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Can someone help me to simplify $$\Bigg( \begin{matrix} 5\cos t &5\sin t \\2\cos t+\sin t & 2\sin t-\cos t\end{matrix} \Bigg) \Bigg( \begin{matrix} u_1 \\u_2\end{matrix} \Bigg) =\Bigg( \begin{matrix} -\cos t \\ \phantom{-} \sin t \end{matrix} \Bigg)$$ to

$${u_1}={1\over5} \big(2-3\cos2t+\sin2t)$$ $$ u_2={1\over5} \big(-1-\cos2t-3\sin2t)?$$

Thank you.

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What have you tried? –  anorton Jan 7 '13 at 13:22
    
this is a part of a problem, –  Miranda Miri Jan 7 '13 at 13:25
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1 Answer 1

$$AX=B$$ $$X=A^{-1}B$$

Where $X$ is the variable/unknown vector, $A$ is the coefficient matrix, and $B$ is the solution vector.

See: http://en.wikipedia.org/wiki/System_of_linear_equations#Matrix_solution

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This is a start, but if you need more direction, let me know--I'm always glad to help! :) –  anorton Jan 7 '13 at 13:27
    
I want to use simplify rows Not inverse. –  Miranda Miri Jan 7 '13 at 13:30
    
Ugh. I've got a class to head to--let me think about that for a bit. –  anorton Jan 7 '13 at 13:34
    
That seem to be the solution to me, you still need to apply some trigonometric identities in the intermediate steps –  xiamx Jan 7 '13 at 13:58
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You could try the effect of multiplying the first row by $\cos t$ and the second row by $2 \cos t - \sin t$ - but whichever way you do it, it is essentially the same problem, and the determinant of the matrix you have comes out as an easy number - so the inverse can be written down. Why use more steps than you need to solve the problem? –  Mark Bennet Jan 7 '13 at 14:09
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