The value(s) of `lambda` for which the system of equations
`(1-lambda)x+3y-4z=0`
`x-(3+lambda)y + 5z=0`
`3x+y-lambdaz=0`
possesses non-trivial solutions .

Options

A. `-1`
B. 0
C. 1
D. 2

Answer

To find the values of \( \lambda \) for which the given system of equations possesses non-trivial solutions, we need to set up the system in matrix form and find the determinant of the coefficient matrix. The system of equations is:

1. \( (1 - \lambda)x + 3y - 4z = 0 \)
2. \( x - (3 + \lambda)y + 5z = 0 \)
3. \( 3x + y - \lambda z = 0 \)

### Step 1: Write the coefficient matrix

The coefficient matrix \( A \) for the system can be written as:

\[ A = \begin{bmatrix} 1 - \lambda & 3 & -4 \\ 1 & -(3 + \lambda) & 5 \\ 3 & 1 & -\lambda \end{bmatrix} \]

### Step 2: Calculate the determinant of the matrix

To find the values of \( \lambda \) for which the system has non-trivial solutions, we need to compute the determinant of matrix \( A \) and set it equal to zero:

\[ \text{det}(A) = 0 \]

Calculating the determinant of \( A \):

\[ \text{det}(A) = (1 - \lambda) \begin{vmatrix} -(3 + \lambda) & 5 \\ 1 & -\lambda \end{vmatrix} - 3 \begin{vmatrix} 1 & 5 \\ 3 & -\lambda \end{vmatrix} - 4 \begin{vmatrix} 1 & -(3 + \lambda) \\ 3 & 1 \end{vmatrix} \]

Calculating the 2x2 determinants:

1. \( \begin{vmatrix} -(3 + \lambda) & 5 \\ 1 & -\lambda \end{vmatrix} = (-3 - \lambda)(-\lambda) - (5)(1) = \lambda(3 + \lambda) - 5 = \lambda^2 + 3\lambda - 5 \)

2. \( \begin{vmatrix} 1 & 5 \\ 3 & -\lambda \end{vmatrix} = (1)(-\lambda) - (5)(3) = -\lambda - 15 \)

3. \( \begin{vmatrix} 1 & -(3 + \lambda) \\ 3 & 1 \end{vmatrix} = (1)(1) - (3)(-(3 + \lambda)) = 1 + 9 + 3\lambda = 10 + 3\lambda \)

Substituting these back into the determinant expression:

\[ \text{det}(A) = (1 - \lambda)(\lambda^2 + 3\lambda - 5) - 3(-\lambda - 15) - 4(10 + 3\lambda) \]

Expanding this:

\[ = (1 - \lambda)(\lambda^2 + 3\lambda - 5) + 3\lambda + 45 - 40 - 12\lambda \]

\[ = (1 - \lambda)(\lambda^2 + 3\lambda - 5) - 9\lambda + 5 \]

### Step 3: Set the determinant to zero

Now we need to set the determinant equal to zero and solve for \( \lambda \):

\[ (1 - \lambda)(\lambda^2 + 3\lambda - 5) - 9\lambda + 5 = 0 \]

### Step 4: Solve the equation

This will yield a polynomial in \( \lambda \). Expanding and simplifying will lead to a cubic equation.

### Step 5: Factor or use the quadratic formula

After simplification, we can find the roots of the polynomial using factoring or the quadratic formula.

### Final Values

After solving the polynomial, we find the values of \( \lambda \) that satisfy the equation.

### Conclusion

The values of \( \lambda \) for which the system has non-trivial solutions are \( \lambda = 0 \) and \( \lambda = -1 \).

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