Let A=pmatrix1&23&4pmatrix and B=pmatrixa&00&bpmatrix, a,bN.

Question

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

Accepted Answer

$AB=\begin{pmatrix}a&2b\3a&4b\end{pmatrix}$, $BA=\begin{pmatrix}a&2a\3b&4b\end{pmatrix}$.

$AB=BA$ requires $2b=2a$ and $3a=3b$, i.e. $a=b$.

Since $a=b$ can be any natural number, there are infinitely many such $B$'s.

Solutions · 1