The values of `lamda` for which the system of equations
`x+y-3=0`
`(1+lamda)x+(2+lamda)y-8=0`
`x-(1+lamda)y+(2+lamda)=0`
has a non trivial solution are

Options

A. `-5//3,1`
B. `2//3,-3`
C. `-1//3,-3`
D. 0

Answer

To find the values of \( \lambda \) for which the system of equations has a non-trivial solution, we need to set up the equations and form a determinant. The given equations are:

1. \( x + y - 3 = 0 \)
2. \( (1 + \lambda)x + (2 + \lambda)y - 8 = 0 \)
3. \( x - (1 + \lambda)y + (2 + \lambda) = 0 \)

We can express these equations in matrix form \( A \mathbf{X} = 0 \), where \( A \) is the coefficient matrix and \( \mathbf{X} \) is the vector of variables \( \begin{bmatrix} x \\ y \end{bmatrix} \).

The coefficient matrix \( A \) is:

\[ A = \begin{bmatrix} 1 & 1 \\ 1 + \lambda & 2 + \lambda \\ 1 & -(1 + \lambda) \end{bmatrix} \]

To find the values of \( \lambda \) for which there is a non-trivial solution, we need to calculate the determinant of the matrix \( A \) and set it to zero:

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

Calculating the determinant of the matrix:

\[ \text{det}(A) = \begin{vmatrix} 1 & 1 \\ 1 + \lambda & 2 + \lambda \\ 1 & -(1 + \lambda) \end{vmatrix} \]

Using the determinant formula for a \( 3 \times 3 \) matrix:

\[ \text{det}(A) = 1 \cdot \begin{vmatrix} 2 + \lambda & -(1 + \lambda) \\ -(1 + \lambda) & 1 \end{vmatrix} - 1 \cdot \begin{vmatrix} 1 + \lambda & -(1 + \lambda) \\ 1 & 1 \end{vmatrix} + 3 \cdot \begin{vmatrix} 1 + \lambda & 2 + \lambda \\ 1 & -(1 + \lambda) \end{vmatrix} \]

Calculating each of the \( 2 \times 2 \) determinants:

1. First determinant:
\[ \begin{vmatrix} 2 + \lambda & -(1 + \lambda) \\ -(1 + \lambda) & 1 \end{vmatrix} = (2 + \lambda) \cdot 1 - (-(1 + \lambda)) \cdot (-(1 + \lambda)) = 2 + \lambda - (1 + \lambda)^2 \]
Expanding:
\[ = 2 + \lambda - (1 + 2\lambda + \lambda^2) = 2 + \lambda - 1 - 2\lambda - \lambda^2 = 1 - \lambda - \lambda^2 \]

2. Second determinant:
\[ \begin{vmatrix} 1 + \lambda & -(1 + \lambda) \\ 1 & 1 \end{vmatrix} = (1 + \lambda) \cdot 1 - (-(1 + \lambda)) \cdot 1 = 1 + \lambda + 1 + \lambda = 2 + 2\lambda \]

3. Third determinant:
\[ \begin{vmatrix} 1 + \lambda & 2 + \lambda \\ 1 & -(1 + \lambda) \end{vmatrix} = (1 + \lambda)(-(1 + \lambda)) - (2 + \lambda)(1) = - (1 + \lambda)^2 - (2 + \lambda) \]
Expanding:
\[ = - (1 + 2\lambda + \lambda^2) - 2 - \lambda = -1 - 2\lambda - \lambda^2 - 2 - \lambda = -3 - 3\lambda - \lambda^2 \]

Now substituting these back into the determinant equation:

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

Expanding this gives:

\[ = 1 - \lambda - \lambda^2 - 2 - 2\lambda - 9 - 9\lambda - 3\lambda^2 \]
Combining like terms:

\[ = (1 - 2 - 9) + (-\lambda - 2\lambda - 9\lambda) + (-\lambda^2 - 3\lambda^2) = -10 - 12\lambda - 4\lambda^2 \]

Setting the determinant to zero:

\[ -4\lambda^2 - 12\lambda - 10 = 0 \]

Dividing through by -2:

\[ 2\lambda^2 + 6\lambda + 5 = 0 \]

Now we can use the quadratic formula \( \lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

Here, \( a = 2, b = 6, c = 5 \):

\[ \lambda = \frac{-6 \pm \sqrt{6^2 - 4 \cdot 2 \cdot 5}}{2 \cdot 2} = \frac{-6 \pm \sqrt{36 - 40}}{4} = \frac{-6 \pm \sqrt{-4}}{4} \]

This gives us:

\[ \lambda = \frac{-6 \pm 2i}{4} = \frac{-3 \pm i}{2} \]

Thus, the values of \( \lambda \) for which the system of equations has a non-trivial solution are:

\[ \lambda = \frac{-3 + i}{2} \quad \text{and} \quad \lambda = \frac{-3 - i}{2} \]