1. Key Concepts and Formulas

• Matrix Multiplication: The product of two matrices $M_1$ and $M_2$ (where the number of columns of $M_1$ equals the number of rows of $M_2$) results in a new matrix. The element in the $i$-th row and $j$-th column of the product matrix is obtained by taking the dot product of the $i$-th row of $M_1$ and the $j$-th column of $M_2$.
• Commutativity of Matrices: Two square matrices $A$ and $B$ of the same order are said to commute if their product is independent of the order of multiplication, i.e., $AB = BA$. It is important to remember that matrix multiplication is generally not commutative.
• Matrix Equality: Two matrices are equal if and only if they have the same dimensions and their corresponding elements are identical.
• Natural Numbers ($\mathbb{N}$): In the context of JEE Mathematics, natural numbers typically refer to the set of positive integers: $\{1, 2, 3, \dots\}$.

2. Step-by-Step Solution

Step 1: Define the given matrices and the condition for commutativity. We are given two $2 \times 2$ matrices: $A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$ $B = \begin{pmatrix} a & 0 \\ 0 & b \end{pmatrix}$ where $a, b \in \mathbb{N}$. Our goal is to determine if there exist any natural numbers $a$ and $b$ such that $A$ and $B$ commute, which means satisfying the condition $AB = BA$.

Step 2: Calculate the matrix product $AB$. We perform the matrix multiplication for $AB$: $AB = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} \begin{pmatrix} a & 0 \\ 0 & b \end{pmatrix}$ $AB = \begin{pmatrix} (1)(a) + (2)(0) & (1)(0) + (2)(b) \\ (3)(a) + (4)(0) & (3)(0) + (4)(b) \end{pmatrix}$ $AB = \begin{pmatrix} a & 2b \\ 3a & 4b \end{pmatrix}$

Step 3: Calculate the matrix product $BA$. Next, we perform the matrix multiplication for $BA$: $BA = \begin{pmatrix} a & 0 \\ 0 & b \end{pmatrix} \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$ $BA = \begin{pmatrix} (a)(1) + (0)(3) & (a)(2) + (0)(4) \\ (0)(1) + (b)(3) & (0)(2) + (b)(4) \end{pmatrix}$ $BA = \begin{pmatrix} a & 2a \\ 3b & 4b \end{pmatrix}$

Step 4: Equate the corresponding elements of $AB$ and $BA$. For $A$ and $B$ to commute, we must have $AB = BA$. By the definition of matrix equality, this requires each corresponding element of the two product matrices to be identical: $\begin{pmatrix} a & 2b \\ 3a & 4b \end{pmatrix} = \begin{pmatrix} a & 2a \\ 3b & 4b \end{pmatrix}$ This equality gives us a system of four scalar equations:

1. $a = a$
2. $2b = 2a$
3. $3a = 3b$
4. $4b = 4b$

Step 5: Solve the system of equations for $a$ and $b$. Equations (1) and (4) are identities ($a=a$ and $4b=4b$), meaning they are always true and do not impose any specific conditions on $a$ or $b$. From equation (2): $2b = 2a$. Dividing both sides by 2 (since $2 \neq 0$), we get $b = a$. From equation (3): $3a = 3b$. Dividing both sides by 3 (since $3 \neq 0$), we get $a = b$. Both informative equations consistently lead to the same condition: $a = b$. This means that for matrices $A$ and $B$ to commute, the elements $a$ and $b$ of matrix $B$ must be equal.

Step 6: Determine the existence and number of such matrices $B$. The problem states that $a, b \in \mathbb{N}$ (natural numbers, i.e., positive integers $\{1, 2, 3, \dots\}$). The condition for commutativity is $a=b$. Since there are infinitely many natural numbers, we can choose any natural number for $a$ (e.g., $1, 2, 3, \dots$), and $b$ will be equal to that chosen value. For example:

• If $a=1, b=1$, then $B = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$ (the identity matrix), which always commutes with any square matrix. Since $1 \in \mathbb{N}$, this is a valid matrix $B$.
• If $a=2, b=2$, then $B = \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix}$. Since $2 \in \mathbb{N}$, this is another valid matrix $B$.
• In general, if $a=k$ and $b=k$ for any $k \in \mathbb{N}$, then $B = \begin{pmatrix} k & 0 \\ 0 & k \end{pmatrix} = kI$ (a scalar matrix). A scalar matrix $kI$ always commutes with any square matrix $A$ of the same order, as $A(kI) = k(AI) = kA$ and $(kI)A = k(IA) = kA$.

Since there are infinitely many natural numbers $k$, there exist infinitely many matrices $B$ of the form $\begin{pmatrix} k & 0 \\ 0 & k \end{pmatrix}$ where $k \in \mathbb{N}$ such that $AB=BA$.

3. Common Mistakes & Tips

• Order of Multiplication: Always remember that matrix multiplication is not generally commutative. $AB$ and $BA$ must be calculated separately.
• Element-wise Equality: When equating matrices, ensure that every corresponding element is set equal, leading to a system of scalar equations.
• Domain of Variables: Pay close attention to the domain of variables (e.g., $a, b \in \mathbb{N}$). This can limit or expand the number of possible solutions.
• Scalar Matrices: Recognize that scalar matrices (matrices of the form $kI$) commute with all square matrices of the same order. This is a useful shortcut if the condition leads to $a=b$.

4. Summary

By performing the matrix multiplications for $AB$ and $BA$ and then equating their corresponding elements, we derived a system of equations. Solving this system revealed that the condition for $AB=BA$ to hold is $a=b$. Since $a$ and $b$ must be natural numbers, and there are infinitely many natural numbers that can be chosen for $a$ (and thus for $b$), there exist infinitely many matrices $B$ that satisfy the given condition. These matrices $B$ are specifically of the form $kI$ where $k$ is any natural number.

The final answer is $\boxed{\text{D}}$.