Let A = [ * 20 c 1&2 3&4 ] and B = [ * 20 c a&0 0&b ] ;,a,b N the… - Vidyadip

MCQ

Let $A = \left[ {\begin{array}{*{20}{c}}1&2\\3&4\end{array}} \right]$ and $B = \left[ {\begin{array}{*{20}{c}}a&0\\0&b\end{array}} \right]\;,a,b \in N$ then

A
there cannot exist any $B$ such that $AB = BA$
B
there exist more then one but finite number of $B's$ such that $AB=BA$
C
there exists exactly one $B$ such that $AB = BA$
✓
there exist infinitely many $B's$ such that $AB = BA$

✓

Answer

Correct option: D.

there exist infinitely many $B's$ such that $AB = BA$

d
$A=\left[\begin{array}{ll}{1} & {2} \\ {3} & {4}\end{array}\right] \quad $ $B=\left[\begin{array}{ll}{a} & {0} \\ {0} & {b}\end{array}\right]$

$A B=\left[\begin{array}{ll}{a} & {2 b} \\ {3 a} & {4 b}\end{array}\right]$

$B A=\left[\begin{array}{ll}{a} & {0} \\ {0} & {b}\end{array}\right]\left[\begin{array}{ll}{1} & {2} \\ {3} & {4}\end{array}\right]$$=\left[\begin{array}{ll}{a} & {2 a} \\ {3 b} & {4 b}\end{array}\right]$

Hence, $A B=B A$ only when $a=b$

$\therefore$ There can be infinitely many $B^{\prime} s$

for which $A B=B A$

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