Consider the system of equations
`{:(2x+lambday+6z=8),(x+2y+muz=5),(x+y+3z=4):}`
The system of equations has :
Infinitely many solutions if :

Options

A. `lambda=2, mu=3`
B. `lambdane 2 , mu=3`
C. `lambda ne 2, mu ne 3 `
D. `lambda =2 , mu in R `

Answer

To determine the conditions under which the given system of equations has infinitely many solutions, we need to analyze the system using determinants. The system of equations is:

1. \( 2x + \lambda y + 6z = 8 \)
2. \( x + 2y + \mu z = 5 \)
3. \( x + y + 3z = 4 \)

### Step 1: Write the system in matrix form

The system can be represented in the form of a matrix equation \( AX = B \), where:

\[ A = \begin{bmatrix} 2 & \lambda & 6 \\ 1 & 2 & \mu \\ 1 & 1 & 3 \end{bmatrix}, \quad X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, \quad B = \begin{bmatrix} 8 \\ 5 \\ 4 \end{bmatrix} \]

### Step 2: Calculate the determinant of matrix \( A \)

To find the conditions for infinitely many solutions, we need to calculate the determinant of matrix \( A \) and set it equal to zero:

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

Calculating the determinant using the formula for a 3x3 matrix:

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

Calculating the 2x2 determinants:

1. \( \begin{vmatrix} 2 & \mu \\ 1 & 3 \end{vmatrix} = (2)(3) - (1)(\mu) = 6 - \mu \)

2. \( \begin{vmatrix} 1 & \mu \\ 1 & 3 \end{vmatrix} = (1)(3) - (1)(\mu) = 3 - \mu \)

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

Putting it all together:

\[ \text{det}(A) = 2(6 - \mu) - \lambda(3 - \mu) + 6(-1) \]

Expanding this gives:

\[ \text{det}(A) = 12 - 2\mu - 3\lambda + \lambda\mu - 6 \]

Thus:

\[ \text{det}(A) = 6 - 2\mu - 3\lambda + \lambda\mu \]

### Step 3: Set the determinant equal to zero

For the system to have infinitely many solutions, we set the determinant equal to zero:

\[ 6 - 2\mu - 3\lambda + \lambda\mu = 0 \]

### Step 4: Rearranging the equation

Rearranging gives:

\[ \lambda\mu - 3\lambda - 2\mu + 6 = 0 \]

### Step 5: Factor the equation

We can factor this equation:

\[ (\lambda - 2)(\mu - 3) = 0 \]

### Step 6: Find the values of \( \lambda \) and \( \mu \)

From the factored equation, we have two conditions:

1. \( \lambda - 2 = 0 \) → \( \lambda = 2 \)
2. \( \mu - 3 = 0 \) → \( \mu = 3 \)

### Conclusion

The system of equations has infinitely many solutions if:

- \( \lambda = 2 \) and \( \mu = 3 \)