For the system of linear equations `2x+3y+5z=9, 7x+3y-2z=8 and 2x+3y+lambda z=mu`
Under what condition does the above system of equations have infinitely many solutions ?

Options

A. `lambda = 5 and mu ne 9`
B. `lambda = 5 and mu = 9`
C. `lambda = 9 and mu = 5`
D. `lambda = 9 and mu ne 5`

Explanation

To determine the conditions under which the given system of linear equations has infinitely many solutions, we will analyze the equations step by step.

### Given Equations:
1. \( 2x + 3y + 5z = 9 \) (Equation 1)
2. \( 7x + 3y - 2z = 8 \) (Equation 2)
3. \( 2x + 3y + \lambda z = \mu \) (Equation 3)

### Step 1: Form the Coefficient Matrix A and the Constant Matrix B
The coefficient matrix \( A \) and the constant matrix \( B \) can be represented as follows:

\[ A = \begin{bmatrix} 2 & 3 & 5 \\ 7 & 3 & -2 \\ 2 & 3 & \lambda \end{bmatrix}, \quad B = \begin{bmatrix} 9 \\ 8 \\ \mu \end{bmatrix} \]

### Step 2: Find the Determinant of Matrix A
To find the condition for infinitely many solutions, we need to compute the determinant of matrix \( A \). The system will have infinitely many solutions if the determinant of \( A \) is zero and the system is consistent.

\[ \text{det}(A) = \begin{vmatrix} 2 & 3 & 5 \\ 7 & 3 & -2 \\ 2 & 3 & \lambda \end{vmatrix} \]

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

\[ \text{det}(A) = 2 \begin{vmatrix} 3 & -2 \\ 3 & \lambda \end{vmatrix} - 3 \begin{vmatrix} 7 & -2 \\ 2 & \lambda \end{vmatrix} + 5 \begin{vmatrix} 7 & 3 \\ 2 & 3 \end{vmatrix} \]

Calculating the 2x2 determinants:

1. \( \begin{vmatrix} 3 & -2 \\ 3 & \lambda \end{vmatrix} = 3\lambda + 6 \)
2. \( \begin{vmatrix} 7 & -2 \\ 2 & \lambda \end{vmatrix} = 7\lambda + 4 \)
3. \( \begin{vmatrix} 7 & 3 \\ 2 & 3 \end{vmatrix} = 21 - 6 = 15 \)

Substituting back into the determinant:

\[ \text{det}(A) = 2(3\lambda + 6) - 3(7\lambda + 4) + 5(15) \]

Expanding this:

\[ = 6\lambda + 12 - 21\lambda - 12 + 75 \]
\[ = -15\lambda + 75 \]

### Step 3: Set the Determinant to Zero
For the system to have infinitely many solutions, we set the determinant to zero:

\[ -15\lambda + 75 = 0 \]

Solving for \( \lambda \):

\[ 15\lambda = 75 \implies \lambda = 5 \]

### Step 4: Check Consistency
Now we need to ensure that the system is consistent when \( \lambda = 5 \). We substitute \( \lambda = 5 \) into the third equation:

\[ 2x + 3y + 5z = \mu \]

Now we need to check if the equations are consistent. We can do this by substituting the values of \( \lambda \) and checking the rank of the augmented matrix.

### Conclusion
The system of equations will have infinitely many solutions if:

\[ \lambda = 5 \quad \text{and} \quad \mu = 9 \]