Question

If \[ A=\begin{bmatrix} 3 & 4\\ 5 & 6 \end{bmatrix} \] and \[ B=\begin{bmatrix} x & 0\\ 0 & y \end{bmatrix}, \quad x,y\in \mathbb{N}, \] then:

• A. There is exactly one such matrix \(B\) such that \(AB=I\)
• B. There is no matrix \(B\) such that \(AB=BA\)
• C. There exist only a finite number of matrices \(B\) such that \(AB=BA\)
• D. There exist infinite number of matrices \(B\) such that \(AB=BA\)

Hint:

For matrix commutativity questions, always compute both \(AB\) and \(BA\), then compare corresponding entries.

Answer 1

The Correct Option is D

Step 1: Write the given matrices.
We have $ A = \begin{bmatrix} 3 & 4 \\ 5 & 6 \end{bmatrix} $ and $ B = \begin{bmatrix} x & 0 \\ 0 & y \end{bmatrix} $ where $ x, y \in \mathbb{N} $. We want to find when $ AB = BA $.
Step 2: Compute AB.
\[ AB = \begin{bmatrix} 3 & 4 \\ 5 & 6 \end{bmatrix} \begin{bmatrix} x & 0 \\ 0 & y \end{bmatrix} = \begin{bmatrix} 3x & 4y \\ 5x & 6y \end{bmatrix} \]
Step 3: Compute BA.
\[ BA = \begin{bmatrix} x & 0 \\ 0 & y \end{bmatrix} \begin{bmatrix} 3 & 4 \\ 5 & 6 \end{bmatrix} = \begin{bmatrix} 3x & 4x \\ 5y & 6y \end{bmatrix} \]
Step 4: Set AB = BA and compare entries.
For $ AB = BA $, every entry must match. Comparing the (1,2) entry: $ 4y = 4x $, which gives $ x = y $. Comparing the (2,1) entry: $ 5x = 5y $, which again gives $ x = y $.
Step 5: Count the solutions.
The condition is simply $ x = y $, with $ x, y \in \mathbb{N} $. We can choose $ x = y = 1, 2, 3, 4, \ldots $ There are infinitely many natural numbers, so there are infinitely many matrices $ B $.
Step 6: State the conclusion.
There exist infinitely many matrices $ B $ such that $ AB = BA $.
\[ \boxed{\text{There exist infinite number of matrices } B \text{ such that } AB = BA} \]