Proving ${\det}(A) = 0$ Prove that if a row or column of an $n\times n$ matrix $A$ consists of entirely zeros, then ${\det}(A)=0$.
I know the definition of determinant and how to calculate the determinant for any $n\times n$ matrix. I'm just somewhat lost on how to bring what I know into a generic proof for any matrix that is $n\times n$.
Any help or guidance would be greatly appreciated. Thanks! 
 A: The trivial case is when $A$ is a $1\times 1$ matrix.
For $n\times n$ matrix $A$, where $n$ is an integer greater than or equal to 2, we can use cofactor expansion in a particular row (or column) to find $\det(A)$. The following link is useful: http://en.wikipedia.org/wiki/Minor_%28linear_algebra%29. The section on "Applications of minors and cofactors" is quite useful.
Now, suppose that the $i$th row consists entirely of $0$s. We choose this row for finding $\det(A)$ using expansion by cofactors. Then, $\det(A)=\displaystyle\sum_{j=1}^{n}a_{ij}C_{ij}=0$ because $a_{ij}=0$ for all $j=1,2,\ldots,n$.
Similarly, suppose that the $j$th column consists entirely of $0$s. We choose this column for finding $\det(A)$ using expansion by cofactors. Then, $\det(A)=\displaystyle\sum_{i=1}^{n}a_{ij}C_{ij}=0$ because $a_{ij}=0$ for all $i=1,2,\ldots,n$.
A: You mention in comments that you know $\det(A)=0$ iff $A$ is uninvertible.  To show $A$ is uninvertible if it has a zero row, find $b$ such that $Ax=b$ has no solution.  To show $A$ is uninvertible if $A$ has a zero column, find nonzero $x$ such that $Ax=0$.
A: Use contradiction. suppose that the matrix $A_{n\times n}$ have a row or column that consists of entirely zeros and $\det(A)\neq0$ then, there is  $A^{-1}$ such that:
$$A\cdot A^{-1}=I_{n\times n}$$
but this not is possible. 
