Linear Algebra: Fastest way of finding the determinant of a matrix?

Question

What is the fastest way of finding the determinant. I know taking the first 2 rows and adding the products and subtracting respectively. Another is a_11 |a_22*a_33 - a_23*a_32| etc... Another is adding the diagonal and subtracting the reverse + products of the corners. Then of course reducing the matrix. But I see my prof do something that's really quick, and trying to figure out. She multiplies out the diagonal then points to the lower triangle, writes a number, the points to the top writes a number and adds them. What did she do?

Update:

@Siths, yeah that's the way I've been doing it, it's apparently faster.

Siths and Giggles · Accepted Answer

Suppose you have the following 3x3 matrix:
[ 1 ... 3 ... -4 ]
[ 2 ... -1 ... 0 ]
[ 1 ... -3 ... 5 ]

(This is the way I remember to find the determinant of a matrix. I'm not sure if it's the same as your prof's.)
Now, imagine extending the matrix by rewriting its contents to the right:
[ 1 ... 3 ... -4 ] ... 1 ... 3 ... -4
[ 2 ... -1 ... 0 ] ... 2 ... -1 ... 0
[ 1 ... -3 ... 5 ] ... 1 ... -3 ... 5

Circle (I've marked with asterisks) the first diagonal set of numbers starting from the left (1, -1, and 5), and multiply them together.
[ 1* ... 3 ... -4 ] ... 1 ... 3 ... -4
[ 2 ... -1* ... 0 ] ... 2 ... -1 ... 0
[ 1 ... -3 ... 5* ] ... 1 ... -3 ... 5
(1)(-1)(5) = -5

Then the next (3, 0, and 1), and multiply them together (0).
And the next (-4, 2, and -3), and multiply again (24).
Only circle as many diagonals as there are numbers in the top row of the matrix (there are three).

Now, you take the products of these diagonals and add them together.
(-5) + (0) + (24) = 19

Next thing I do is circle the diagonal sets starting from the right, and move left from there:
[ 1 ... 3 ... -4 ] ... 1 ... 3 ... -4*
[ 2 ... -1 ... 0 ] ... 2 ... -1* ... 0
[ 1 ... -3 ... 5 ] ... 1* ... -3 ... 5

First set (-4, -1, and 1), and multiply (4).
Second set: (3, 2, and 5), product = 30
Third set: (1, 0, -3), product = 0

Add these products together:
(4) + (30) + (0) = 34

Next, you subtract the second sum of the products from the first:
19 - 34 = -15

Check this link for more info: http://en.wikipedia.org/wiki/Determinant
- - -
Might as well add the general case for any 3x3 matrix:
[ a ... b ... c ] ... a ... b ... c
[ d ... e ... f  ] ... d ... e ... f
[ g ... h ... i  ] ... g ... h ... i

Determinant =  (aei + bfg + cdh) - (ceg + bdi + afh)

drought · Answer

enable A be the orthogonal matrix then, det(A)=det(A^T){if the rows and columns of a determinant is interchanged then the cost of determinamt is comparable} concern for orthogonality of matrix is, A*A^T=I,I is the identity matrix det(A*A^T)=det(I) det(A)*det(A^T)=a million det(A)^2=a million

Linear Algebra: Fastest way of finding the determinant of a matrix?

2 Answers