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 unique solutions ?

Options

A. `lambda = 5 and mu = 9`
B. `lambda ne 9 and mu = 7` only
C. `lambda ne 5 and mu` has any real value
D. `lambda` has any real value and `mu ne 9`

Explanation

To determine the condition under which the given system of linear equations has a unique solution, we need to analyze the determinant of the coefficient matrix. The system of equations is:

1. \( 2x + 3y + 5z = 9 \)
2. \( 7x + 3y - 2z = 8 \)
3. \( 2x + 3y + \lambda z = \mu \)

### Step 1: Form the Coefficient Matrix
The coefficient matrix \( A \) for the system can be represented as:

\[ A = \begin{bmatrix} 2 & 3 & 5 \\ 7 & 3 & -2 \\ 2 & 3 & \lambda \end{bmatrix} \]

### Step 2: Calculate the Determinant of the Coefficient Matrix
To find the condition for a unique solution, we need to calculate the determinant of matrix \( A \) and set it not equal to zero:

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

Using the determinant formula for a 3x3 matrix:

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

where \( a, b, c \) are the elements of the first row, and \( d, e, f, g, h, i \) are the elements of the remaining rows.

Substituting the values:

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

Calculating each of these 2x2 determinants:

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

2. \( \begin{vmatrix} 7 & -2 \\ 2 & \lambda \end{vmatrix} = 7\lambda - (-2)(2) = 7\lambda + 4 \)

3. \( \begin{vmatrix} 7 & 3 \\ 2 & 3 \end{vmatrix} = 7 \cdot 3 - 3 \cdot 2 = 21 - 6 = 15 \)

Now substituting back into the determinant:

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

Expanding this:

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

Combining like terms:

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

### Step 3: Set the Determinant Not Equal to Zero
For the system to have a unique solution, we require:

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

Solving for \( \lambda \):

\[ -15\lambda \neq -75 \implies \lambda \neq 5 \]

### Conclusion
Thus, the condition under which the system of equations has a unique solution is:

\[ \lambda \neq 5 \]