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October 24th, 2014, 10:19 AM   #1
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determinant

Can someone help me to solve this determinant? I have no idea how to start...

залить картинку
thanks for the help!!

Last edited by robertson; October 24th, 2014 at 10:34 AM.
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October 24th, 2014, 10:47 AM   #2
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Zero? All columns contain zero, right?
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October 24th, 2014, 10:53 AM   #3
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Originally Posted by Monox D. I-Fly View Post
Zero? All columns contain zero, right?
But
$\displaystyle \left | \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right | = 1$

and there is a 0 in every column.

It has to be all zeros in at least one column or all zeros in at least one row to give a determinant of zero.

-Dan
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October 24th, 2014, 11:24 AM   #4
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I guess, to solve this determinant we need to extract (don't know if it's the right word) by first row and column. Or something like that. What do You think?
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October 24th, 2014, 11:51 AM   #5
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Originally Posted by robertson View Post
I guess, to solve this determinant we need to extract (don't know if it's the right word) by first row and column. Or something like that. What do You think?
Sounds like a plan to me. Give it a try and let us know how it works.

-Dan
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October 24th, 2014, 02:09 PM   #6
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hmmm .... nop, I'm not so good at this one... Doesn't work :/
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October 24th, 2014, 03:05 PM   #7
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Try a couple of simple cases:

$\displaystyle Det \left [ \begin{matrix} a & b & c \\ -x & x & 0 \\ 0 & -x & x \end{matrix} \right ] = a \left | \begin{matrix} x & 0 \\ -x & x \end{matrix} \right | - b \left | \begin{matrix} -x & 0 \\ 0 & x \end{matrix} \right | + c \left | \begin{matrix} -x & x \\ 0 & -x \end{matrix} \right |$

$\displaystyle = a ( x^2 ) - b ( -x^2 ) + c (x^2 ) = (a + b + c)x^2$

Try the 4 x 4 yourself. There's an obvious pattern that emerges. Post your the work of your 4 x 4 and we'll take a look at it.

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October 25th, 2014, 01:49 AM   #8
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What do you think? So if I'm right answer will be (a0+a1+a2+...+an)x^(n-1).
Btw how do you do these formulas?
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October 25th, 2014, 11:05 PM   #9
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Originally Posted by robertson View Post

What do you think? So if I'm right answer will be (a0+a1+a2+...+an)x^(n-1).
Btw how do you do these formulas?
Well done. I didn't know the formula. I derived it, saw the pattern and passed that on to you.

I had originally thought to give you the idea of expanding out by the second row, not the first, and I forgot to do so. I apologize. That's the only simple way to spot a useful pattern. (Well, you can do it the way I presented it, but it's harder.)

-Dan
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