If the system of equations 2x+ay+6z=8, x+2y+z=5, 2x+ay+3z=4 has a unique solution then 'a' cannot be equal to :

Options

A. 2
B. 3
C. 4
D. 5

Answer

To determine the value of 'a' for which the system of equations has a unique solution, we need to analyze the coefficient matrix and its determinant. The given equations are:

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

### Step 1: Write the Coefficient Matrix
The coefficient matrix \( A \) can be represented as:
\[ A = \begin{bmatrix} 2 & a & 6 \\ 1 & 2 & 1 \\ 2 & a & 3 \end{bmatrix} \]

### Step 2: Calculate the Determinant of the Coefficient Matrix
To find the determinant of matrix \( A \), we can use the formula for the determinant of a 3x3 matrix:
\[ \text{det}(A) = a_{11}(a_{22}a_{33} - a_{23}a_{32}) - a_{12}(a_{21}a_{33} - a_{23}a_{31}) + a_{13}(a_{21}a_{32} - a_{22}a_{31}) \]

Substituting the values from our matrix:
\[ \text{det}(A) = 2(2 \cdot 3 - 1 \cdot a) - a(1 \cdot 3 - 1 \cdot 2) + 6(1 \cdot a - 2 \cdot 2) \]

### Step 3: Simplify the Determinant
Calculating each term:
1. First term: \( 2(6 - a) = 12 - 2a \)
2. Second term: \( -a(3 - 2) = -a \)
3. Third term: \( 6(a - 4) = 6a - 24 \)

Combining these:
\[ \text{det}(A) = (12 - 2a) - a + (6a - 24) \]
\[ = 12 - 2a - a + 6a - 24 \]
\[ = (12 - 24) + (-2a - a + 6a) \]
\[ = -12 + 3a \]

### Step 4: Set the Determinant Not Equal to Zero
For the system to have a unique solution, the determinant must not be equal to zero:
\[ -12 + 3a \neq 0 \]

### Step 5: Solve for 'a'
Setting the determinant equal to zero to find the critical value:
\[ -12 + 3a = 0 \]
\[ 3a = 12 \]
\[ a = 4 \]

### Conclusion
Thus, for the system of equations to have a unique solution, 'a' cannot be equal to 4.