if the system of equations
`{:(ax+y+2z=0),(x+2y+z=b),(2x+y+az=0):}`
has no solution then (a+b) can be equals to :

Options

A. A)`-1`
B. B) 2
C. C) 3
D. D) 4

Answer

To determine the values of \( a + b \) for the system of equations given that it has no solution, we will follow these steps:

### Step 1: Write down the system of equations
The given equations are:
1. \( ax + y + 2z = 0 \)
2. \( x + 2y + z = b \)
3. \( 2x + y + az = 0 \)

### Step 2: Formulate the coefficient matrix and calculate the determinant
The coefficient matrix \( A \) is:
\[ A = \begin{bmatrix} a & 1 & 2 \\ 1 & 2 & 1 \\ 2 & 1 & a \end{bmatrix} \]
To find the determinant \( \Delta \) (denoted as \( |A| \)), we calculate:
\[ \Delta = \begin{vmatrix} a & 1 & 2 \\ 1 & 2 & 1 \\ 2 & 1 & a \end{vmatrix} \]

### Step 3: Calculate the determinant using cofactor expansion
Expanding along the first row:
\[ \Delta = a \begin{vmatrix} 2 & 1 \\ 1 & a \end{vmatrix} - 1 \begin{vmatrix} 1 & 1 \\ 2 & a \end{vmatrix} + 2 \begin{vmatrix} 1 & 2 \\ 2 & 1 \end{vmatrix} \]
Calculating the 2x2 determinants:
1. \( \begin{vmatrix} 2 & 1 \\ 1 & a \end{vmatrix} = 2a - 1 \)
2. \( \begin{vmatrix} 1 & 1 \\ 2 & a \end{vmatrix} = a - 2 \)
3. \( \begin{vmatrix} 1 & 2 \\ 2 & 1 \end{vmatrix} = 1 - 4 = -3 \)

Substituting back into the determinant:
\[ \Delta = a(2a - 1) - (a - 2) - 6 \]
Simplifying:
\[ \Delta = 2a^2 - a - a + 2 - 6 = 2a^2 - 2a - 4 \]
Setting the determinant to zero for no solution:
\[ 2a^2 - 2a - 4 = 0 \]
Dividing by 2:
\[ a^2 - a - 2 = 0 \]

### Step 4: Factor the quadratic equation
Factoring:
\[ (a - 2)(a + 1) = 0 \]
Thus, the solutions for \( a \) are:
\[ a = 2 \quad \text{or} \quad a = -1 \]

### Step 5: Check the second condition for no solution
For the system to have no solution, at least one of \( \Delta_x, \Delta_y, \Delta_z \) must not be zero. We will compute \( \Delta_x, \Delta_y, \Delta_z \) for both values of \( a \).

1. **For \( a = 2 \)**:
\[ \Delta_x = \begin{vmatrix} 0 & 1 & 2 \\ b & 2 & 1 \\ 0 & 1 & 2 \end{vmatrix} \]
This determinant will yield zero since two rows are identical.

2. **For \( a = -1 \)**:
We will compute \( \Delta_x \) again:
\[ \Delta_x = \begin{vmatrix} 0 & 1 & 2 \\ b & 2 & 1 \\ 0 & 1 & -1 \end{vmatrix} \]
Expanding this determinant will yield a non-zero value for \( b \neq 0 \).

### Step 6: Conclusion on values of \( a + b \)
If \( a = -1 \), then:
\[ a + b = -1 + b \]
Since \( b \) can take any value except 0 (to ensure at least one of \( \Delta_x, \Delta_y, \Delta_z \) is non-zero), the possible values of \( a + b \) can be:
- If \( b = 2 \), then \( a + b = 1 \)
- If \( b = 3 \), then \( a + b = 2 \)
- If \( b = 4 \), then \( a + b = 3 \)

Thus, the possible values for \( a + b \) are 2, 3, and 4.

### Final Answer:
The values of \( a + b \) can be \( 2, 3, \) or \( 4 \).