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 October 24th, 2014, 09:19 AM #1 Newbie   Joined: Oct 2014 From: Lithuania Posts: 4 Thanks: 0 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 09:34 AM.
 October 24th, 2014, 09:47 AM #2 Senior Member     Joined: Nov 2010 From: Indonesia Posts: 2,000 Thanks: 132 Math Focus: Trigonometry Zero? All columns contain zero, right?
October 24th, 2014, 09:53 AM   #3
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Quote:
 Originally Posted by Monox D. I-Fly 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

 October 24th, 2014, 10:24 AM #4 Newbie   Joined: Oct 2014 From: Lithuania Posts: 4 Thanks: 0 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?
October 24th, 2014, 10:51 AM   #5
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Quote:
 Originally Posted by robertson 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

 October 24th, 2014, 01:09 PM #6 Newbie   Joined: Oct 2014 From: Lithuania Posts: 4 Thanks: 0 hmmm .... nop, I'm not so good at this one... Doesn't work :/
 October 24th, 2014, 02:05 PM #7 Math Team     Joined: May 2013 From: The Astral plane Posts: 1,889 Thanks: 769 Math Focus: Wibbly wobbly timey-wimey stuff. 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. -Dan
 October 25th, 2014, 12:49 AM #8 Newbie   Joined: Oct 2014 From: Lithuania Posts: 4 Thanks: 0 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?
October 25th, 2014, 10:05 PM   #9
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Quote:
 Originally Posted by robertson 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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