Question

Let $A=\begin{bmatrix}1 & 2 \\ 3 & 4\end{bmatrix}$ and $B=\begin{bmatrix}a & 0 \\ 0 & b\end{bmatrix} a, b \in N$. Then,

• A. there exists more than one but finite number of $B^{\prime}$ s such that $A B=B A$
• B. there exists exactly one $B$ such that $A B=B A$
• C. there exists infinitely many $B^{\prime}$ s such that $A B=B A$
• D. there cannot exist any $B$ such that $A B=B A$

Answer 1

Answer: (C)

Given that,
$A =\begin{bmatrix}1 & 2 \\3 & 4\end{bmatrix}$
and $B B=\begin{bmatrix}a & 0 \\0 & b\end{bmatrix}$
Now, $ A B=\begin{bmatrix}1 & 2 \\ 3 & 4\end{bmatrix}\begin{bmatrix}a & 0 \\ 0 & b\end{bmatrix}=\begin{bmatrix}a & 2 b \\ 3 a & 4 b\end{bmatrix}$
and $ B A=\begin{bmatrix}a & 0 \\ 0 & b\end{bmatrix}\begin{bmatrix}1 & 2 \\ 3 & 4\end{bmatrix}=\begin{bmatrix}a & 2 a \\ 3 b & 4 b\end{bmatrix}$
If $A B=B A$, then $a=b$
Hence, $A B=B A$ is possible for infinitely many values of $B$.