Show if $A$ has a zero row, then $AB$ has a zero row. Let $A$ and $B$ be $n \times n$ matrices. Show that if the $i$th row of $A$ has all zero entries, then the $i$th row of $AB$ will have all zero entries. Also give and example using $2 \times 2$ matrices to show the converse is not true.
 A: Since, for $ \ \mathbf{C} \ = \ \mathbf{AB} \ $ ,
$$ \sum_{k=1}^n \ a_{ik} \ b_{kj} \ = \ c_{ij} \ \ , $$
having every entry of the $ \ i$th row of $ \ \mathbf{A} \ $ equal to zero will make every entry $ \ c_{ij} \ $ equal to zero.
For $ \ \mathbf{D} \ = \ \mathbf{BA} \ $ , we have instead
$$ \sum_{k=1}^n \ b_{ik} \ a_{kj} \ = \ d_{ij} \ \ , $$
which means that only one term of the sum which produces the entries in $ \ D \ $ is guaranteed to be zero.  
Counterexamples for the converse are therefore numerous: all we need to do is contrive for the product of the matrices to produce a zero row by having entries in one row of the first matrix and the entries along each column of the second matrix have suitable proportions, such as
$$\left[ \begin{array}{cc}
-1 &  5 \\ 6 & -4 \end{array} \right] \ \left[ \begin{array}{cc}
2 & -4 \\ 3 & 6 \end{array} \right] \ = \ \left [ \begin{array}{cc}
13 & 34 \\ 0 & 0 \end{array} \right] \ \ . $$
A: Notice:
$$ \begin{bmatrix} 1 & -1 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix}$$
A: The $i$th row of $AB$ is $\sum_{j=1}^n a_{ij}$Row$_j(B)=\sum_{j=1}^n 0 \cdot $Row$_{j}(B)$, and consequently is the zero vector.
A: The $i$'th row of $A$ is $0$ if and only if the $i$'th entry of $Ax$ is $0$ for all vectors $x$.  Similarly for $AB$.  Since  $ABx = Ay$ where $y = Bx$, we conclude that if the $i$'th row of $A$ is $0$ then so is the $i$'th row of $AB$.
