Matrix multiplication commutative?

2.7.2 Matrix multiplication commutative?

\[\left.\begin{array}{ccc} \left(\begin{array}{ccc}2&1&0\\1&2&0\\1&1&1\end{array}\right)\left(\begin{array}{ccc}1&0&-1\\0&1&0\\1&1&1\end{array}\right)&=&\left(\begin{array}{ccc}2&1&-2\\1&2&-1\\2&2&0\end{array}\right)\\\\ \left(\begin{array}{ccc}1&0&-1\\0&1&0\\1&1&1\end{array}\right)\left(\begin{array}{ccc}2&1&0\\1&2&0\\1&1&1\end{array}\right)&=&\left(\begin{array}{ccc}1&0&-1\\1&2&0\\4&4&1\end{array}\right)\end{array}\right\}\neq\]

$\Rightarrow$ This clearly means that the multiplication of matrices is not commutative, i.e.

\[\tilde A\tilde B\neq\tilde B\tilde A \mbox{ in general!!}\]

But at least: \begin{eqnarray*}\tilde A(\tilde B\tilde C)&=&(\tilde A\tilde B)\tilde C \qquad\mbox{associative rule}\\ \mbox {and } \tilde A(\tilde B+\tilde C)&=&\tilde A\tilde B+\tilde A\tilde C\qquad\mbox{ distributive rule}\\ \mbox{but: }(\tilde A+\tilde B)(\tilde A-\tilde B)&=&\tilde A^2-\tilde A\tilde B+\tilde B\tilde A-\tilde B^2\neq\tilde A^2-\tilde B^2\mbox{ in general since $\tilde A\tilde B\neq\tilde B\tilde A$} \end{eqnarray*}

Example:

\[ \begin{array}{cccc} 2\times4&4\times3& &2\times3\\\\ \left(\begin{array}{cccc}5&2&-2&3\\9&2&-3&4\end{array}\right)&\left(\begin{array}{ccc}2&2&2\\-1&3&-5\\16&8&24\\8&0&16\end{array}\right)&=&\left(\begin{array}{ccc}0&0&0\\0&0&0\end{array}\right)\end{array}\]

$\Rightarrow$ Matrices do have zero divisors (see example)!! Thus, if $\tilde A\cdot\tilde B=\tilde 0$ and $\tilde A\neq\tilde 0$ then in general not $\tilde B=\tilde 0$, and if $\tilde A\tilde B=\tilde A\tilde C, \tilde A\neq\tilde0$ than not $\tilde B = \tilde C$!!! (Example in homework)
Note, that this would be trivial for real or complex numbers $\Rightarrow$ Matrices are different from numbers (which makes them relevant for quantum mechanics)

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Matrices