Let $\mathrm{A}=\left(\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right)$ and $\mathrm{B}=\left(\begin{array}{ll}\mathrm{a} & 0 \\ 0 & \mathrm{~b}\end{array}\right), \mathrm{a}, \mathrm{b} \in \mathrm{N}$. Then

(D) there exist infinitely many $B$'s such that $A B=B A$

(A) there cannot exist any $B$ such that $A B=B A$

(B) there exist more than one but finite number of $B$'s such that $A B=B A$

(C) there exists exactly one $B$ such that $A B=B A$

JEE Main · 2006 · 30 Apr 2006 · Offline · Matrices

Let

A = (1324)

and

B = (a 0 0 b), a, b \in N

. Then

Answer: (D)

Compute both products.

A B = (1324) (a 0 0 b) = (a 3 a 2 b 4 b)

, while

B A = (a 0 0 b) (1324) = (a 3 b 2 a 4 b)

. Equating corresponding entries forces

2 b = 2 a

and

3 a = 3 b

, both of which say

a = b

. Since

a

and

b

may be any equal natural numbers

1, 2, 3, \dots

, there are infinitely many matrices

B

that commute with

A

JEE Main · 2006 · 30 Apr 2006 · Offline · Single correct option