# Ways of Proving invertibility of $3 \times 3$ matrices

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

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

• look here – Bman72 Jan 22 '14 at 11:50

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.
• 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
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.