The values of a and b, for which the system of equations
`2x+3y+6z=8`
`x+2y+az=5`
`3x+5y+9z=b`
has no solution ,are :

Options

A. `a=3,bne13`
B. `ane3,bne13`
C. `ane3,b=3`
D. `a=3,b=13`

Answer

To find the values of \( a \) and \( b \) for which the system of equations has no solution, we need to analyze the system of equations given:

1. \( 2x + 3y + 6z = 8 \) (Equation 1)
2. \( x + 2y + az = 5 \) (Equation 2)
3. \( 3x + 5y + 9z = b \) (Equation 3)

### Step 1: Form the Coefficient Matrix and Find the Determinant

We first form the coefficient matrix from the coefficients of \( x, y, z \) in the equations:

\[ A = \begin{bmatrix} 2 & 3 & 6 \\ 1 & 2 & a \\ 3 & 5 & 9 \end{bmatrix} \]

To find the condition for no solution, we need to calculate the determinant \( \Delta \) of this matrix \( A \).

### Step 2: Calculate the Determinant \( \Delta \)

Using the determinant formula for a \( 3 \times 3 \) matrix, we have:

\[ \Delta = 2 \begin{vmatrix} 2 & a \\ 5 & 9 \end{vmatrix} - 3 \begin{vmatrix} 1 & a \\ 3 & 9 \end{vmatrix} + 6 \begin{vmatrix} 1 & 2 \\ 3 & 5 \end{vmatrix} \]

Calculating each of these \( 2 \times 2 \) determinants:

1. \( \begin{vmatrix} 2 & a \\ 5 & 9 \end{vmatrix} = (2)(9) - (5)(a) = 18 - 5a \)

2. \( \begin{vmatrix} 1 & a \\ 3 & 9 \end{vmatrix} = (1)(9) - (3)(a) = 9 - 3a \)

3. \( \begin{vmatrix} 1 & 2 \\ 3 & 5 \end{vmatrix} = (1)(5) - (2)(3) = 5 - 6 = -1 \)

Substituting these back into the determinant formula:

\[ \Delta = 2(18 - 5a) - 3(9 - 3a) + 6(-1) \]

Expanding this:

\[ \Delta = 36 - 10a - 27 + 9a - 6 \]

Combining like terms:

\[ \Delta = (36 - 27 - 6) + (-10a + 9a) = 3 - a \]

### Step 3: Set the Determinant to Zero for No Solution

For the system to have no solution, we need \( \Delta = 0 \):

\[ 3 - a = 0 \implies a = 3 \]

### Step 4: Find the Condition for \( b \)

Next, we need to find the value of \( b \) such that the system still does not have a solution. We can do this by checking the determinant of the modified matrix where we replace the last column with the constants from the equations.

The modified matrix is:

\[ B = \begin{bmatrix} 2 & 3 & 8 \\ 1 & 2 & 5 \\ 3 & 5 & b \end{bmatrix} \]

Calculating the determinant \( \Delta_3 \):

\[ \Delta_3 = 2 \begin{vmatrix} 2 & 5 \\ 5 & b \end{vmatrix} - 3 \begin{vmatrix} 1 & 5 \\ 3 & b \end{vmatrix} + 8 \begin{vmatrix} 1 & 2 \\ 3 & 5 \end{vmatrix} \]

Calculating each of these determinants:

1. \( \begin{vmatrix} 2 & 5 \\ 5 & b \end{vmatrix} = (2)(b) - (5)(5) = 2b - 25 \)

2. \( \begin{vmatrix} 1 & 5 \\ 3 & b \end{vmatrix} = (1)(b) - (3)(5) = b - 15 \)

3. \( \begin{vmatrix} 1 & 2 \\ 3 & 5 \end{vmatrix} = -1 \) (as calculated before)

Substituting these back into the determinant formula:

\[ \Delta_3 = 2(2b - 25) - 3(b - 15) + 8(-1) \]

Expanding this:

\[ \Delta_3 = 4b - 50 - 3b + 45 - 8 \]

Combining like terms:

\[ \Delta_3 = (4b - 3b) + (-50 + 45 - 8) = b - 13 \]

### Step 5: Set the Condition for \( b \)

For the system to have no solution, we need \( \Delta_3 \neq 0 \):

\[ b - 13 \neq 0 \implies b \neq 13 \]

### Final Answer

Thus, the values of \( a \) and \( b \) for which the system of equations has no solution are:

\[ \boxed{(3, b \neq 13)} \]