Show that a matrix with (I) a row of zeros and (II) a column of zeros cannot be invertible (respectively) Show that a matrix with a row of zeros cannot be invertible. Show that
a matrix with a column of zeros cannot be invertible.
What I tried: I tried to show that a matrix $A \in M_n (\mathbb{R})$ such that $(A)_{ij} = 0 \forall j \in \mathbb{N}, j\leq n$ and then show that $A A^{-1} \neq I$ but I got stuck.
 A: Win Vineeth has given the quickest way to see this. Of course, you can also see this using basic properties of matrix multiplication. If $A$ has a row of all $0$'s, then for any matrix $B$ (where $A$ and $B$ are square matrices of the same size), it's not hard to show that $AB$ has a row of all $0$'s as well. Similarly, if $A$ has a column of all $0$'s, then $BA$ has a column of all $0$'s. In either case, there is no matrix $B$ such that $AB = BA = I$. 
A: If a matrix has a row of zeroes or a column of zeros, the determinant of the matrix is $0$. Hence, they are not invertible.
Why? 
You calculate the determinant of the matrix by choosing a row or column and multiplying each element of it with the adjoint of the smaller matrix obtained after removing the respective row and column. When each element is zero, the total determinant is also zero. If determinant is zero, $A^{-1} =$$adj(A)\over det(A)$ is not defined.
A: If row $i$ of $A$ is zero, then for any matrix $B$ the row $i$ of the product $AB$ (if defined at all) will also be zero, whence $AB\neq I$. Similarly if column $j$ of $A$ is zero, then for any matrix $B$ the column $j$ of the product $BA$ will also be zero, so $BA\neq I$. So under either condition, $A$ is not invertible.
(Note that for $B$ to be the inverse of $A$, both $AB$ and $BA$ must be identity matrices).
A: Another way to see it (however in the spirit of @Marc van Leeuwn's and @Alex Wertheim's answers) is to do the following:
Suppose the $i$th-column of $A$ is zero. Then fix a family of vectors $v(t) \in \mathbb{R}^n$, where $t \in \mathbb{R}$ is the $i$th-entry of the vector $v(t)$ and all other entries are equal (e.g. all $1$). Then since the $i$th-column of $A$ is zero all this vectors get mapped to the same vector under $A$. So $A$ is not injective, in particular not invertible (not bijective).
The case if the $i$th-row is zero can be treaten analogously.
A: Yet another way of looking at it is:

The system of homogeneous equations AX = 0 has only the trivial
  solution X = 0 if and only if A is invertible.

Now, if there is a row/column of all zeroes, at least one of the variables will be free in the said system of homogeneous equations, resulting in non-trivial solution. Thus, A is not invertible.
