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I wanted to know if I did this problem right or not.

$A$ and $B$ are the following matrices:




$$\frac{\langle A,B\rangle}{\|A\|\|B\|} =\frac{\left\langle\begin{bmatrix}2&6\\1&-3\end{bmatrix}, \begin{bmatrix}3&2\\1&0\end{bmatrix}\right\rangle}{\sqrt{\left\langle\begin{bmatrix}2&6\\1&-3\end{bmatrix},\begin{bmatrix}2&6\\1&-3\end{bmatrix}\right\rangle}\sqrt{\left\langle\begin{bmatrix}3&2\\1&0\end{bmatrix},\begin{bmatrix}3&2\\1&0\end{bmatrix}\right\rangle}}.$$

I ended up getting:

$\arccos(19/184) = 84.07^\circ$

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Justin: When I made the math prettier, I also took the liberty of correcting your denominator. I hope this wasn't going too far. I figured that it was a typo, because the expression you had was equal to $1$. But perhaps that was part of the problem. – Jonas Meyer Apr 5 '11 at 1:37
up vote 0 down vote accepted

$a\cdot b=|a||b|\cos\theta$ so $$\theta=\arccos\Big(\frac{a\cdot b}{|a||b|}\Big).$$ we have $a\cdot b=6+12+1=19, |a||b|=\sqrt{4+36+1+9}\sqrt{9+4+1}=10\sqrt{7}$.
so the angle is $\arccos(19/(10\sqrt{7}))$

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It's amazing that you could understand Justin's question. – GWu Apr 5 '11 at 1:30

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