Does an $n\times n$ matrix $A$ only have an inverse if $rank(A)=n$? If so, why? I'm currently learning about the rank and inverses of matrices, and every time I attempt to find the inverse of a matrix with a rank smaller than it's number of rows, I find I am unable. One example of this is as follows:
\begin{bmatrix}1&1&0\\1&0&1\\1&0&1\end{bmatrix}
The rank of this matrix is 2 while it has 3 rows, and it doesn't have an inverse. Have I just run into examples where this is true, or is this always true? If it is true, why?
 A: The rank of a matrix is equivalent to the number of nonzero rows after Gaussian elimination. If the rank is not maximal, then there is at least one nonzero row after elimination. Do the determinant about this row, and it will be $0$, i.e., the matrix is not invertible.
A: Let's look at the $3 \times 3$ case in general. If $A = [A_1|A_2|A_3]$ then note for $c = [c_1,c_2,c_3]^T$:
$$ Ac = [A_1|A_2|A_3]\left[ \begin{array}{c} c_1 \\ c_2 \\ c_3 \end{array} \right] =c_1A_1+c_2A_2+c_3A_3.$$
Thus, the equation $Ac=0$ is equivalent to the condition $c_1A_1+c_2A_2+c_3A_3=0$. Now, if $\{A_1,A_2,A_3 \}$ is a linearly independent set then we know the condition $c_1A_1+c_2A_2+c_3A_3=0$ implies $c_1=c_2=c_3=0$. On the other hand, the existence of an inverse $A^{-1}$ also allows us to trade $Ac=0$ for $A^{-1}Ac=A^{-1}0$ or $Ic=0$ or more to the point $c=0$. Hence, if the inverse exists, the only solution to $Ac=0$ is just $c=0$. But, as I explained above, this is equivalent to insisting the only solution to $c_1A_1+c_2A_2+c_3A_3=0$ is the zero solution $c_1=c_2=c_3=0$.
A: The rank of a matrix is the number of linearly independent column vectors of this matrix, i.e. the dimension of the subspace $\,\operatorname{Im}A$. Furthermore, $A$ can be seen as the matrix associated, in a certain basis,   to a linear map $f\colon K^n\to K^n$ ($K$ is the base field).
For an $n\times n$-matrix, this means that 
\begin{align*}\operatorname{rank}A=n &\iff \operatorname{rank}f=n \iff f\enspace\text{is surjective}\\ &\iff f\enspace\text{is bijective}\iff A\enspace\text{is invertible.}\end{align*}
A: If a matrix $A$ is invertible, then $A\vec x=\vec b$ has a unique solution $\vec x=A^{-1}\vec b$. By the rank theorem we know that $rank(A)+nullity(A)=n$. If $rank(A) < n$, then $nullity(A) \geq 1$. But this means that $A\vec x = \vec 0$ has infinitely many solutions. This is not possible if $A$ is invertible, because we expect a unique solution.
