$A$ is a matrix of 3 rows and 2 columns while $B$ is a matrix of 2 rows and three columns. If $AB=C$ and $BA=D$, then pick the correct option

Question

Options

A) determinant of C and D are always equal

B) determinant of C is zero

C) determinant of D is zero

The reasoning should be really simple.

$$|C|=|A||B|$$ $$|D|=|B||A|$$

So $$|D|=|C|$$

But the right answer is option b. I have the written explanation for this answer, but I a found it ridiculous and ultimately unsatisfactory. What could be the reason for b) being the answer?

Given solution

Let $A’$ be a 3x3 matrix with first two columns identical to A and third column has all elements equal to zero. Similarly, construct $B’$ as a 3x3 matrix

$AB=A’B’=C$ $\implies$ $|C|=|A’||B’|=0$

Answer 1

Your identity for determinant multiplication is only true for square matrices, since other matrices do not have a proper determinant.

What you can do to check if the determinant of a matrix is $0$ is ask - does this matrix have a non-trivial kernel? And indeed, the kernel of $C$ is non-trivial - $B$ sends a $3$ dimensional space into a $2$ dimensional one (which means that some non-zero vector must be mapped to $0$, and no matter what you do later with $A$, you already have a non-trivial kernel (which can only grow with $A$), and so $C$ must have determinant $0$.

Answer 2

For non-square matrices, the determinant does not make sense, so your statement is false.

But think of $A$ and $B$ as linear homomorphism : $a : \mathbb{K}^2 \to \mathbb{K}^3$ and $b : \mathbb{K}^3 \to \mathbb{K}^2$. Then $b$ cannot be injective, and $a$ cannot be surjective. Thus $c=a\circ b : \mathbb{K}^3 \to \mathbb{K}^3$ cannot be injective or surjective : as it is an endomorphism, it as a determinant which is forced to be $0$. Consequently $C = AB$ is the matrix of a non-bijective linear endomorphism, so has determinant zero.

Edit Here is an example where $C$ has determinant $0$ but not $D$ : \begin{align} A &= \begin{pmatrix}1 & 0 \\ 0 & 1 \\ 0 & 0 \end{pmatrix} \\ B &= \begin{pmatrix}1 & 0& 0 \\ 0 & 1 &0 \end{pmatrix} \\ C &= \begin{pmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{pmatrix} \\ D &= \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \end{align}