Consider the system of equations
`2x+lambday+6z=8`
`x+2y+muz=5`
`x+y+3z=4`
The system of equations has :
Exactly one solution if :

Options

A. `lambda=2,mu=3`
B. `lambdane2,mu=3`
C. `lambdane2,mune3`
D. `lambda=2,mu in R`

Answer

To determine the conditions under which the given system of equations has exactly one solution, we need to analyze the determinant of the coefficient matrix. 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 coefficient matrix
The coefficient matrix \( A \) for the system can be represented as:

\[ A = \begin{bmatrix} 2 & \lambda & 6 \\ 1 & 2 & \mu \\ 1 & 1 & 3 \end{bmatrix} \]

### Step 2: Calculate the determinant of the matrix
To find the determinant of matrix \( A \), we can use the formula for the determinant of a 3x3 matrix:

\[ \text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \]

For our matrix:

- \( a = 2, b = \lambda, c = 6 \)
- \( d = 1, e = 2, f = \mu \)
- \( g = 1, h = 1, i = 3 \)

Thus, the determinant becomes:

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

### Step 3: Simplify the determinant expression
Calculating each term:

1. \( 2(6 - \mu) = 12 - 2\mu \)
2. \( -\lambda(3 - \mu) = -3\lambda + \lambda\mu \)
3. \( 6(1 - 2) = 6(-1) = -6 \)

Combining these, we have:

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

This simplifies to:

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

### Step 4: Set the determinant to zero for one solution
For the system to have exactly one solution, the determinant must be non-zero. Thus, we set:

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

### Step 5: Rearranging the equation
Rearranging gives us:

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

### Step 6: Factor the equation
We can factor this equation as follows:

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

### Step 7: Find the conditions
From the factored form, we can derive the conditions:

1. \( \lambda = 2 \)
2. \( \mu = 3 \)

Thus, the system of equations has exactly one solution when \( \lambda = 2 \) and \( \mu = 3 \).

### Final Answer
The values of \( \lambda \) and \( \mu \) for which the system has exactly one solution are:

\[ \lambda = 2 \quad \text{and} \quad \mu = 3 \]

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