$\det(AB) \not = \det(A)\det(B)$?

Question

What’s wrong with my reasoning? $$\det\left(\left[\begin{matrix} 2&-2\\-2&2\end{matrix}\right]\right)=2\times2-(-2)(-2)=0$$ $$\left[\begin{matrix} 2&-2\\-2&2\end{matrix}\right] \left[\begin{matrix} 2&-2\\-2&2\end{matrix}\right]= \left[\begin{matrix} 4&0\\0&4\end{matrix}\right]$$ $$\det\left(\left[\begin{matrix} 4&0\\0&4\end{matrix}\right]\right)= 4\times 4 - 0 = 16$$ $$16=\det\left(\left[\begin{matrix} 4&0\\0&4\end{matrix}\right]\right)=\det\left(\left[\begin{matrix} 2&-2\\-2&2\end{matrix}\right] \left[\begin{matrix} 2&-2\\-2&2\end{matrix}\right] \right)=\det\left(\left[\begin{matrix} 2&-2\\-2&2\end{matrix}\right] \right) \det\left(\left[\begin{matrix} 2&-2\\-2&2\end{matrix}\right] \right) =0 \times 0 = 0$$

Answer 1

$$\left[\begin{matrix} 2&-2\\-2&2\end{matrix}\right] \left[\begin{matrix} 2&-2\\-2&2\end{matrix}\right]= \left[\begin{matrix} 8&-8\\-8&8\end{matrix}\right]$$

You can check the resultant determinant is zero.

Answer 2

It turns out that$$\begin{bmatrix}2&-2\\-2&2\end{bmatrix}.\begin{bmatrix}2&-2\\-2&2\end{bmatrix}=\begin{bmatrix}8&-8\\-8&8\end{bmatrix}.$$