Math Is Fun Forum

EPhillips1989

Member

Registered: 2007-11-03

Posts: 29

matrices

can anyone prove that if ad-bc=0 then the matrix A is singular(not invertible)

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: matrices

I'm assuming you're talking about a 2x2 matrix with entries:

( a  b )
( c  d )

If so, then ad-bc gives the determinant of that matrix.
And if the determinant of a matrix is 0, then it is singular.

---

Why did the vector cross the road?
It wanted to be normal.

mikau

Member

Registered: 2005-08-22

Posts: 1,504

Re: matrices

another way to do it, suppose a and c are nonzero. then
multiply the top row by c and the bottom row by a

(ac bc)
(ac da)

subtract the top row from the bottom row

(ac       bc)
(0  bc - da)

now clearly of ad - bc = 0 then bc - da = -(ac - bc) = -0 = 0 which implies the second row is zero and thus having a nonzero row, the matrix is not invertible.

using a similar procedure, we can show that if a or b is zero, and b or c is non zero, then the matrix can be shown to have a zero row. And if all entries are zero, its trivial.

Mathsy's proof is just as good if you are allowed to use the determinant rule, however I suspect you haven't learned that yet or it would have been obvious.

Last edited by mikau (2007-11-26 03:30:06)

---

A logarithm is just a misspelled algorithm.