If $A$ is a $3\times 2$ matrix and $B$ is a $2\times 3$ matrix why is $AB$ not invertible I am getting nowhere with this question. I found a similar question on this site but I need help in a more step by step way. I would be very grateful.
 A: Let $A=\begin{bmatrix}
a_1 & a_2 \\ 
b_1 & b_2 \\ 
c_1 & c_2
\end{bmatrix}$ and $B=\begin{bmatrix}
a'_1 & b'_1 & c'_1\\ 
a'_2 & b'_2 & c'_2
\end{bmatrix}$.
Also let $A'=\begin{bmatrix}
a_1 & a_2 & 0 \\ 
b_1 & b_2 &  0\\ 
c_1 & c_2 & 0
\end{bmatrix}$ and $B'=\begin{bmatrix}
a'_1 & b'_1 & c'_1\\ 
a'_2 & b'_2 & c'_2\\ 
0 & 0 & 0
\end{bmatrix}$. So we have $A\times B = A' \times B'$, but we have:
$$det(A\times B)=det(A'\times B')=det(A')\cdot det(B')=0$$, therefore $A\times B$ is not invertible.
A: There are at most 2 columns which are not linearly independent in matrix $A$. Therefore, as multiplication with $B$ is just a linear combination of the columns of $A$, then it cannot be that now we have 3 linearly independent columns. Hence the matrix $AB$ cannot be invertible.
A: Consider matrices as models of linear maps. Then the product of two matrices stands for the composition of the maps. The first matrix (right hand side of the multiplication) maps $\mathbb R^3$ to $\mathbb R^2.$ The second matrix (left hand side of the multiplication) maps $\mathbb R^2$ to $\mathbb R^3.$ The latter map can never be onto, so neither can the composed map be invertible.
A: HINT:
Each of the $3$ columns of $AB$ is a linear combinations of the $2$ columns of $A$. Use the multilinearity properties of the determinant we conclude that $\det(AB) = 0$, and so $AB$ cannot be invertible.
Obs: This proof works for matrices over a commutative ring with $1$. 
A: AB is linearly dependent by A's columns. and A has only 2 columns, which means AB will have 2 columns which are linearly dependent. Thus, AB can't be linearly independent. Which means, $|AB| = 0$. and from here it is pretty clear it can't be invertible.
