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lecture_8

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lecture_8 [2016/02/18 10:17] – rupertlecture_8 [2016/02/18 10:20] (current) – rupert
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   - If $\newcommand{\mat}[1]{\begin{bmatrix}#1\end{bmatrix}} A=\mat{1&0&5\\2&-1&3}$ and  $B=\mat{1&2\\3&4\\5&6}$, then $AB=\mat{26&32\\14&18}$ and $BA=\mat{5&-2&11\\11&-4&27\\17&-6&43}$. Note that $AB$ and $BA$ are both defined, but $AB\ne BA$ since $AB$ and $BA$ don't even have the [[same size]].   - If $\newcommand{\mat}[1]{\begin{bmatrix}#1\end{bmatrix}} A=\mat{1&0&5\\2&-1&3}$ and  $B=\mat{1&2\\3&4\\5&6}$, then $AB=\mat{26&32\\14&18}$ and $BA=\mat{5&-2&11\\11&-4&27\\17&-6&43}$. Note that $AB$ and $BA$ are both defined, but $AB\ne BA$ since $AB$ and $BA$ don't even have the [[same size]].
-  - If $A=\mat{1&2\\3&4\\5&6}$, $B=\mat{2&1&1\\1&2&0\\1&0&2\\1&0&2\\2&2&1}$ and $C=\mat{1&3&0&7\\0&4&6&8}$, then $A$ is $3\times 2$, $B$ is $4\times 3$ and $C$ is $2\times 4$, so +  - If $A=\mat{1&2\\3&4\\5&6}$, $B=\mat{2&1&1\\1&2&0\\1&0&2\\2&2&1}$ and $C=\mat{1&3&0&7\\0&4&6&8}$, then $A$ is $3\times 2$, $B$ is $4\times 3$ and $C$ is $2\times 4$, so 
     * $AB$, $CA$ and $BC$ don't exist (i.e., they are undefined);     * $AB$, $CA$ and $BC$ don't exist (i.e., they are undefined);
     * $AC$ exists and is $3\times 4$;     * $AC$ exists and is $3\times 4$;
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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 $B$.
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 $\text{row}_i(I_n)\cdot \text{col}_j(A)$, by the definition of [[matrix multiplication]], which is therefore $\text{row}_i(I_n)\cdot \text{col}_j(A)$, by the definition of [[matrix multiplication]], which is therefore
 \begin{align*} [0~0~\dots~0~1~0~\dots~0]\begin{bmatrix}a_{1j}\\a_{2j}\\\vdots\\a_{nj}\end{bmatrix} &= 0a_{1j}+0a_{2j}+\dots+0a_{i-1,j}+1a_{ij}+0a_{i+1,j}+\dots+0a_{nj} \\&= a_{ij}.\end{align*} \begin{align*} [0~0~\dots~0~1~0~\dots~0]\begin{bmatrix}a_{1j}\\a_{2j}\\\vdots\\a_{nj}\end{bmatrix} &= 0a_{1j}+0a_{2j}+\dots+0a_{i-1,j}+1a_{ij}+0a_{i+1,j}+\dots+0a_{nj} \\&= a_{ij}.\end{align*}
-So the matrices $I_nA$ and $A$ have the same $(i,j)$ entries, for any $(i,j)$. So $I_nA=A$.+So the matrices $I_nA$ and $A$ have the same size, and the same $(i,j)$ entries, for any $(i,j)$. So $I_nA=A$.
  
lecture_8.1455790644.txt.gz · Last modified: by rupert
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