Help me find the determinant of the 5x5 matrix.?

Question

6 -1 3 4 6
8 1 6 9 6
1 2 5 6 6
9 7 7 7 1
1 2 7 8 10

I don't need the answer, but I want to know the formula. I have no idea how to do this ^^;.

Answer 1

this will be hell if you do it the normal way...
The NORMAL way:
to find the determinant of a matrix larger than 2x2, pick a row or column... for each of it's members find:
(1) (-1)^(m+n) where m and n tell the position of that member in the matrix... eg if it is the left upper corner then m=1, n=1
the value of (1) is either -1 or 1
(2) find the DETERMINANT of another matrix that is acquired by deleting the m and n row and colum from the original matrix... the matrix acquired should be smaller, eg if you start out with 5x5 you'll get 4x4 matrices in this step
Next, multiply (1) and (2) AND the actual value of the number in that position...

Do these steps for all the members in the row/column and add them up...

As you can see this method works best for 3x3 matrices, for obvious reasons that you only have to do it once cuz it reduces to 2x2 matrices in step 2

OR

ALTERNATE METHOD: cross multiplying (do an internet search or ask someone on this one cuz it's hard to type...)
This one is much, much easier and should take around 2 minutes tops

Answer 2

Z_o_r_r_o's method works, but takes many calculations.

Helmut's method only works for 3x3 determinants. For 5x5 determinants you would need to add/subtract 120 products of 5 numbers, not just the 10 diagonals.

----------- QUICK METHOD --------------------
(Requires only 60 multiplications and 30 divisions.)

Let M(i,j) stand for the element in the i-th row, j-th column.

a := 1
REPEAT for i = 1 .. 4
.. p := M(i,i)
.. REPEAT for j = i+1 .. 5
.. .. q := M(j,i)
.. .. REPEAT for k = i .. 5
.. .. .. REPLACE M(j,k) by (p*M(j,k) - q*M(i,k)) / a
.. a := p
Det = M(5,5)

-------------------- EXAMPLE --------------------

First stage: a = 1; i = 1; p = 6
======================
* Row j = 2: q = 8
(6 * 8 - 8 * 6)/1 = 0
(6 * 1 - 8 * (-1))/1 = 14
(6 * 6 - 8 * 3)/1 = 12
(6 * 9 - 8 * 4)/1 = 22
(6 * 6 - 8 * 6)/1 = -12

* Row j = 3: q = 1
(6 * 1 - 1 * 6)/1 = 0
(6 * 2 - 1 * (-1))/1 = 13
(6 * 5 - 1 * 3)/1 = 27
etc.

At the end of stage 1, the matrix looks like
6 -1 3 4 6
0 14 12 22 -12
0 13 27 32 30
0 51 15 6 -48
0 13 39 44 54

Second stage: a = 6; i = 2; p = 14
=========================
* Row j = 3: q = 13
(14 * 13 - 13 * 14)/6 = 0
(14 * 27 - 13 * 12)/6 = 37
(14 * 32 - 13 * 30)/6 = 27
etc.

The matrix becomes
6 * * * *
* 14 * * *
* 0 37 27 96
* 0 -67 -173 -10
* 0 65 55 152

Third stage: a = 14; i = 3; p = 37
==========================
The matrix becomes
6 * * * *
* 14 * * *
* * 37 27 96
* * 0 -328 433
* * 0 20 -44

Fourth stage: a = 37; i = 4; p = -328
==========================
* Row j = 5; q = 20
(-328 * (-44) - 20 * 433)/37 = 156

The matrix becomes
6 * * * *
* 14 * * *
* * 37 * *
* * * -328 433
* * * 0 156

The determinant is 156.

Answer 3

+6*1*5*7*10...It's kind of hard to see, but you take the
-1*6*6*1*1......products of n diagonals going down-
+3*9*6*2*9 ....and-left, and subtract the products of n
+4*6*7*7*1 ....diagonals going up and right. Each
+6*8*7*2*8 ....product must have n terms in it, where n
-1*7*5*9*6 .....is the size of the matrix. Note the wrap-
-9*2*6*4*10....around effect.
-1*1*3*1*8
+8*1*6*7*7
-6*6*6*7*2 = determinant

The differing signs are due to the single -1 in row 1.

These days I set these up in a spreadsheet & solve the simultaneous equations by eliminating columns & rows 1 by 1.

Answer 4

I'm sorry, this sort of problem always ends up being a lot of work.

In the most general sense you find the determinant of 5 4x4 matrices and then their sum is the determinate of the 5x5. To find the determinate of a 4x4 you reduce it to a 3x3 and etc.

Here is the general rule to reduce a matrix.
|a b c|
|d e f| = a*|e f| -d*|b c| + g *|b c|
|g h i|____|h i|___|h i|_____|e f |

So what I did is go down a column and each number in that column by the determinate of a matrix found by omitting the row and column the number is in from the original. Also you have to alternate between plus and minus.

Answer 5

Look at this link:

http://i2.photobucket.com/albums/y16/zorro1267/determ.gif

This shows how to break a 5x5 determinant into 4x4 determinants.

The 4x4 determinants are evaluated the same way by picking a row and then crossing out row and columns of a particular element in the chosen row.

The sign always alternates as you move along the elements of the row. The first element is always positive. Then the 2nd negative, then positive, etc.

Answer 6

try here