# 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.

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.

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.

• Well, not strong objection, but $B = 2x3$ Commented Dec 11, 2015 at 12:01
• @IlanAizelmanWS: Thank you for comment. I edited my post. Commented Dec 11, 2015 at 12:05
• I like your answer. But it would be much easier to understand if you kept the OP's notation. The original given matrices were called $A$ and $B$, not $A$ and $A'$. Commented Dec 11, 2015 at 13:36
• @bubba: Yes, you are right. thank you for your comment. I change their names. Commented Dec 11, 2015 at 13:44

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.

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 $1$.

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.