Determinants Matrices Question 57
Question: sLet $ A= \begin{bmatrix} 1 & 2 \\ 3 & 4 \\ \end{bmatrix} $ and $ B= \begin{bmatrix} a & 0 \\ 0 & b \\ \end{bmatrix} $ where a, b are natural numbers, then which one of the following is correct-
Options:
A) There exist more than one but finite number of B’s such that AB = BA
B) There exists exactly one B such that AB = BA
C) There exist infinitely many B’s such that AB=BA
D) There cannot exist any B such that AB = BA
Show Answer
Answer:
Correct Answer: C
Solution:
- [c] $ 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} $ and $ 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} $ If $ AB=BA $
$ \Rightarrow \begin{bmatrix} a & 2b \\ 3a & 4b \\ \end{bmatrix} = \begin{bmatrix} a & 2a \\ 3b & 4b \\ \end{bmatrix} \Rightarrow a=b $ From the above it is clear that there exist infinitely many B’s such that AB = BA.
BETA
