Is there any trick for finding the determinant of this 4x4 matrix?
[ 1, 1, 1, 1 ]
[ 1+a, 1, 1, 1 ]
[ 1, 1+b,1, 1 ]
[ 1, 1, 1+c, 1 ]
2006-10-24 19:35:05 · 6 answers · asked by Amir E 1 in Science & Mathematics ➔ Mathematics
Yes.
1,1,1,1
a,0,0,0 2nd row - 1st
0,b,0,0 3rd row - 1st
0,0,c,0 4th row - 1st
Now it's a lot easier for you to calculate the determinant.
2006-10-24 22:45:18 · answer #1 · answered by Anonymous · 0⤊ 0⤋
Brute force seems to be the easiest:
[.. 1,.. 1,.. 1,.. 1 ]
[ 1+a, 1,.. 1,.. 1 ]
[.. 1, 1+b,.1,.. 1 ]
[.. 1,.. 1, 1+c, 1 ]
Î = 1 + 1 + 1 + (1 + a )( 1 + b)(1 + c) - 1 - b - 1 - c - 1 - a -1
Î = (1 + a )( 1 + b)(1 + c) - 1 - b - c - a
Î = (1 + b + a + ab)(1 + c) - 1 - b - c - a
Î = 1 + b + a + ab + c + bc +ac + abc - 1 - b - c - a
Î = ab+ bc +ac + abc
Given the solution, it LOOKS like there should be a quick way of getting there, but I think that's an illusion.
2006-10-25 03:54:37 · answer #2 · answered by Helmut 7 · 0⤊ 0⤋
Break it down into 4 3x3 matrices by using the first row and last column, that way all the 3x3 matrices will be multiplied by one, to make it easier for computation. Also don't forget the plus and minus locations of the matrix.
2006-10-25 02:50:10 · answer #3 · answered by bloop87 4 · 0⤊ 0⤋
try to reduce it to a 3*3 matrics. than solve it.
2006-10-25 02:43:54 · answer #4 · answered by pramod bhat 1 · 1⤊ 0⤋
Use the criss-cross trick..
2006-10-25 02:40:51 · answer #5 · answered by avon_manu1009 2 · 0⤊ 0⤋
Sorry! I really hate mathematics!
2006-10-25 04:28:11 · answer #6 · answered by Anonymous · 0⤊ 0⤋