The values of `lambda` and b for which the equations `x+y+z=3`, `x+3y+2z=6`, and `x+lambday+3z=b` have

Options

A. (a) a uniqe solution if `lambda ne 5, b in R`
B. (b) no solution if `lambda ne 5,b=9`
C. (c) infinite many solution `lambda=5,b=9`
D. (d) None of the above

Answer

To find the values of \( \lambda \) and \( b \) for which the equations

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

have unique solutions, no solutions, or infinitely many solutions, we can analyze the system using the concept of determinants.

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

The system of equations can be represented in matrix form as:

\[ \begin{bmatrix} 1 & 1 & 1 \\ 1 & 3 & 2 \\ 1 & \lambda & 3 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 3 \\ 6 \\ b \end{bmatrix} \]

### Step 2: Calculate the determinant of the coefficient matrix

The coefficient matrix is

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

To find the determinant of \( A \):

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

Calculating the minors:

1. \( \begin{vmatrix} 3 & 2 \\ \lambda & 3 \end{vmatrix} = 3 \cdot 3 - 2 \cdot \lambda = 9 - 2\lambda \)
2. \( \begin{vmatrix} 1 & 2 \\ 1 & 3 \end{vmatrix} = 1 \cdot 3 - 2 \cdot 1 = 3 - 2 = 1 \)
3. \( \begin{vmatrix} 1 & 3 \\ 1 & \lambda \end{vmatrix} = 1 \cdot \lambda - 3 \cdot 1 = \lambda - 3 \)

Putting it all together:

\[ \text{det}(A) = 1(9 - 2\lambda) - 1 + 1(\lambda - 3) = 9 - 2\lambda - 1 + \lambda - 3 = 5 - \lambda \]

### Step 3: Analyze conditions for unique solutions, no solutions, and infinitely many solutions

1. **Unique Solutions**: The system has a unique solution if \( \text{det}(A) \neq 0 \).

\[ 5 - \lambda \neq 0 \implies \lambda \neq 5 \]

For any value of \( b \), the system will have a unique solution.

2. **No Solutions**: The system has no solution if \( \text{det}(A) = 0 \) and the augmented matrix has a different rank.

\[ 5 - \lambda = 0 \implies \lambda = 5 \]

For no solution, we need to check the augmented matrix:

- The third equation becomes \( x + 5y + 3z = b \).
- The rank of the coefficient matrix is 2 (since one row becomes dependent).
- The rank of the augmented matrix must be 3, which requires \( b \neq 9 \) (as shown in the video).

3. **Infinitely Many Solutions**: The system has infinitely many solutions if \( \text{det}(A) = 0 \) and the rank of the augmented matrix equals the rank of the coefficient matrix.

\[ 5 - \lambda = 0 \implies \lambda = 5 \]
\[ b = 9 \]

### Summary of Conditions

- **Unique Solution**: \( \lambda \neq 5 \) and \( b \in \mathbb{R} \)
- **No Solution**: \( \lambda = 5 \) and \( b \neq 9 \)
- **Infinitely Many Solutions**: \( \lambda = 5 \) and \( b = 9 \)