Matrices

Take two arrows $\vec u$ and $\vec v$ and two numbers $a$ and $b$. The combination $a\vec u+b\vec v$ is the only move linear algebra knows: stretch, then add. Drag the arrowheads, push the sliders, and watch where the result can go. As long as $\vec u$ and $\vec v$ point in different directions, that result can reach any point of the plane. Line them up — and the whole plane collapses into a single line.

Combination

$a\vec u+b\vec v$ — every point you can build from two arrows

Span

$\operatorname{Vect}(\vec u,\vec v)$, the set of all such combinations

Test

$\det(\vec u,\vec v)=u_xv_y-u_yv_x$. Zero means collinear, and the span is only a line.

Here is the whole secret. To know a linear map, it is enough to know where $\vec e_1=(1,0)$ and $\vec e_2=(0,1)$ go: everything else follows by combination. Stack those two images as columns and you have written the matrix. Drag them here: the entire grid bends along, the unit square becomes a parallelogram, and the four numbers write themselves.

A linear map makes only two promises: $A(\vec x+\vec y)=A\vec x+A\vec y$ and $A(\lambda\vec x)=\lambda A\vec x$. That is why straight lines stay straight, zero stays at zero, and an even grid stays even — only tilted.

The unit square has area $1$. After $A$ it becomes a parallelogram — and $\det A$ is exactly the area of that parallelogram, with a sign. The sign says whether the plane was flipped: emeraude when the orientation holds, camellia when it reversed. Drag one column past the other and watch the colour switch. At the instant of crossing, the area is zero: the plane has been flattened onto a line, and nothing will unfold it.

2 × 2

$\det\begin{pmatrix}a&b\\c&d\end{pmatrix}=ad-bc$

Product

$\det(AB)=\det A\cdot\det B$ — areas multiply, one map after the other

Invertible

$A^{-1}$ exists $\iff\det A\neq0$. A flattened plane has no way back.

$AB$ means: $B$ first, then $A$. The row-by-column rule, $(AB)_{ij}=\sum_k a_{ik}b_{kj}$, is only the bookkeeping of that composition. Here $R$ rotates by the angle you choose and $S$ shears. Rotate then shear, then do it the other way round, and compare the two squares: they are not the same. $AB\neq BA$ is not a warning to memorise — it is what you are looking at.

Every equation $ax+by+cz=d$ is a plane in space. The whole system asks a single question: which points belong to all three at once? There are only four possible answers — a point, a line, a whole plane, or nothing. Gaussian elimination only swaps and combines rows; each step swings the planes before your eyes, but their intersection never moves by a hair. That is the entire justification of the method.

Rouché

Kronecker–Capelli: solvable $\iff\operatorname{rank}A=\operatorname{rank}[A|\mathbf b]$

Unique

and that common rank equals $n$, the number of unknowns

Freedom

otherwise the solutions form a space of dimension $n-\operatorname{rank}A$

Cramer

$x_i=\dfrac{\Delta_i}{\Delta}$ when $\Delta=\det A\neq0$ — elegant on paper, slow in practice

The letter F is there for a reason: it has no symmetry, so it confesses everything — the stretch, the rotation, and above all the flip. Apply $A$, then $A^{-1}$, and it returns exactly where it was: that is what $A^{-1}A=I$ means. Now flatten the matrix by bringing one column onto the other. The letter falls onto a segment. Two different points now share one image — no rule could ever tell them apart, and the inverse does not exist. That is why $\det A=0$ forbids the way back.

2 × 2

$A^{-1}=\dfrac{1}{ad-bc}\begin{pmatrix}d&-b\\-c&a\end{pmatrix}$

Order

$(AB)^{-1}=B^{-1}A^{-1}$ — undo the last gesture first

Solving

$A\mathbf x=\mathbf b\Rightarrow\mathbf x=A^{-1}\mathbf b$, when the inverse exists

Continued in ESPACES — Le Spectre, where the question is no longer what a matrix does, but in which frame it becomes simple.