# Non-zero determinant iff matrix is invertible.

1. Jan 14, 2010

### ksm100

1. The problem statement, all variables and given/known data
Given that A is any 2x2 matrix show that it is invertible if and only if det(A) $$\neq$$ 0.

2. Relevant equations

3. The attempt at a solution
If A is invertible then we know there exists an inverse matrix, say B, such that AB = BA = I.
It follows that det(AB)=det(BA)=det(I), and we know that det(I) = (1*1) - (0*0) = 1, so
det(AB) = det(A)det(B) = 1 implies both det(A) and det(B) are both nonzero.

However, I'm unsure how to show the converse.

If we suppose det(A) is not equal to 0, we know that no rows/columns of A are all zero, no two rows/columns are equal, and one row/column is not a multiple of the other.

I'm stuck here.. if anyone could help I'd really appreciate it!
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Jan 14, 2010

### rock.freak667

How would you write A-1 in terms of det(A) and adj(A)?

3. Jan 14, 2010

### Hurkyl

Staff Emeritus
Do you have any explicit formulas* for the inverse?

If not, then if you wrote down a generic matrix (its entries are variables), do you have an algorithm to compute* the inverse?

If not, can you write down another generic matrix at least solve* the system of equations that says "this matrix is the inverse of the other one"?

*: Paying careful attention to when it does and doesn't work? e.g. if you would divide by the expression (b-a), you should keep track of the fact you're assuming b-a is nonzero. (And then consider the case b=a separately)

4. Jan 14, 2010

### ksm100

So I can just say that because I know that det(A) isn't zero, 1/det(A) is defined and therefore A^(-1) exists?

5. Jan 14, 2010

### rock.freak667

Well I think that would work. I am not sure what sort of proof you are expected to give though.

6. Jan 15, 2010

### JG89

Use the fact that an n*n matrix is invertible if and only if its rank is n.