How do you find the determinant of a 3 by 3 matrix by using 2 by 2 matrices?

Answer 1

your matrice :


a b c
d e f
g h i


det = a*[ei-fh] - d*[bi-ch] + g*[bf-ce]

in fact, there's a lot of solution .. remenber the rule for + & -

your matrice for multiplication is :

+ - +
- + -
+ - +

Answer 2

If you have a Ti-89, then it will do it for you.

Otherwise take the top row and do + - +, those are your base numbers. In Calc III these are often the unit vectors i, j, and k. Anyways, multiply these by the determinants of the smaller 2x2 matrices x-ing out the column the number is in to get the 2x2.

IE

1 2 3
4 5 6
7 8 9

for the first part, 1's 2x2 would be

56
89

and the determinant of that is
5*9-6*8
which is a little backwards alpha if you draw it out

Here is how I remember, and how i do it on homework. i use 3 lines and go all the way across the page.

|1 2 3| = 1(5*9-6*8)
|4 5 6| = -2(4*9-6*7)
|7 8 9| = +3(4*8-5*7)

Answer 3

If you have a matrix:

a b c
d e f
g h i

Pick a row or column.
Take the first element in that row (or column) and multiply it by the remaining 2x2 matrix of elements that are not in the same row or column. repeat for the 2nd element in the row you picked then the the 3rd.

In addition you have to change the signs of every other term based on the following.

+ - +
- + -
+ - +

so from our original matrix if we take the first row
a * |e f| - b * |d f| + c |d e|
. |h i| |g i| |g h|

if you picked the second row, the signs on your terms would be
- + - just like the second row of the matrix of + and - above

You can use this procedure on any size matrix to find the determinant although it gets pretty tedious for anything larger than 3 x 3.

Answer 4

The method is annoying and involved. But you'll get used to it right away. I suggest you block out the first row and column, and multiply the number that shares both with the determinant of a matrix formed of the remaining rows and columns. Call this product "A". Then block out the first row and second column, and multiply the number that shares both with the determinant of a matrix formed of the remaining rows and columns. Call this product "B". Then block out the first row and the third column, and multiply the number that shares both with the determinant of a matrix formed of the remaining rows and columns. Call this product "C". The 3x3 determinant will be A - B + C.

Answer 5

It's called 'expansion by minors and cofactors' an it's the *only* way to find the determinant of a matrix larger than 3X3.

There's a good article on the subject at http://www.richland.edu/james/lecture/m116/matrices/determinant.html


Doug

Answer 6

..but Doug, don't forget the graphing calculator!