More Problems on Linear Systems and Matrices

Set

\begin{displaymath}B = \left(\begin{array}{rrrr}
a & b \\
c & d\\
\end{array} \right) \;.\end{displaymath}

Then we have

\begin{displaymath}A\; B = \left(\begin{array}{rrrr}
a-c & b-d \\
2a+c &2b+d\\
\end{array} \right) \end{displaymath}

and

\begin{displaymath}B\; A = \left(\begin{array}{rrrr}
a+2b & -a+b \\
c+2d &-c+d\\
\end{array} \right) \;.\end{displaymath}

So the equation $A\;B = B\; A$ implies

\begin{displaymath}\left\{\begin{array}{lcl}
a-c &=& a+2b \\
b-d &=& -a+b \\
2a+c &=& c+2d \\
2b+d &=& -c+d \\
\end{array} \right.\end{displaymath}

This clearly implies

\begin{displaymath}c = -2b \;\;\mbox{and}\;\; a=d\end{displaymath}

So

\begin{displaymath}B = \left(\begin{array}{rrrr}
a & b \\
-2b & a\\
\end{array} \right) \end{displaymath}

where $a$ and $b$ are two arbitrary numbers. Consider the matrix

\begin{displaymath}C = \left(\begin{array}{rrrr}
1 & 1 \\
1 & 1\\
\end{array} \right)\;.\end{displaymath}

Since $C$ is not of the form found before, then we must have

\begin{displaymath}A\;C \neq C\; A\;.\end{displaymath}

The reader may want to check that is the case.


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