The system of equations `x+y+z=6, x+2y + 3z= 10, x+2y + lamdaz=mu` has no solution for

Options

A. `lamda ne 3 , mu = 10`
B. `lamda = 3 , mu ne 0`
C. `lamda ne 3 , mu ne 0`
D. none of these

Answer

To determine the values of \( \lambda \) and \( \mu \) for which the system of equations has no solution, we will analyze the given equations step by step.

### Given Equations:
1. \( x + y + z = 6 \) (Equation 1)
2. \( x + 2y + 3z = 10 \) (Equation 2)
3. \( x + 2y + \lambda z = \mu \) (Equation 3)

### Step 1: Form the Coefficient Matrix
We can express the system of equations in matrix form. The coefficient matrix \( A \) for the system is:
\[ A = \begin{bmatrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 2 & \lambda \end{bmatrix} \]

### Step 2: Calculate the Determinant
To find the values of \( \lambda \) for which the system has no solution, we need to calculate the determinant of the coefficient matrix \( A \) and set it to zero.

The determinant \( |A| \) can be calculated as follows:
\[ |A| = \begin{vmatrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 2 & \lambda \end{vmatrix} \]

Using the determinant formula:
\[ |A| = 1 \begin{vmatrix} 2 & 3 \\ 2 & \lambda \end{vmatrix} - 1 \begin{vmatrix} 1 & 3 \\ 1 & \lambda \end{vmatrix} + 1 \begin{vmatrix} 1 & 2 \\ 1 & 2 \end{vmatrix} \]

Calculating the 2x2 determinants:
1. \( \begin{vmatrix} 2 & 3 \\ 2 & \lambda \end{vmatrix} = 2\lambda - 6 \)
2. \( \begin{vmatrix} 1 & 3 \\ 1 & \lambda \end{vmatrix} = \lambda - 3 \)
3. \( \begin{vmatrix} 1 & 2 \\ 1 & 2 \end{vmatrix} = 0 \)

Putting it all together:
\[ |A| = 1(2\lambda - 6) - 1(\lambda - 3) + 0 \]
\[ |A| = 2\lambda - 6 - \lambda + 3 = \lambda - 3 \]

### Step 3: Set the Determinant to Zero
For the system to have no solution, we set the determinant equal to zero:
\[ \lambda - 3 = 0 \]
Thus,
\[ \lambda = 3 \]

### Step 4: Find the Value of \( \mu \)
Now we substitute \( \lambda = 3 \) into the third equation and find \( \mu \) such that the system remains inconsistent.

The modified third equation becomes:
\[ x + 2y + 3z = \mu \]

### Step 5: Form the New Determinant
Now we need to find the determinant of the new system:
\[ \begin{vmatrix} 6 & 10 & \mu \\ 1 & 2 & 3 \\ 1 & 2 & 3 \end{vmatrix} \]

Calculating this determinant:
\[ |B| = 6 \begin{vmatrix} 2 & 3 \\ 2 & 3 \end{vmatrix} - 10 \begin{vmatrix} 1 & 3 \\ 1 & 3 \end{vmatrix} + \mu \begin{vmatrix} 1 & 2 \\ 1 & 2 \end{vmatrix} \]

The 2x2 determinants are:
1. \( \begin{vmatrix} 2 & 3 \\ 2 & 3 \end{vmatrix} = 0 \)
2. \( \begin{vmatrix} 1 & 3 \\ 1 & 3 \end{vmatrix} = 0 \)
3. \( \begin{vmatrix} 1 & 2 \\ 1 & 2 \end{vmatrix} = 0 \)

Thus,
\[ |B| = 6(0) - 10(0) + \mu(0) = 0 \]

This means \( \mu \) can be any value except for the case where the equations are consistent.

### Conclusion
For the system to have no solution, we have:
- \( \lambda = 3 \)
- \( \mu \) can be any value except for values that would make the equations consistent.