If the system of linear equations `x+ay+z=3 and x+2y+2z=6 and x+5y+3z=b` has no solution, then (a) `a=-1,b=9 (2) a=-1,b != 9` (3) `a != -1 , b =9` (4) `a=1,b != 9`

Options

A. a=-1,b=9
B. `a ne -1, b=9`
C. `a=1, b ne9`
D. `a=-1 , b ne 9`

Answer

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

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

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

The coefficient matrix \( A \) for the system of equations is:

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

To find the condition for no solution, we need to calculate the determinant of this matrix and set it equal to zero.

\[ \text{det}(A) = \begin{vmatrix} 1 & a & 1 \\ 1 & 2 & 2 \\ 1 & 5 & 3 \end{vmatrix} \]

### Step 2: Calculate the Determinant

Using the determinant formula for a 3x3 matrix:

\[ \text{det}(A) = 1 \cdot \begin{vmatrix} 2 & 2 \\ 5 & 3 \end{vmatrix} - a \cdot \begin{vmatrix} 1 & 2 \\ 1 & 3 \end{vmatrix} + 1 \cdot \begin{vmatrix} 1 & 2 \\ 1 & 5 \end{vmatrix} \]

Calculating the 2x2 determinants:

1. \( \begin{vmatrix} 2 & 2 \\ 5 & 3 \end{vmatrix} = (2)(3) - (2)(5) = 6 - 10 = -4 \)
2. \( \begin{vmatrix} 1 & 2 \\ 1 & 3 \end{vmatrix} = (1)(3) - (2)(1) = 3 - 2 = 1 \)
3. \( \begin{vmatrix} 1 & 2 \\ 1 & 5 \end{vmatrix} = (1)(5) - (2)(1) = 5 - 2 = 3 \)

Now substituting these values back into the determinant:

\[ \text{det}(A) = 1(-4) - a(1) + 1(3) = -4 - a + 3 = -a - 1 \]

Setting the determinant to zero for the system to have no solution:

\[ -a - 1 = 0 \implies a = -1 \]

### Step 3: Analyze the Third Equation

Now we substitute \( a = -1 \) into the third equation and analyze the condition for \( b \).

The modified equations become:

1. \( x - y + z = 3 \)
2. \( x + 2y + 2z = 6 \)
3. \( x + 5y + 3z = b \)

We need to check the consistency of these equations. We can express the first two equations in terms of \( z \):

From Equation 1:

\[ z = 3 - x + y \]

Substituting \( z \) into Equation 2:

\[ x + 2y + 2(3 - x + y) = 6 \]
\[ x + 2y + 6 - 2x + 2y = 6 \]
\[ - x + 4y + 6 = 6 \implies -x + 4y = 0 \implies x = 4y \]

Substituting \( x = 4y \) into the expression for \( z \):

\[ z = 3 - 4y + y = 3 - 3y \]

Now substituting \( x = 4y \) and \( z = 3 - 3y \) into Equation 3:

\[ 4y + 5y + (3 - 3y) = b \]
\[ 4y + 5y + 3 - 3y = b \implies 6y + 3 = b \]

For the system to have no solution, \( b \) must not equal \( 9 \):

\[ 6y + 3 \neq 9 \implies 6y \neq 6 \implies y \neq 1 \]

### Conclusion

Thus, we conclude that:

- \( a = -1 \)
- \( b \neq 9 \)

The correct option is \( (2) \, a = -1, b \neq 9 \).