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I need to "demonstrate that a larger n x n matrix is invertible". From Google, I see that a matrix is only invertible if its row reduced echelon form is an identity matrix.

Is this true? Does it have to be an identity matrix similar to \begin{pmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{pmatrix}

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    $\begingroup$ There are many equivalent conditions to a matrix being invertible. See Invertible Matrix Theorem. Yes, among the equivalent conditions is that the RREF of the matrix is an identity matrix which is a square matrix who have $1$'s along the main diagonal (where the row number is equal to the column number) and zeroes everywhere else. $\endgroup$
    – JMoravitz
    Mar 3 '20 at 19:31
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Yes, it is true.

If a matrix is non-singular, we can perform row operations to reduce it to the identity matrix and the RREF is indeed the identity matrix. That is we can find elementary matrices such that

$$E_n \ldots E_1 A = I_n$$

A non-singular matrix can look like $\begin{bmatrix}1 & 2 \\ -1 & 3\end{bmatrix}$ and its RREF is the identity matrix.

I noticed that you use the word "similar" , of which it carries a special meaning. Two matrices $A$ and $B$ are similar if there exists a non-singular matrix $P$. I am not sure if you use it with this definition in mind or not but if a matrix $A$ is similar to the identity matrix, then we have $A=P^{-1}IP=I$. That is the only matrix that is similar to the identity matrix is the identity matrix.

You might like to check if your matrix has any special structure such as is it a triangular matrix or a diagonally dominant matrix to check if a matrix is invertible.

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