An invertible matrix Given Matrix $A$, checking that its diagonal elements are nonzero or whether its determinant is nonzero, can we say the matrix is invertible for sure? Are there other properties that by looking at the matrix you can tell whether it's invertible or not?
 A: If you notice that two rows are scalar multiples of one another, then that would lead the determinant being zero, and therefore the matrix would not be invertible.
Example: $\begin{bmatrix}1&2&3\\2&4&6\\12&2&3\end{bmatrix}$ is not invertible because rows $1$ and $2$ are scalar multiples of each other. Also, as for checking if the elements on the diagonal are non-zero, this only works if the matrix $A$ is already simplified to Reduced Row Echelon Form, because if two rows in the matrix were linearly dependent, then the RREF of the matrix would have a row of zeros.
If the determinant is non-zero or if the matrix has $n$ pivot points (for $A_{n\times n})$, you can say for certain that the matrix is invertible.
Also, for your question about other properties that a matrix might have that would indicate invertibility: the following are equivalent, so if you notice any of the properties below describe your matrix, then your matrix is invertible.
$(a) A$ is an invertible matrix.
$(b) A$ is row equivalent to the $n \times n$ identity matrix, $I_n$. 
$(c) A$ has $n$ pivot positions. 
$(d)$ The equation $Ax=0$ has only the trivial solution. 
$(e)$ The columns of $A$ form a linearly independent set. 
$(f)$ The linear transformation $x\to Ax$ is one-to-one. 
$(g)$ The equation $Ax=b$ has at least one solution for each $b\in\mathbb{R}^n$. 
$(h)$ The columns of $A$ span $\mathbb{R}^n$.  
$(i)$ There is an $n \times n$ matrix $C$ such that $CA=I.$ 
$(j)$ There is an $n \times n$ matrix $D$ such that $AD=I$. 
$(k)$ $A^T$ is an invertible matrix. 
