Determinant of a 4x4 matrix?

Question

-7 0 0 3
-5 -5 0 0
0 -5 0 0
-9 -4 5 2

find the determinant of this matrix!
thank you!

Answer 1

Without a calculator? I guess you'd expand along the third row or the third column since they each have 3 zeros. Say you use the third column. You would have -5

(negative because of the
+ - + -
- + - +
+ - + -
- + - + pattern)

times the 3x3 determinant formed by crossing out the third column and the 4th row which is

-7 0 3
-5 -5 0
0 -5 0

Which now could be expanded along the third column to be

3 times the determinant formed by crossing out the third column and fist row:

-5 -5
0 -5

which is 25

So -5 (3)(25) = -375

Answer 2

So we want to use the thrid row down because it has a lot of zeros and everything will cancel out except for the -5 times a new determinate.

We are going to use the method of co factors... the best way is by example

so say i had a 3X3

1 0 0
0 1 0
0 0 1

this i obviously easy but say i didn't know. Take the first row (you can pick any row but you need to make an adjustment that i will show you in a bit.

now i take the first element of the first row (one) and cross out its column and row. Notice how i get a new square, but smaller, matrix from the numbers that are not crossed out. Don't see it? well here

(* * *)
(*1 0)
(*0 1)

now we take one and multiply times the determinate of the new matrix

1*det(new matrix)

we are not done because we need to cross out every column like this so,

0*det

(***)
(0*0)
(0*1)

and the next

0*det...

(* * *)
(0 1*)
(0 0 *)

Now we need to get the proper sings of the cofactors (the number from the row we picked). The sings of the numbers go like this

(+ - + - +....)
(- + - +.......)
(+ - +.........)

so it kinda looks like a chess board with positives and negatives. Another way to do it is to know the i and j of the indivdual cofactor.... then it's sign is (-1)^(i+j).

so for your matrix

-(-5)*det...
(-7 0 3
-5 0 0
-9 5 2)

use the second row now because we have a lot of 0's again.

-(-5)*-(-5)*det...
(0 3
5 2)

so we have an easy determinate now

-(-5)*-(-5)*(-15) = -375

Answer 3

You can define the determinant recursivley: consider the matrix a b c d e f g h i j k l m n o p the determinant of this matrix is a multiplied my the determinant of the submatrix: f g h j k l n o p minus b time the det of the submatrix: g h e k l i o p m plus c times the det of the submatrix; h e f l i j p m n minus d times the det of the submatrix: e f g i j k m n o so a*det(submatrix1)-b*det(submatrix2)+c*de... so if you can find the determinant of the 4 3x3 sub matricies, you can find the 4x4 determinant. It turns out you can define the 3x3 determinant in a similar way, same as a 2x2, and the det of a 1x1 is just the single element. Hope this helps!

Answer 4

= - -5 * det ( 7 0 3)
................. (-5 0 0)
................. (-9 5 2)
= + 5 * 3 * -25
= -375

Answer 5

long drawn out process. look it up on wikipedia they have a pretty good walkthru on determinites. Also you can just input that into a graphing calculator and get the determiite.