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I know only two way in proving invertibility of matrices which is by using determinant and making the matrices into row echelon form. Are there any other ways to prove invertibility of matrices?

Example of invertible $3 \times 3$ matrices
$$\left( \begin{matrix} 1& 0& 2 \\ 3& 1& 6\\ 8& 9& 2 \end{matrix} \right)$$

Example of non-invertible $3 \times 3$ matrices

$$\left( \begin{matrix} 1& 2& 3\\ 4& 5& 6\\ 1& 2& 3 \end{matrix} \right)$$

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look here –  Ale Jan 22 at 11:50
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2 Answers

Wikipedia has a nice list of answers to your questions.

Letting $A$ be a square matrix, I think some of the nicer ones on the list are,

  • The equation $Ax = 0$ has only the trivial solution $x = 0$.
  • The number $0$ is not an eigenvalue of $A$.
  • The columns of $A$ are linearly independent.
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The statement about eigen-values is just a re-statement of the first point, right? –  awllower Jul 21 '13 at 13:48
    
All of the statements in the list are actually equivalent so in that sense the answer is "yes". The proof of equivalency is pretty straightforward: If $Ax=0$ has only the solution $x=0$ then $0$ is not an eigenvalue $A$ (since $0$ would have no corresponding eigenvector). On the other hand, if $0$ is not an eigenvalue of $A$ then, that means the kernel is trivial, which means the only solution to $Ax=0$ is $x=0$. –  Christian Bueno Jul 21 '13 at 20:03
    
And it is because that the proof is so straightforward that I thought it would be trivial to add that point. But it is not a big matter in fact. –  awllower Jul 22 '13 at 1:53
    
Ah now I understand what you mean. Nevertheless, I felt the two statements had distinct enough "feels" that it would be nice to state both, even if the equivalence was trivial. –  Christian Bueno Jul 22 '13 at 14:00
    
OK. Thanks for clarifying. –  awllower Jul 23 '13 at 7:17
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In addition one should note that for square matrices in $M_n(K)$, in addition to the determinant, the field is important, too. For example, the first matrix $$ A=\begin{pmatrix} 1 & 0 & 2 \cr 3 & 1 & 6 \cr 8 & 9 & 2 \end{pmatrix} $$ is not invertible over a field of characteristic $2$ or $7$, because in this case the determinant $\det(A)=-14=0$. This arises, for example, in coding theory.

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