If the system of equations
x+y+az = b
2x + 5y +2z = 6
x + 2y + 3z = 3
has infinitely many solutions, then 2a + 3b is equal to

Options

A. 25
B. 20
C. 23
D. 28

Answer

To solve the given system of equations for the condition of infinitely many solutions, we will follow these steps:

### Step-by-Step Solution:

1. **Write the system of equations:**
The equations given are:
\[ (1) \quad x + y + az = b \]
\[ (2) \quad 2x + 5y + 2z = 6 \]
\[ (3) \quad x + 2y + 3z = 3 \]

2. **Form the coefficient matrix and the augmented matrix:**
The coefficient matrix \(A\) is:
\[ A = \begin{pmatrix} 1 & 1 & a \\ 2 & 5 & 2 \\ 1 & 2 & 3 \end{pmatrix} \]
The augmented matrix \([A|B]\) is:
\[ [A|B] = \begin{pmatrix} 1 & 1 & a & | & b \\ 2 & 5 & 2 & | & 6 \\ 1 & 2 & 3 & | & 3 \end{pmatrix} \]

3. **Calculate the determinant of the coefficient matrix \(A\):**
For the system to have infinitely many solutions, the determinant of the coefficient matrix must be zero:
\[ \Delta = \begin{vmatrix} 1 & 1 & a \\ 2 & 5 & 2 \\ 1 & 2 & 3 \end{vmatrix} \]

Calculate the determinant:
\[ \Delta = 1 \cdot (5 \cdot 3 - 2 \cdot 2) - 1 \cdot (2 \cdot 3 - 1 \cdot 2) + a \cdot (2 \cdot 2 - 5 \cdot 1) \]
\[ = 1 \cdot (15 - 4) - 1 \cdot (6 - 2) + a \cdot (4 - 5) \]
\[ = 1 \cdot 11 - 1 \cdot 4 - a \]
\[ = 11 - 4 - a = 7 - a \]

Set the determinant to zero for infinitely many solutions:
\[ 7 - a = 0 \implies a = 7 \]

4. **Calculate the determinant for the augmented matrix:**
Now we need to find the value of \(b\) such that the determinant of the augmented matrix is also zero:
\[ \Delta_x = \begin{vmatrix} b & 1 & a \\ 6 & 5 & 2 \\ 3 & 2 & 3 \end{vmatrix} \]

Substitute \(a = 7\):
\[ \Delta_x = \begin{vmatrix} b & 1 & 7 \\ 6 & 5 & 2 \\ 3 & 2 & 3 \end{vmatrix} \]

Calculate the determinant:
\[ \Delta_x = b \cdot (5 \cdot 3 - 2 \cdot 2) - 1 \cdot (6 \cdot 3 - 2 \cdot 7) + 7 \cdot (6 \cdot 2 - 5 \cdot 3) \]
\[ = b \cdot (15 - 4) - (18 - 14) + 7 \cdot (12 - 15) \]
\[ = b \cdot 11 - 4 + 7 \cdot (-3) \]
\[ = 11b - 4 - 21 = 11b - 25 \]

Set this determinant to zero:
\[ 11b - 25 = 0 \implies 11b = 25 \implies b = \frac{25}{11} \]

5. **Calculate \(2a + 3b\):**
Now substitute \(a\) and \(b\) into \(2a + 3b\):
\[ 2a + 3b = 2(7) + 3\left(\frac{25}{11}\right) \]
\[ = 14 + \frac{75}{11} \]
\[ = \frac{154}{11} + \frac{75}{11} = \frac{229}{11} \]

### Final Answer:
\[ 2a + 3b = \frac{229}{11} \]