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| lecture_7b [2015/02/17 10:27] – rupert | lecture_7b [2016/02/09 12:25] (current) – rupert |
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| * $\begin{bmatrix}3\\2\\4\\0\\-1\end{bmatrix}\ne \begin{bmatrix}3&2&4&0&-1\end{bmatrix}$, since these matrices have different sizes: the first is $5\times 1$ but the second is $1\times 5$. | * $\begin{bmatrix}3\\2\\4\\0\\-1\end{bmatrix}\ne \begin{bmatrix}3&2&4&0&-1\end{bmatrix}$, since these matrices have different sizes: the first is $5\times 1$ but the second is $1\times 5$. |
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| * $\begin{bmatrix}1\\2\end{bmatrix}\ne\begin{bmatrix}1 &0\\2&0\end{bmatrix}$ since these matrices are not the same size. | * $\begin{bmatrix}1\\2\end{bmatrix}\ne\begin{bmatrix}1 &0\\2&0\end{bmatrix}$ since these matrices are not the same size. |
| * $\begin{bmatrix}1&0\\0&1\end{bmatrix}\ne \begin{bmatrix}1&0\\1&0\end{bmatrix}$ because even though they have the same size, the $(2,1)$ entries are different. | * $\begin{bmatrix}1&0\\0&1\end{bmatrix}\ne \begin{bmatrix}1&0\\1&0\end{bmatrix}$ because even though they have the same size, the $(2,1)$ entries are different. |
| * If $\begin{bmatrix}3x&7y+2\\8z-3&w^2\end{bmatrix}=\begin{bmatrix}1&2z\\\sqrt2&9\end{bmatrix}$ then we know that all the corresponding entries are equal, so we get four equations:\begin{align*}3x&=1\\7y+2&=2z\\8z-3&=\sqrt2\\w^2&=9\end{align*} | * If $\begin{bmatrix}3x&7y+2\\8z-3&w^2\end{bmatrix}=\begin{bmatrix}1&2z\\\sqrt2&9\end{bmatrix}$ then we know that all the corresponding entries are equal, so we get four equations:\begin{align*}3x&=1\\7y+2&=2z\\8z-3&=\sqrt2\\w^2&=9\end{align*} |
