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lecture_8

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lecture_8 [2016/02/18 10:19] – rupertlecture_8 [2016/02/18 10:20] (current) – rupert
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     * In particular, $AB\ne BA$ and $AC\ne CA$ and $BC\ne CB$, since in each case one of the matrices doesn't exist.     * In particular, $AB\ne BA$ and $AC\ne CA$ and $BC\ne CB$, since in each case one of the matrices doesn't exist.
   - If $A=\mat{0&1\\0&0}$ and $B=\mat{0&0\\1&0}$, then $AB=\mat{1&0\\0&0}$ and $BA=\mat{0&0\\0&1}$. So $AB$ and $BA$ are both defined and have the [[same size]], but they are not [[equal matrices]]: $AB\ne BA$.   - If $A=\mat{0&1\\0&0}$ and $B=\mat{0&0\\1&0}$, then $AB=\mat{1&0\\0&0}$ and $BA=\mat{0&0\\0&1}$. So $AB$ and $BA$ are both defined and have the [[same size]], but they are not [[equal matrices]]: $AB\ne BA$.
 +  - If $A=\mat{1&0\\0&0}$ and $B=\mat{0&0\\0&1}$, then $AB=\mat{0&0\\0&0}$ and $BA=\mat{0&0\\0&0}$. So $AB=BA$ in this case.
   - If $A=0_{n\times n}$ is the $n\times n$ zero matrix and $B$ is any $n\times n$ matrix, then $AB=0_{n\times n}$ and $BA=0_{n\times n}$. So in this case, we do have $AB=BA$.   - If $A=0_{n\times n}$ is the $n\times n$ zero matrix and $B$ is any $n\times n$ matrix, then $AB=0_{n\times n}$ and $BA=0_{n\times n}$. So in this case, we do have $AB=BA$.
   - If $A=\mat{1&2\\3&4}$ and $B=\mat{7&10\\15&22}$, then $AB=\mat{37&54\\81&118}=BA$, so $AB=BA$ for these particular matrices $A$ and $B$.   - If $A=\mat{1&2\\3&4}$ and $B=\mat{7&10\\15&22}$, then $AB=\mat{37&54\\81&118}=BA$, so $AB=BA$ for these particular matrices $A$ and $B$.
lecture_8.1455790768.txt.gz · Last modified: by rupert
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