To find the inverse of \( 9a \) modulo \( 44 \) in terms of \( b \), we start with the given information.
We know that \( a \cdot b \equiv 1 \mod 44 \). This means that \( b \) is the multiplicative inverse of \( a \) modulo \( 44 \).
We want to find the inverse of \( 9a \) modulo \( 44 \). We denote the inverse we want to find as \( c \), then we want:
\[ 9a \cdot c \equiv 1 \mod 44. \]
To solve for \( c \), we can express \( c \) in terms of \( b \). Since \( 9a \cdot c \equiv 1 \mod 44 \), we can rewrite \( c \) as follows:
\[ c \equiv \frac{1}{9a} \mod 44. \]
We can relate \( 9a \) to \( a \) and its inverse \( b \). First, we can multiply \( 9 \) by \( a \) and then use \( b \):
\[ c \equiv \frac{b}{9} \mod 44. \]
To express this correctly, we need to find the multiplicative inverse of \( 9 \) modulo \( 44 \). Let's find that inverse.
We need \( x \) such that:
\[ 9x \equiv 1 \mod 44. \]
To find \( x \), we can use the Extended Euclidean Algorithm. We calculate \( \gcd(9, 44) \):
\[ 44 = 4 \times 9 + 8, \] \[ 9 = 1 \times 8 + 1, \] \[ 8 = 8 \times 1 + 0. \]
From here, we see that \( \gcd(9, 44) = 1 \), indicating that 9 has an inverse modulo 44.
Now we backtrack the Extended Euclidean algorithm:
- From \( 9 = 1 \cdot 8 + 1 \), we have: \[ 1 = 9 - 1 \cdot 8. \]
- Replace \( 8 \) using \( 44 = 4 \times 9 + 8 \) to get \( 8 \): \[ 8 = 44 - 4 \cdot 9. \]
- Substituting back gives: \[ 1 = 9 - 1 \cdot (44 - 4 \cdot 9) = 9 - 44 + 4 \cdot 9 = 5 \cdot 9 - 44. \]
Thus, we have: \[ 5 \cdot 9 \equiv 1 \mod 44. \]
Therefore, the inverse of \( 9 \) modulo \( 44 \) is \( 5 \).
Now we relate this back to \( a \) and \( b \): Using \( c \equiv \frac{b}{9} \mod 44 \): \[ c \equiv 5b \mod 44. \]
Thus, the inverse of \( 9a \) modulo \( 44 \) in terms of \( b \) is:
\[ \boxed{5b}. \]