Question

Let $A=\left(\begin{array}{cc}1& 2\\ 3& 4\end{array}\right)andB=\left(\begin{array}{cc}a& 0\\ 0& b\end{array}\right),a,b\in N.$ Then:

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

Answer 1

The correct option is B there exist exactly infinitely many B's such that $AB=BA$

Given

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

Finding AB

$AB=\left[\begin{array}{cc}1& 2\\ 3& 4\end{array}\right]\left[\begin{array}{cc}a& 0\\ 0& b\end{array}\right]=\left[\begin{array}{cc}a+2\times 0& 1\times 0+2\times b\\ 3\times a+4\times 0& 3\times 0+4\times b\end{array}\right]$

$AB=\left[\begin{array}{cc}a& 2b\\ 3a& 4b\end{array}\right]$

Finding BA

$BA=\left[\begin{array}{cc}a& 0\\ 0& b\end{array}\right]\left[\begin{array}{cc}1& 2\\ 3& 4\end{array}\right]=\left[\begin{array}{cc}a+0\times 3& 2a+0\times 4\\ 0\times 1+3b& 0\times 2+4b\end{array}\right]$

$BA=\left[\begin{array}{cc}a& 2a\\ 3b& 4b\end{array}\right]$

If $AB=BA$

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

Comparing each element we get

$a=a,2b=2a$

$\Rightarrow a=b$

$\therefore$There are infinitely many b's

for which $AB=BA$

Answer=B