of this matrix to zero:
\[ \text{det}(A) = 0 \]

Calculating the determinant using the formula for a \(3 \times 3\) matrix:
\[ \text{det}(A) = 4 \begin{vmatrix} 2 & 1 \\ 2 & 3 \end{vmatrix} - (-\lambda) \begin{vmatrix} 2 & 1 \\ \mu & 3 \end{vmatrix} + 2 \begin{vmatrix} 2 & 2 \\ \mu & 2 \end{vmatrix} \]

Calculating the \(2 \times 2\) determinants:
1. \(\begin{vmatrix} 2 & 1 \\ 2 & 3 \end{vmatrix} = (2 \cdot 3 - 1 \cdot 2) = 6 - 2 = 4\)
2. \(\begin{vmatrix} 2 & 1 \\ \mu & 3 \end{vmatrix} = (2 \cdot 3 - 1 \cdot \mu) = 6 - \mu\)
3. \(\begin{vmatrix} 2 & 2 \\ \mu & 2 \end{vmatrix} = (2 \cdot 2 - 2 \cdot \mu) = 4 - 2\mu\)

Substituting these back into the determinant:
\[ \text{det}(A) = 4(4) + \lambda(6 - \mu) + 2(4 - 2\mu) \]
\[ = 16 + \lambda(6 - \mu) + 8 - 4\mu \]
\[ = 24 + \lambda(6 - \mu) - 4\mu \]

### Step 3: Set the Determinant to Zero
Now, we set the determinant to zero for a non-trivial solution:
\[ 24 + \lambda(6 - \mu) - 4\mu = 0 \]

### Step 4: Analyze the Equation
Rearranging gives:
\[ \lambda(6 - \mu) = 4\mu - 24 \]

### Step 5: Find Values of \(\lambda\) and \(\mu\)
To find specific values, we can test values for \(\mu\):
1. If we let \(\mu = 6\):
\[ \lambda(6 - 6) = 4(6) - 24 \implies 0 = 24 - 24 \implies 0 = 0 \]
This holds true for any value of \(\lambda\). Thus, if \(\mu = 6\), \(\lambda\) can be any real number.

### Conclusion
The values of \(\lambda\) and \(\mu\) for which the system has a non-trivial solution are:
\[ \mu = 6 \quad \text{and} \quad \lambda \in \mathbb{R} \]