# Two proofs about invertible Matrix and row equivalent to the identity matrix.

2 proofs: 1. A matrix n*n is invertible if and only if it's row equivalent to the identity matrix. 2. A matrix n*n is not invertible if and only if it is row equivalent to a matrix with zero row.

(proofs without determinant of course)

I know it's a simple question but I would like to see a simple and formalic proof for this issue because I could only write one with words and it looks terrible..

Thanks

First off, let me mention that there is nothing wrong with a wordy proof as long as your arguments are clear. In fact, trying to avoid words will often be counterproductive. You'll find that my proofs are quite wordy as well; it's a bit unavoidable given the nature of the statements you're trying to prove.

As for the proof, here is one of many ways to prove the statement (I will prove both statements together).

A matrix $A$ is row equivalent to a matrix in Reduced Row Echelon Form (RREF) $R$. Therefore there exists a sequence of elementary matrices $\{E_i\}$ such that $$E_k\cdots E_2E_1A = R$$ Since each elementary matrix $E_i$ is invertible, it follows that $A$ will be invertible if and only if $R$ is invertible.

But $R$ is in RREF, so either $R$ has an all zero row or is equal to the identity. Therefore $R$ is invertible if and only if it is the identity and $R$ will be non-invertible if and only if it has an all zero row. Putting all this together, you have your two statements:

1. $A$ will be invertible if and only if it RREF $R$ is invertible if and only if $R=I$.

2. $A$ will be non-invertible if and only if its RREF form $R$ is non-invertible if and only if $R$ has an all zero row.

A n*n is an invertible matrix. -> Ax=b has a unique solution for each b belonging to Rn. This means that we have n pivots in A and no free variables. Thus, when transformed to it's RREF , we get the nth order identity matrix.