Let `A=[(2,3),(5,7)] and B=[(a, 0), (0, b)]` where `a, b in N`. The number of matrices B such that `AB=BA`, is equal to

Options

A. 0
B. 1
C. 2
D. infinite

Answer

To solve the problem of finding the number of matrices \( B \) such that \( AB = BA \), where \( A = \begin{pmatrix} 2 & 3 \\ 5 & 7 \end{pmatrix} \) and \( B = \begin{pmatrix} a & 0 \\ 0 & b \end{pmatrix} \) with \( a, b \in \mathbb{N} \), we will follow these steps:

### Step 1: Compute the product \( AB \)
To find \( AB \), we multiply matrix \( A \) by matrix \( B \):

\[ AB = \begin{pmatrix} 2 & 3 \\ 5 & 7 \end{pmatrix} \begin{pmatrix} a & 0 \\ 0 & b \end{pmatrix} \]

Calculating this gives:

\[ AB = \begin{pmatrix} 2a + 3 \cdot 0 & 2 \cdot 0 + 3b \\ 5a + 7 \cdot 0 & 5 \cdot 0 + 7b \end{pmatrix} = \begin{pmatrix} 2a & 3b \\ 5a & 7b \end{pmatrix} \]

### Step 2: Compute the product \( BA \)
Next, we compute \( BA \):

\[ BA = \begin{pmatrix} a & 0 \\ 0 & b \end{pmatrix} \begin{pmatrix} 2 & 3 \\ 5 & 7 \end{pmatrix} \]

Calculating this gives:

\[ BA = \begin{pmatrix} a \cdot 2 + 0 \cdot 5 & a \cdot 3 + 0 \cdot 7 \\ 0 \cdot 2 + b \cdot 5 & 0 \cdot 3 + b \cdot 7 \end{pmatrix} = \begin{pmatrix} 2a & 3a \\ 5b & 7b \end{pmatrix} \]

### Step 3: Set \( AB = BA \)
Now we set the two products equal to each other:

\[ \begin{pmatrix} 2a & 3b \\ 5a & 7b \end{pmatrix} = \begin{pmatrix} 2a & 3a \\ 5b & 7b \end{pmatrix} \]

From this equality, we derive the following equations:

1. \( 2a = 2a \) (always true)
2. \( 3b = 3a \)
3. \( 5a = 5b \)
4. \( 7b = 7b \) (always true)

### Step 4: Solve the equations
From equations 2 and 3, we can simplify:

- From \( 3b = 3a \), we get \( b = a \).
- From \( 5a = 5b \), we also get \( a = b \).

Thus, we conclude that:

\[ a = b \]

### Step 5: Determine the number of matrices \( B \)
Since \( a \) and \( b \) are natural numbers, and they must be equal, we can denote \( a = b = n \) where \( n \in \mathbb{N} \). Therefore, the possible matrices \( B \) can be represented as:

\[ B = \begin{pmatrix} n & 0 \\ 0 & n \end{pmatrix} \]

where \( n \) can take any natural number value (1, 2, 3, ...).

### Conclusion
Since there are infinitely many natural numbers, the number of matrices \( B \) such that \( AB = BA \) is infinite.