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a)

| -1 -2 -3 ..... -k |   | -1 -2 -3 ..... -k |
|  1  0  3 .....  k |   |  0 -2  0 .....  0 |
|  1  2  0 .....  k | = |  0  0 -3 .....  0 | = 
|  .............  k |   |  ................ |
|  1  2  3 .....  0 |   |  0  0  0 ..... -k |

= (-1)(-2)(-3)...(-k) =

= (-1)^k*1*2*...*k =

= (-1)^k * k!

b)

|  0     0    .....  0       a_1k |   | a_k1 a_k2 ....... a_kk-1 a_kk |
|  0     0    .....  a_2k-1  a_2k | = | ............................. | *(-1)^k =
|  .............................. |   | 0    0    ....... a_2k-1 a_2k |
|  a_k1  a_k2 .....  a_kk-1  a_kk |   | 0    0    ....... 0      a_1k |

Please note that _k1, _k2 and all symbols after the _ are in subscript of a (I don't know how to make it).

Any hints how to solve the b) (I'm totally stuck, please help) and is a) correct?

user95523
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1 Answers1

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For b), the determinant of a triangular matrix is the product of the diagonals. But b) isn't quite triangular in the sense we understand triangular matrices usually. So, normally the step is to switch the rows around until you get a triangular matrix and every time you do multiply the determinant by -1 but...

A much simpler approach: the determinant is just the product of the diagonals. If your matrix has an odd dimension you multiply that determinant by -1 or else you're fine.

Here are two examples i just typed on wolframalpha. They're basically the same-ish matrix except one is 3x3 and the other is 4x4 and you can look at the results

3x3: http://www.wolframalpha.com/input/?i=determinant+%7B%7B0%2C0%2C1%7D%2C%7B0%2C2%2C3%7D%2C%7B-3%2C1%2C2%7D%7D

4x4: http://www.wolframalpha.com/input/?i=determinant+%7B%7B0%2C0%2C0%2C1%7D%2C%7B0%2C0%2C2%2C3%7D%2C%7B0%2C-3%2C1%2C2%7D%2C%7B1%2C1%2C1%2C1%7D%7D

Hassan Hayat
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  • I do know how to calculate the matrix if there are given numeric values, but I'm having some problems understanding the general concept in this example. Any hints for that? – user95523 Oct 14 '13 at 23:32
  • Ok, so the idea is that you can find the determinant of a matrix by recursion. (a good primer on determinants: http://www.youtube.com/watch?v=H9BWRYJNIv4 ) So, if you watch the video, you can see how a determinant is calculated by splitting it up into smaller determinants and constants and so on until you get a 1x1 determinant. The cool thing with triangular matrices is that only one of those constants ends up being non-zero at every step of the recursion, thus the whole determinant being the product of the diagonals. – Hassan Hayat Oct 14 '13 at 23:38
  • Now, the problem with your matrix is that it's not triangular in the sense triangular matrices are usually understood. You can get a triangular matrix out of that by shuffling the rows around. Shuffling the rows around is almost never a problem except with determinants. The reason that is a problem is it's kinda like you're playing with your frame of reference and flipping everything. To get a "real" triangular matrix, you want your rows to be listed in reverse order. Every time you switch two rows you have to multiply your determinant by -1. So if you play with this concept you'll get to – Hassan Hayat Oct 14 '13 at 23:44
  • the result I told you where you just check if your matrix has an even dimension and only then multiply by -1 – Hassan Hayat Oct 14 '13 at 23:45
  • Well, we aren't really studying so deep and we usually solve not-so-advanced problems, and I think where I'm going is way too deep for the knowledge I have of matrix. Is there any other simplier solution as in the other example, where I just multiply one row and add it to other one to get zeros and then calculate the triangular matrix? – user95523 Oct 14 '13 at 23:46
  • your matrix looks like this: $\begin{matrix} 0 & 0 & 1\ 0 & 2 & 3\ 4 & 5 & 6 \end{matrix}$

    And a triangular matrix looks like this:

    $\begin{matrix} 4 & 5 & 6\ 0 & 2 & 3\ 0 & 0 & 1 \end{matrix}$

    So, you just flip the order of the rows. Nothing deep. Super simple. You want the zeros to flow from the bottom left to the top right instead of the opposite.

     – Hassan Hayat Oct 14 '13 at 23:53
  • So, in this case, you'd flip the top row with the bottom row. So, the determinant gets multiplied by -1. Then boom you have your triangular matrix. The determinant is 4 x 2 x 1 = 8 then x -1 = -8 – Hassan Hayat Oct 14 '13 at 23:55
  • (oops, i just realized i said multiply by -1 when the dimension is even all along. huge typo, it's the other way around. just edited the answer) – Hassan Hayat Oct 14 '13 at 23:57
  • I also edited the answer, could you check if I'm on the right track? – user95523 Oct 14 '13 at 23:58
  • yup, perfect just one little detail. you have to multiply the right side by $-1^k$ (that solves the whole multiplying negative numbers thing) – Hassan Hayat Oct 15 '13 at 00:00
  • Shall I multiply the whole determinant or just the last element of the right side? – user95523 Oct 15 '13 at 00:02
  • the whole determinant – Hassan Hayat Oct 15 '13 at 00:02
  • Spoiler alert:

    The answer will end up being $determinant = a_{11} * a_{22} * ... * a_{kk} * -1^k$

     – Hassan Hayat Oct 15 '13 at 00:04
  • I multiplied the determinant, check the edit. Just a question about the answer, why is it the first row? And not the multiplication of the main diagonal? – user95523 Oct 15 '13 at 00:06
  • Sorry, it was a typo – Hassan Hayat Oct 15 '13 at 00:08
  • Ok, well, that wasn't so hard to be honest, except I don't quite get the -1^k part, but I'll check it out again and see if I can learn it. – user95523 Oct 15 '13 at 00:12