Determinant always equal to zero? I just finished writing a computer program that takes as input a number of matrices and computes the inverse of the product of matrices. 
To test this program, I wanted to input a $3 \times 2$ matrix followed by a $2\times 3$ matrix so that the product would be a $3\times 3$ matrix. No matter how hard I try, the determinant of the product turns out to be zero and so the product is non-invertible. Is there a theorem in linear algebra that implies that the product of a $3\times 2$ matrix and $2\times 3$ matrix will always have determinant zero?
 A: Yes.
If you consider a matrix $A$ as a linear mapping $v \mapsto Av$ then a square matrix has determinant different from zero if and only if the mapping is invertible. The product of matrices represents the composition of linear mappings. But for a composition to be invertible, it is required that the first mapping is injective. However a linear mapping from $3$D space to $2$D space cannot be injective... 
added: alternative proof. Let $A$ be your $3\times 2$ matrix and $B$ the $2\times 3$ matrix so that $AB$ is a $3\times 3$ matrix. Now consider the $3\times 3$ matrices $A'$ and $B'$ constructed by adding a column of zeros to $A$ and a row of zeros to $B$. Then notice that $A'B'=AB$ because in the row-by-column multiplication in $A'B'$ you get the row-by-column multiplication in $AB$ plus a $0\cdot 0$. Now you know that $\det A'=\det B'=0$ hence $\det AB = \det A'B' = \det A' \det B' = 0$.
A: Yes.
Let's call $A$,$B$ and $M$ your matrix, $M=AB$.
$B$ is not square so it exists $(\lambda_{i})$ non all null such that $\sum\limits_{i} \lambda_{i} C_{i}(B) =0$
You have $\sum\limits_{i} \lambda_{i} C_{i}(M) = \sum\limits_{i} \lambda_{i} A* C_{i}(B) = A*(\sum\limits_{i} \lambda_{i} C_{i}(B)) = 0 $ 
A: Classical explanation is based on the fact that if we have $m \times n$ matrix  $A$ (with more rows than columns - number for rows can be treated as the number of components of a vector, number of columns as the number of vectors) then vector $w=Av$ , where $v$ is $n \times 1$ vector, must lie in the column space of $A$ $        ($denoted $C(A)$). 
In the case of matrix $3 \times 2$ it is a plane in $R^3$.   
Now we have $2 \times 3$ matrix  $B$ consisting of three columns $b_1, b_2, b_3$ so the matrix $AB=[ Ab_1 \ \  Ab_2 \ \ Ab_3]$ where every column of the $3 \times 3$ $AB$ lies in the mentioned earlier plane.
So we have three vectors lying in the 2-d plane located in 3-d space (evidently they can't be linearly independent) therefore the matrix $AB$ must be singular.
