Proof on invertible matrices

[dim&dimmer]

Homework Statement

Show that the nxn matrix A is invertible iff its determinant is non-zero.
I think I can do this, but would like the validity checked.

Homework Equations

I would use |A| = the product of diagonal entries, because I don't know how to prove the non-diagonal entries of zero for $$A{C^T} =|A|I$$, where C is the cofactor matrix.(shown on Gil's videos)
$$AA^{-1}=I$$

The Attempt at a Solution

Using eliminationto show A in rref, That is, all non diagonal entries of zero, then the det is the product of diagonals. If det=0 then at least one entry of diagonals (a11,a22,...,ann) = 0. Then this 0('s) entry cannot be multiplied to produce the entry in the identity matrix needed for $$AA^{-1}=I$$
Is this ok?, if not , some direction would be most appreciated.
DIM

 

[mighty2000]

Your solution is on the right track, but there are a few things that could be clarified or expanded upon.

Firstly, the statement "If det=0 then at least one entry of diagonals (a11,a22,...,ann) = 0" is not necessarily true. The determinant being 0 means that the product of diagonal entries is 0, but that doesn't necessarily mean that every diagonal entry is 0. For example, in a 2x2 matrix with a11 = 2 and a22 = 0, the determinant is 0 but only one of the diagonal entries is 0.

To prove that the determinant being 0 implies at least one diagonal entry is 0, you could use contradiction. Assume that all diagonal entries are non-zero, and then show that this leads to a contradiction with the determinant being 0.

Secondly, you mention using elimination to show A in rref, but you should also mention that the rref must have a pivot in every row. This is important because if there is a row of all 0's in the rref, then the determinant will be 0 regardless of the diagonal entries.

Finally, you could expand on your explanation of why a non-zero determinant implies invertibility. You could mention that a non-zero determinant means that the matrix is not singular, meaning it has a unique solution when solving a system of linear equations. This is necessary for the existence of an inverse. You could also mention that the inverse can be found using the adjugate matrix, which is essentially the transpose of the cofactor matrix. This ties in with the cofactor matrix mentioned in the Homework Equations section.

Overall, your solution is good but could benefit from some more clarification and expansion. Good job!