How to find detA if A is 4X4 matrix

[ranoo]

How to find detA if A is 4x4 matrix or more than 4x4.

 

[zhermes]

Do you know how to take the determinant of a 3x3 matrix?
Its the exact same idea: you multiply the first row, i'th column value by the determinant of all of the rows and columns besides the first row and i'th column. I.e. a |4x4| is found from a series of four |3x3|.

 

[Anonymous217]

Look up "Cofactor expansion".

 

[ranoo]

thank you

 

[HallsofIvy]

The simplest way, in my opinion, to find the determinant of a large matrix is NOT "cofactor expansion" but row- reduction. Row reduce the given matrix to the identity matrix using the three row- operations:
1) multiply an entire row by a number
2) swap two rows.
3) Add a multiple of one row to another row.

The determinant of the identity matrix is, of course, 1. Then remember that
1) multiplying an entire row by a number multiplies the determinant by that number.
2) swapping two rows multiplies the determinant by -1.
3) adding a multiple of one row to another row does NOT change the determinant.

So if you keep track of the numbers you multiplied the rows by, you can divide 1 by them. If you swapped rows an odd number of times the determinant is the negative of that quotient, if an even number of times, the determinant is that quotient.

(If you cannot row reduce the matrix to the identity matrix, then it is not invertible and its determinant is 0.)

 

[quietrain]

hi, i still don't really understand,

lets say i have a 4x4 matrix of

0 0 0 a
0 b 0 0
0 0 -b 0
a 0 0 0

so i did the cofactor expansion way and i ended up with E² (b² + a² - E² ) = (ab)²
where E = eigenvalue

which is complicated like crazy...

did i do something wrong or is there an easy way to see straight away what are the eigen values and eigen vectors for this 4x4 matrix? becos my lecturer just gave the eigen values as a , b , -a , -b. how did she arrive at this?

thanks a million!

 

[jasonRF]

The simplest way, in my opinion, to find the determinant of a large matrix is NOT "cofactor expansion" but row- reduction. Row reduce the given matrix to the identity matrix using the three row- operations:
1) multiply an entire row by a number
2) swap two rows.
3) Add a multiple of one row to another row.

The determinant of the identity matrix is, of course, 1. Then remember that
1) multiplying an entire row by a number multiplies the determinant by that number.
2) swapping two rows multiplies the determinant by -1.
3) adding a multiple of one row to another row does NOT change the determinant.

So if you keep track of the numbers you multiplied the rows by, you can divide 1 by them. If you swapped rows an odd number of times the determinant is the negative of that quotient, if an even number of times, the determinant is that quotient.

(If you cannot row reduce the matrix to the identity matrix, then it is not invertible and its determinant is 0.)

I just wanted to emphasize that this approach has WAY FEWER computations than a cofactor expansion for even modestly large matrices. If I have a numerical matrix this is the way I do it every time. For an NxN matrix, this takes roughly N^3 operations. The cofactor approach takes something like N!. So for a 10x10 matrix, the fast way takes the order of 1,000 operations, and the slow way takes more than 1,000,000 operations. Of course both of these would be done on a computer.

Cheers!

jason

edit: a furhter example: a 20x20 matrix takes about 8,000 operations the fast way, and more than 10^{18} the slow way. A 1 GHz computer that can do one operation per clock cycle would take more than 50 years the slow way!

 

[mathwonk]

just rearrange the columns until it is diagonal, then multiply the diagonal terms (and take (-1) to the number of column changes.)

i.e. just interchange first and last column. then take minus the product of the diagonal entries, namely a^2.b^2.