For what values of a and b the system of equations
2 x + ay + 6 z = 8
x+2y+bz=5
x + y + 3 z = 4 , has a unique solution ?

Options

A. a = 2 , b = 3
B. `a != 2, b != 3 `
C. a = - 2, b = - 3
D. None of these

Answer

To find the values of \( a \) and \( b \) for which the system of equations has a unique solution, we need to analyze the determinant of the coefficient matrix formed by the system of equations. The system of equations is:

1. \( 2x + ay + 6z = 8 \)
2. \( x + 2y + bz = 5 \)
3. \( x + y + 3z = 4 \)

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

\[ A = \begin{bmatrix} 2 & a & 6 \\ 1 & 2 & b \\ 1 & 1 & 3 \end{bmatrix} \]

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

\[ \text{det}(A) = 2 \begin{vmatrix} 2 & b \\ 1 & 3 \end{vmatrix} - a \begin{vmatrix} 1 & b \\ 1 & 3 \end{vmatrix} + 6 \begin{vmatrix} 1 & 2 \\ 1 & 1 \end{vmatrix} \]

Calculating the minors:

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

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

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

Substituting these back into the determinant formula:

\[ \text{det}(A) = 2(6 - b) - a(3 - b) + 6(-1) \]

Expanding this gives:

\[ \text{det}(A) = 12 - 2b - 3a + ab - 6 \]

Simplifying further:

\[ \text{det}(A) = 6 - 2b - 3a + ab \]

### Step 3: Set the determinant not equal to zero
For the system to have a unique solution, the determinant must not be equal to zero:

\[ 6 - 2b - 3a + ab \neq 0 \]

### Step 4: Rearranging the equation
Rearranging gives:

\[ ab - 3a - 2b + 6 \neq 0 \]

### Step 5: Finding values for \( a \) and \( b \)
To ensure that the determinant is non-zero, we can analyze specific cases. For instance, if we set \( a = 2 \) and \( b = 3 \):

\[ 2(3) - 3(2) - 2(3) + 6 = 6 - 6 - 6 + 6 = 0 \]

Thus, for \( a = 2 \) and \( b = 3 \), the determinant equals zero, which means these values are not allowed.

### Conclusion
The values of \( a \) and \( b \) for which the system has a unique solution are:

\[ a \neq 2 \quad \text{and} \quad b \neq 3 \]