The system of equations 6x+5y+`lamda`z= 0, 3z-y+4z= 0, x+2y-3z=0 has non-trivial solutions for

Options

A. `lamda=0`
B. `lamda=1`
C. `lamda=-5`
D. None of these

Explanation

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

1. \(6x + 5y + \lambda z = 0\)
2. \(3z - y + 4z = 0\) (This equation seems to have a typo; it should be \(3z - y + 4z = 0\) which simplifies to \(7z - y = 0\) or \(y = 7z\))
3. \(x + 2y - 3z = 0\)

Now, we can express the system in matrix form \(A\mathbf{X} = 0\), where \(A\) is the coefficient matrix and \(\mathbf{X} = \begin{bmatrix} x \\ y \\ z \end{bmatrix}\).

The coefficient matrix \(A\) is:

\[ A = \begin{bmatrix} 6 & 5 & \lambda \\ 0 & -1 & 7 \\ 1 & 2 & -3 \end{bmatrix} \]

To find the values of \(\lambda\) for which there are non-trivial solutions, we need to calculate the determinant of \(A\) and set it equal to zero:

\[ \text{det}(A) = \begin{vmatrix} 6 & 5 & \lambda \\ 0 & -1 & 7 \\ 1 & 2 & -3 \end{vmatrix} \]

Calculating the determinant using the rule of Sarrus or cofactor expansion:

1. Expand along the first row:

\[ \text{det}(A) = 6 \begin{vmatrix} -1 & 7 \\ 2 & -3 \end{vmatrix} - 5 \begin{vmatrix} 0 & 7 \\ 1 & -3 \end{vmatrix} + \lambda \begin{vmatrix} 0 & -1 \\ 1 & 2 \end{vmatrix} \]

2. Calculate the 2x2 determinants:

\[ \begin{vmatrix} -1 & 7 \\ 2 & -3 \end{vmatrix} = (-1)(-3) - (7)(2) = 3 - 14 = -11 \]

\[ \begin{vmatrix} 0 & 7 \\ 1 & -3 \end{vmatrix} = (0)(-3) - (7)(1) = 0 - 7 = -7 \]

\[ \begin{vmatrix} 0 & -1 \\ 1 & 2 \end{vmatrix} = (0)(2) - (-1)(1) = 0 + 1 = 1 \]

3. Substitute these values back into the determinant equation:

\[ \text{det}(A) = 6(-11) - 5(-7) + \lambda(1) \]

\[ = -66 + 35 + \lambda \]

\[ = \lambda - 31 \]

4. Set the determinant equal to zero for non-trivial solutions:

\[ \lambda - 31 = 0 \]

5. Solve for \(\lambda\):

\[ \lambda = 31 \]

Thus, the value of \(\lambda\) for which the system has non-trivial solutions is \(\lambda = 31\).