how to find determinant of a 4x4 matrix?

Question

i know only upto 3x3 matrix.

Answer 1

Determinants for matrices up to 3x3 can be found just by multiplying the right diagonals (\)and adding them together then by multiplying the left diagonals (/) and subtracting them from the total.



For a 2x2 matrix there is 1 right and 1 left diagonal and for a 3x3 matrix there are 3 right and 3 left diagonals. However, for larger matrices this method will not usually result in the determinant. Instead you have to use cofactors to calculate the matrix.



The cofactor of element (i,j) where i is the row and j is the column is the determinant of the matrix excluding the ith row and jth column.



The determinant of a matrix is then the sum along any row or column of the matrix of each cofactor, multiplied by its corresponding element multiplied by -1^(i+ j)



Suppose you have the 2x2 matrix



|a b|

|c d|



The cofactor of a is



|- -|

|- d|



which is just d (you don't subsisute 0 for the excluded row and column you just assume they don't exist.



Similarly, the cofactor of b is



|- -|

|c -|



which is just c.



So the determinant along row 1 is



D = -1^(1+1)*a*d + (-1)^(1+2)*b*c

D = ad-bc



which is just what you'd expect.



If I wanted to calculate the determinant over column 2 (b d) then the cofactor of b is c and the cofactor of d is a and



D = -1^(1+2)*b*c + -1^(2+2)*d*a

D = -bc+da

D = ad-bc



again. Now when you use this method to calculate the determinant of a 3x3 matrix, the cofactors will be determinants of 2x2 matrices (since a full row and column of a 3x3 matrix are excluded) which you know how to compute. Similarly, the cofactors of a 4x4 matrix are determinants of 3x3 matrices and you will see this is sort of a recursive relation.



Once you understand how to use cofactors, calculating a determinant of any matrix will become though tedious.



One tip is to calculate the determinant over a row or column with the most 0s since you don't need to calculate the cofactor of any 0 element.

Answer 2

You have to use "Laplace expansion".

It works just like written here:



http://en.wikipedia.org/wiki/Laplace_expansion



You will get your 4x4 matrix determinant divided into a sum of four 3x3 determinants, which you already know how to find.



Hovever, there is a "trick" that allows you to find only one 3x3 deteminant not four.

There is a rule saying:

"If to any column(row) in a square matrix you add another column(row) multiplicated by any number, the determinant for that matrix won't change it's value".



It's very useful. I'll show an example. Let's say I have a matrix:



1 2 3 4

2 2 6 4

2 2 3 8

4 2 3 4 ( I have used a simple one to avoid fractions )



According to the above rule I do the folowing:

1. I add 1st row multiplicated by -2 to second row

2. I add 1st row multiplicated by -2 to third row

3. i add 1st row multiplicated by -4 to fourth row

What do I get now?

1 2 3 4

0 -2 0 -4

0 -2 -3 0

0 -6 -9 -12



Now I'm going to make a Laplace expansion along the first column. I get the four smaller determinants in my sum as usual but three of them will be multiplicated by zero! And there's only one left to determine equal to:

-2 0 4

-2 -3 0

-6 -9 -12

And it's plain simple to solve. There can be an error in my equetations but the rule is true. It is also true that it is always possible to bring your matrix to the form in which it has all zeroes or one "1" and other zeroes in one column(row). it doesn't have to be easy if there are are some "unfriendly" numbers but it's possible.

You can use it for a matrix of any size countinuesly decreasing it's size until you can find the determinant.



I hope my answer was helpful:-)

Answer 3

It is the same as 3x3. just cofactor A11, and treat the rest as a 3x3..then do the same for A22..etc.

Answer 4

The matrix is everything. It is all around you. you must wake up, and realise the truth. YOU MUST UNPLUG YOURSELF FROM THE MATRIX LIFE ISN'T REAL THE GOVERNMENT IS REPTILES!!!!!!