Find the value of `lamda and mu` for which the system of equations
`x+y+z=6`
`3x+5y+5z=26`
`x+2y+lamdaz=mu`
has no solution

Options

A. `lamda=2 , mu ne 10`
B. `lamda ne 2 , mu = 10`
C. `lamda ne 3 , mu = 10`
D. `lamda ne 2 , mu ne 10`

Answer

To find the values of \( \lambda \) and \( \mu \) for which the given system of equations has no solution, we will analyze the system using determinants.

The system of equations is:
1. \( x + y + z = 6 \) (Equation 1)
2. \( 3x + 5y + 5z = 26 \) (Equation 2)
3. \( x + 2y + \lambda z = \mu \) (Equation 3)

### Step 1: Form the coefficient matrix and find the determinant

The coefficient matrix \( A \) of the system is:
\[ A = \begin{bmatrix} 1 & 1 & 1 \\ 3 & 5 & 5 \\ 1 & 2 & \lambda \end{bmatrix} \]

We need to find the determinant \( D \) of this matrix:
\[ D = \begin{vmatrix} 1 & 1 & 1 \\ 3 & 5 & 5 \\ 1 & 2 & \lambda \end{vmatrix} \]

### Step 2: Calculate the determinant

Using the determinant formula for a 3x3 matrix:
\[ D = 1 \cdot \begin{vmatrix} 5 & 5 \\ 2 & \lambda \end{vmatrix} - 1 \cdot \begin{vmatrix} 3 & 5 \\ 1 & \lambda \end{vmatrix} + 1 \cdot \begin{vmatrix} 3 & 5 \\ 1 & 2 \end{vmatrix} \]

Calculating the minors:
1. \( \begin{vmatrix} 5 & 5 \\ 2 & \lambda \end{vmatrix} = 5\lambda - 10 \)
2. \( \begin{vmatrix} 3 & 5 \\ 1 & \lambda \end{vmatrix} = 3\lambda - 5 \)
3. \( \begin{vmatrix} 3 & 5 \\ 1 & 2 \end{vmatrix} = 6 - 5 = 1 \)

Substituting these back into the determinant:
\[ D = 1(5\lambda - 10) - 1(3\lambda - 5) + 1(1) \]
\[ D = 5\lambda - 10 - 3\lambda + 5 + 1 \]
\[ D = 2\lambda - 4 \]

### Step 3: Set the determinant to zero

For the system to have no solution, the determinant must be zero:
\[ 2\lambda - 4 = 0 \]
\[ 2\lambda = 4 \]
\[ \lambda = 2 \]

### Step 4: Find \( \mu \) such that the system has no solution

Now we need to find \( \mu \) such that the system has no solution. We will calculate the determinants \( D_1 \) and \( D_2 \) and check their conditions.

#### Calculate \( D_1 \)

Replacing the first column of the coefficient matrix with the constants from the right-hand side:
\[ D_1 = \begin{vmatrix} 6 & 1 & 1 \\ 26 & 5 & 5 \\ \mu & 2 & 2 \end{vmatrix} \]

Calculating \( D_1 \):
\[ D_1 = 6 \cdot \begin{vmatrix} 5 & 5 \\ 2 & 2 \end{vmatrix} - 1 \cdot \begin{vmatrix} 26 & 5 \\ \mu & 2 \end{vmatrix} + 1 \cdot \begin{vmatrix} 26 & 5 \\ \mu & 2 \end{vmatrix} \]

Since the second and third minors are the same, we can simplify:
\[ D_1 = 6(0) - 0 + 0 = 0 \]

#### Calculate \( D_2 \)

Now we calculate \( D_2 \):
\[ D_2 = \begin{vmatrix} 1 & 1 & 1 \\ 3 & 5 & 5 \\ 1 & 2 & \mu \end{vmatrix} \]

Calculating \( D_2 \):
\[ D_2 = 1 \cdot \begin{vmatrix} 5 & 5 \\ 2 & \mu \end{vmatrix} - 1 \cdot \begin{vmatrix} 3 & 5 \\ 1 & \mu \end{vmatrix} + 1 \cdot \begin{vmatrix} 3 & 5 \\ 1 & 2 \end{vmatrix} \]

Calculating the minors:
1. \( \begin{vmatrix} 5 & 5 \\ 2 & \mu \end{vmatrix} = 5\mu - 10 \)
2. \( \begin{vmatrix} 3 & 5 \\ 1 & \mu \end{vmatrix} = 3\mu - 5 \)
3. \( \begin{vmatrix} 3 & 5 \\ 1 & 2 \end{vmatrix} = 1 \)

Putting it all together:
\[ D_2 = 1(5\mu - 10) - 1(3\mu - 5) + 1(1) \]
\[ D_2 = 5\mu - 10 - 3\mu + 5 + 1 \]
\[ D_2 = 2\mu - 4 \]

### Step 5: Set conditions for no solution

For the system to have no solution, we need \( D_2 \neq 0 \):
\[ 2\mu - 4 \neq 0 \]
\[ \mu \neq 2 \]

### Conclusion

Thus, the values of \( \lambda \) and \( \mu \) for which the system of equations has no solution are:
\[ \lambda = 2 \quad \text{and} \quad \mu \neq 2 \]