Matrices move space
The shear matrix $\begin{pmatrix}1 & 1\\ 0 & 1\end{pmatrix}$ slides the top of the unit square one full unit sideways. What happens to the square's area?
The shear leans the square over, yet det A still reads $1.00$ — same base, same height, same area. The sliders reveal who is in charge: a and c steer the e1 arrow, b and d steer the e2 arrow, and the whole lattice follows those two arrows. From the shear, drag d up to $2.00$: the parallelogram doubles in height and det A climbs to $2.00$ with it.
The four numbers are two landing spots. The first column $(a, c)$ is where $\mathbf{e}_1 = (1,0)$ lands; the second column $(b, d)$ is where $\mathbf{e}_2 = (0,1)$ lands. Linearity forces everything else — the point $(x, y)$ must land on $x$ copies of the first column plus $y$ copies of the second:
The determinant $\det A = ad - bc$ is the area report. The unit square becomes a parallelogram of area $|\det A|$, and every region in the plane scales by that same factor; the sign records a flip. The shear's columns are $(1, 0)$ and $(1, 1)$ — base $1$, height $1$ — so the lean costs no area at all.
In the instrument, dragging a and c moves one arrow and the lattice follows. Under $A = \begin{pmatrix} a & b\\ c & d\end{pmatrix}$, where does $\mathbf{e}_1 = (1,0)$ land?
Press Scale: the matrix is $\begin{pmatrix}1.5 & 0\\ 0 & 0.7\end{pmatrix}$ and $\det A$ reads $1.05$. A region of area $2$ goes in. What area comes out?
Transfer: press Reflect, which loads $\begin{pmatrix}1 & 0\\ 0 & -1\end{pmatrix}$, and $\det A$ reads $-1.00$. What does the minus sign say about the motion?
A matrix is two landing spots plus the linearity rule. Read the columns to see where the basis arrows go; read $\det A = ad - bc$ to see what happens to every area — magnitude for size, sign for a flip.