If the system of equations
`x-2y+3z=9`
`2x+y+z=b`
`x-7y+az=24` , has infinitely many solutions, then a - b is equal to ...........

Answer

To solve the given system of equations for the condition of infinitely many solutions, we need to analyze the determinant of the coefficients of the equations. The equations are:

1. \( x - 2y + 3z = 9 \) (Equation 1)
2. \( 2x + y + z = b \) (Equation 2)
3. \( x - 7y + az = 24 \) (Equation 3)

### Step 1: Set up the coefficient matrix and calculate the determinant

The coefficient matrix for the system of equations is:

\[ \begin{bmatrix} 1 & -2 & 3 \\ 2 & 1 & 1 \\ 1 & -7 & a \end{bmatrix} \]

To find the condition for infinitely many solutions, we need to set the determinant of this matrix to zero:

\[ \text{det} = \begin{vmatrix} 1 & -2 & 3 \\ 2 & 1 & 1 \\ 1 & -7 & a \end{vmatrix} \]

### Step 2: Calculate the determinant

Using the formula for the determinant of a 3x3 matrix:

\[ \text{det} = 1 \cdot \begin{vmatrix} 1 & 1 \\ -7 & a \end{vmatrix} - (-2) \cdot \begin{vmatrix} 2 & 1 \\ 1 & a \end{vmatrix} + 3 \cdot \begin{vmatrix} 2 & 1 \\ 1 & -7 \end{vmatrix} \]

Calculating the minors:

1. \( \begin{vmatrix} 1 & 1 \\ -7 & a \end{vmatrix} = 1 \cdot a - 1 \cdot (-7) = a + 7 \)

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

3. \( \begin{vmatrix} 2 & 1 \\ 1 & -7 \end{vmatrix} = 2(-7) - 1(1) = -14 - 1 = -15 \)

Substituting these back into the determinant:

\[ \text{det} = 1(a + 7) + 2(2a - 1) + 3(-15) \]

Expanding this:

\[ = a + 7 + 4a - 2 - 45 = 5a - 40 \]

### Step 3: Set the determinant to zero

For the system to have infinitely many solutions, we set the determinant to zero:

\[ 5a - 40 = 0 \]

Solving for \( a \):

\[ 5a = 40 \implies a = 8 \]

### Step 4: Find the value of \( b \)

Next, we need to find \( b \) using the condition that the determinant formed by replacing the last column with the constants also equals zero. The new matrix is:

\[ \begin{bmatrix} 1 & -2 & 9 \\ 2 & 1 & b \\ 1 & -7 & 24 \end{bmatrix} \]

Calculating this determinant:

\[ \text{det} = 1 \cdot \begin{vmatrix} 1 & b \\ -7 & 24 \end{vmatrix} - (-2) \cdot \begin{vmatrix} 2 & b \\ 1 & 24 \end{vmatrix} + 9 \cdot \begin{vmatrix} 2 & 1 \\ 1 & -7 \end{vmatrix} \]

Calculating the minors:

1. \( \begin{vmatrix} 1 & b \\ -7 & 24 \end{vmatrix} = 1 \cdot 24 - b \cdot (-7) = 24 + 7b \)

2. \( \begin{vmatrix} 2 & b \\ 1 & 24 \end{vmatrix} = 2 \cdot 24 - b \cdot 1 = 48 - b \)

3. \( \begin{vmatrix} 2 & 1 \\ 1 & -7 \end{vmatrix} = -14 - 1 = -15 \)

Substituting these back into the determinant:

\[ \text{det} = 1(24 + 7b) + 2(48 - b) + 9(-15) \]

Expanding this:

\[ = 24 + 7b + 96 - 2b - 135 = 5b - 15 \]

Setting this determinant to zero:

\[ 5b - 15 = 0 \implies 5b = 15 \implies b = 3 \]

### Step 5: Calculate \( a - b \)

Now that we have \( a = 8 \) and \( b = 3 \):

\[ a - b = 8 - 3 = 5 \]

Thus, the final answer is:

\[ \boxed{5} \]