The system of linear equations
`(lamda+3)x+(lamda+2)y+z=0`
`3x+(lamda+3)y+z=0`
`2x+3y+z=0`
has a non trivial solution

Options

A. if `lamda=1`
B. if `lamda=-1`
C. for no real value of `lamda`
D. if `lamda=0`

Answer

To determine the values of \(\lambda\) for which the given system of linear equations has a non-trivial solution, we need to analyze the determinant of the coefficient matrix. The system of equations is:

1. \((\lambda + 3)x + (\lambda + 2)y + z = 0\)
2. \(3x + (\lambda + 3)y + z = 0\)
3. \(2x + 3y + z = 0\)

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

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

We can calculate the determinant using the first row:
\[ \text{det}(A) = (\lambda + 3) \begin{vmatrix} \lambda + 3 & 1 \\ 3 & 1 \end{vmatrix} - (\lambda + 2) \begin{vmatrix} 3 & 1 \\ 2 & 1 \end{vmatrix} + 1 \begin{vmatrix} 3 & \lambda + 3 \\ 2 & 3 \end{vmatrix} \]

### Step 3: Calculate the 2x2 Determinants
Calculating the 2x2 determinants:
1. \(\begin{vmatrix} \lambda + 3 & 1 \\ 3 & 1 \end{vmatrix} = (\lambda + 3)(1) - (3)(1) = \lambda + 3 - 3 = \lambda\)

2. \(\begin{vmatrix} 3 & 1 \\ 2 & 1 \end{vmatrix} = (3)(1) - (2)(1) = 3 - 2 = 1\)

3. \(\begin{vmatrix} 3 & \lambda + 3 \\ 2 & 3 \end{vmatrix} = (3)(3) - (2)(\lambda + 3) = 9 - 2\lambda - 6 = 3 - 2\lambda\)

### Step 4: Substitute Back into the Determinant Expression
Substituting these values back into the determinant expression:
\[ \text{det}(A) = (\lambda + 3)(\lambda) - (\lambda + 2)(1) + 1(3 - 2\lambda) \]
\[ = \lambda^2 + 3\lambda - \lambda - 2 + 3 - 2\lambda \]
\[ = \lambda^2 + (3\lambda - \lambda - 2\lambda) + (3 - 2) \]
\[ = \lambda^2 + 0\lambda + 1 = \lambda^2 + 1 \]

### Step 5: Set the Determinant to Zero
Now, we set the determinant equal to zero:
\[ \lambda^2 + 1 = 0 \]

### Step 6: Solve for \(\lambda\)
This gives us:
\[ \lambda^2 = -1 \]
\[ \lambda = \pm i \]

### Conclusion
The system of equations has a non-trivial solution when \(\lambda\) takes on imaginary values \(\lambda = i\) or \(\lambda = -i\). Therefore, there are no real values of \(\lambda\) for which the system has a non-trivial solution.