Making a matrix singular, i have to make the determinant 0, why isn't this right?>

1. Oct 16, 2005

mr_coffee

Hello everyone i'm suppose to find a k, that will make this matrix singular, well i looked up what that ment, and it means i have to make the determinant 0. So i the matrix is:
-3 2 1
-9 8 3
(9+k) 2 0

so i found a random minor, say M13..
so i would have
-9 8
(9+k) 2

det = (-9)(2) - [8(9+k)] = -18 -72 -8k = -90 -8k;
so i want k to be 0, so i solve for k
-90 -8k = 0;
k = -90/8, i submitted it, and it was incorrect, any ideas why?

2. Oct 16, 2005

AKG

Why would making one of the minors 0 make the whole determinant 0? Did you even calculuate the determinant of your matrix? Basically, what you've done is taken:

a*det(A) - b*det(B) + c*det(C)

and said that if as long as one of det(A), det(B), or det(C) is zero, the whole thing is somehow zero. Clearly that's wrong. If the determinants of all the minors were zero, then you'd be fine, but you can't just show one of them to be zero and think that it does anything.

A matrix is singular when its rows and/or columns are linearly dependent. The easiest way to do this is to make two rows or two columns scalar multiples of each other.

3. Oct 16, 2005

Muzza

You need to work out the determinant of the entire matrix, not just one of its minors.

4. Oct 16, 2005

Fermat

The determinant to be zero is the whole determinant. The det of A.

Work out the det of A with the elements given and equate that to zero.

5. Oct 16, 2005

mr_coffee

ohh my bad, i don't know why I thought that, too much coffee. Thanks it worked out great!