The system of linear equations
`lambda`x + y + z = 3
x - y - 2z = 6
`-x + y + z = mu` has

Options

A. infinite number of solutions for `lambda ne - 1 ` and all `mu`
B. infinite number of solutions for `lambda` = -1 and `mu` = 3
C. no solution for `lambda ne - 1`
D. unique solution for `lambda` = -1 and `mu` = 3

Answer

To solve the system of linear equations given by:

1. \( \lambda x + y + z = 3 \)
2. \( x - y - 2z = 6 \)
3. \( -x + y + z = \mu \)

we will analyze the conditions for the existence of solutions: infinite solutions, no solution, or a unique solution.

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

We can express the system in the form \( AX = B \), where:

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

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

To find the conditions for the solutions, we first need to calculate the determinant of matrix \( A \):

\[ \Delta = \begin{vmatrix} \lambda & 1 & 1 \\ 1 & -1 & -2 \\ -1 & 1 & 1 \end{vmatrix} \]

Expanding this determinant, we have:

\[ \Delta = \lambda \begin{vmatrix} -1 & -2 \\ 1 & 1 \end{vmatrix} - 1 \begin{vmatrix} 1 & -2 \\ -1 & 1 \end{vmatrix} + 1 \begin{vmatrix} 1 & -1 \\ -1 & 1 \end{vmatrix} \]

Calculating the 2x2 determinants:

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

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

3. \( \begin{vmatrix} 1 & -1 \\ -1 & 1 \end{vmatrix} = (1)(1) - (-1)(-1) = 1 - 1 = 0 \)

Substituting back into the determinant:

\[ \Delta = \lambda(1) - 1(-1) + 1(0) = \lambda + 1 \]

### Step 3: Conditions for solutions

1. **Infinite Solutions**: For the system to have infinite solutions, we need \( \Delta = 0 \) and at least one of the modified determinants \( \Delta_1, \Delta_2, \Delta_3 \) must also equal zero.
\[ \lambda + 1 = 0 \implies \lambda = -1 \]

2. **No Solution**: For the system to have no solution, we need \( \Delta = 0 \) and at least one of \( \Delta_1, \Delta_2, \Delta_3 \) must not equal zero.

3. **Unique Solution**: For a unique solution, we need \( \Delta \neq 0 \).
\[ \lambda + 1 \neq 0 \implies \lambda \neq -1 \]

### Step 4: Calculate \( \Delta_1 \)

To find \( \Delta_1 \), we replace the first column of \( A \) with \( B \):

\[ \Delta_1 = \begin{vmatrix} 3 & 1 & 1 \\ 6 & -1 & -2 \\ \mu & 1 & 1 \end{vmatrix} \]

Calculating \( \Delta_1 \):

\[ \Delta_1 = 3 \begin{vmatrix} -1 & -2 \\ 1 & 1 \end{vmatrix} - 1 \begin{vmatrix} 6 & -2 \\ \mu & 1 \end{vmatrix} + 1 \begin{vmatrix} 6 & -1 \\ \mu & 1 \end{vmatrix} \]

Calculating the 2x2 determinants:

1. \( \begin{vmatrix} -1 & -2 \\ 1 & 1 \end{vmatrix} = 1 \) (as calculated previously)

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

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

Substituting back into \( \Delta_1 \):

\[ \Delta_1 = 3(1) - 1(6 + 2\mu) + 1(6 + \mu) = 3 - 6 - 2\mu + 6 + \mu = 3 - \mu \]

### Step 5: Conditions for infinite solutions

For infinite solutions, we need:

1. \( \lambda = -1 \)
2. \( \Delta_1 = 0 \) which gives us \( 3 - \mu = 0 \implies \mu = 3 \)

### Conclusion

- For infinite solutions: \( \lambda = -1 \) and \( \mu = 3 \).
- For no solution: \( \lambda = -1 \) and \( \mu \neq 3 \).
- For a unique solution: \( \lambda \neq -1 \).