<p>Let \(A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}\) and \(B = \begin{bmatrix} a & 0 \\ 0 & b \end{bmatrix}\) where \(a, b \in \mathbb{N}\). Then \(AB = BA\) implies which of the following?</p>
<p>There is exactly one \(B\) for which \(AB = BA\)</p>
<p>There are exactly two \(B\)'s for which \(AB = BA\)</p>
<p>There exist infinitely many \(B\)'s for which \(AB = BA\)</p>
<p>There is no \(B\) for which \(AB = BA\)</p>
Step-by-Step Solution
Key Concept: For AB = BA to hold, the diagonal matrix B must commute with A, which severely constrains a and b. This requires comparing the (1,2) and (2,1) entries of AB and BA, which immediately forces a = b.
<p><strong>Step 1:</strong> Compute $AB$:</p><p>$$AB = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \begin{bmatrix} a & 0 \\ 0 & b \end{bmatrix} = \begin{bmatrix} a & 2b \\ 3a & 4b \end{bmatrix}$$</p><p><strong>Step 2:</strong> Compute $BA$:</p><p>$$BA = \begin{bmatrix} a & 0 \\ 0 & b \end{bmatrix} \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} = \begin{bmatrix} a & 2a \\ 3b & 4b \end{bmatrix}$$</p><p><strong>Step 3:</strong> Apply condition $AB = BA$. Comparing entries:</p><p>\begin{align} \text{(1,2) entry: } &2b = 2a \implies a = b \\ \text{(2,1) entry: } &3a = 3b \implies a = b \text{ (confirms same condition)} \\ \text{(1,1) entry: } &a = a \,\checkmark \\ \text{(2,2) entry: } &4b = 4b \,\checkmark \end{align}</p><p><strong>Step 4:</strong> Conclusion: Since $a, b \in \mathbb{N}$, we have $a = b$ where $a$ and $b$ are any natural numbers.</p><p>Therefore, there exist infinitely many matrices $B$ for which $AB = BA$ (one for each choice of natural number $a = b$).</p><p>$$\therefore \text{Answer: C}$$</p>
Correct Answer: C