What is the determinant of this 5x5 matrix and how can it be calculated?

mexion
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Homework Statement


so my problem is to calculate the determinant of this matrix
\left[\begin{array}{ccccc}<br /> 1 &amp; 2 &amp; 3 &amp; 3 &amp; 5 \\<br /> 3 &amp; 2 &amp; 1 &amp; 2 &amp; 2 \\<br /> 1 &amp; 2 &amp; 3 &amp; 4 &amp; 5 \\<br /> -1 &amp; 0 &amp; -8 &amp; 1 &amp; 2 \\<br /> 7 &amp; 2 &amp; 1 &amp; 3 &amp; 2<br /> \end{array}\right]



Homework Equations





The Attempt at a Solution


my calculation -> http://img29.imageshack.us/img29/1120/21102009124.th.jpg
i know to use Laplace, but my teacher said me "at first you should reduce matrix to 3x3 or even 2x2 - it's easier to calculate the determinant"
but i don't understand this methode at all.
i've done this problem and my solution is -379 and i know it's wrong (correct is -224)
please help.
 
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The third line of your second determinant is incorrect.
 
Your last determinant should be

\left|\begin{array}{cc} -18 &amp; 1\\ -50 &amp;9\end{array}\right|

not

\left|\begin{array}{cc} -18 &amp; 0\\ -50 &amp;9\end{array}\right|
 
yeah, my bad
but thanks now it's correct
thanks a lot :)
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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