1. A matrix is a transformation
A $2\times2$ matrix sends every vector $\mathbf v=(x,y)$ to $A\mathbf v$. Because the map is linear ($A(\alpha\mathbf u+\beta\mathbf v)=\alpha A\mathbf u+\beta A\mathbf v$) it is completely fixed by where it sends the two basis vectors $\hat{\imath}=(1,0)$ and $\hat{\jmath}=(0,1)$, the unit arrows along the axes. Those images are the columns:
That is why dragging the green and pink arrows in the lab edits the matrix directly. Grid lines stay straight, parallel and evenly spaced, and the origin stays fixed.
| Transformation | Matrix | det |
|---|---|---|
| Rotation by θ | $\begin{bmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{bmatrix}$ | 1 |
| Scaling | $\begin{bmatrix}s_x&0\\0&s_y\end{bmatrix}$ | $s_xs_y$ |
| Shear (horizontal) | $\begin{bmatrix}1&k\\0&1\end{bmatrix}$ | 1 |
| Reflection in the x-axis | $\begin{bmatrix}1&0\\0&-1\end{bmatrix}$ | −1 |
| Projection onto the x-axis | $\begin{bmatrix}1&0\\0&0\end{bmatrix}$ | 0 |
2. The determinant: signed area (or volume) scale factor
The unit square (area 1) is mapped to a parallelogram of area $|\det A|$, and every other region is scaled by the same factor. A negative determinant means orientation has flipped (the letter F appears mirrored). $\det A=0$ means space is squashed onto a line or a point, so information is lost and $A$ has no inverse. In 3D, $\det$ is the signed volume of the image of the unit cube, and $\det(AB)=\det A\,\det B$.
3. Inverse
$A^{-1}$ undoes the transformation. It exists exactly when $\det A\ne 0$.
4. Eigenvectors and eigenvalues
An eigenvector is a non-zero vector whose direction is unchanged: $A\mathbf v=\lambda\mathbf v$. It stays on its own span (the line through the origin along it) and is only stretched by the eigenvalue $\lambda$ (flipped if $\lambda<0$). Eigenvalues solve the characteristic equation
- Discriminant $>0$: two real eigenvalues and two eigen-directions (dashed lines in the lab).
- Discriminant $=0$: a repeated eigenvalue, with either every direction (a pure scaling) or only one direction (a shear).
- Discriminant $<0$: a complex pair $\alpha\pm\beta i$. No real direction is preserved. In a suitable (generally non-orthogonal) basis the map is a rotation by $\arg\lambda$ combined with scaling by $|\lambda|$; in ordinary coordinates repeated application spirals points around ellipses rather than circles, unless $A$ is itself a scaled rotation.
Useful checks: $\lambda_1+\lambda_2=\operatorname{tr}A$ and $\lambda_1\lambda_2=\det A$.
5. Composition: order matters
Applying $B$ first and then $A$ is the single matrix $AB$ (read right to left, like function composition $A(B\mathbf v)$). In general $AB\neq BA$. In 2D, rotations about the same origin commute with each other (in 3D they generally do not); a rotation and a shear do not.
6. Homogeneous coordinates
Translation $\mathbf v\mapsto\mathbf v+\mathbf t$ is not linear (it moves the origin), so no $2\times2$ matrix can do it. The trick is to write a 2D point as $(x,y,1)$ and use $3\times3$ matrices:
Now rotations, scalings and translations are all matrix products and can be chained. This is exactly how computer graphics pipelines and robot forward kinematics work: each robot joint contributes one transform, and the gripper pose is the product $T_{02}=T_{01}\,T_{12}$.
7. Linear systems and Gaussian elimination
The system $A\mathbf x=\mathbf b$ is solved by row operations on the augmented matrix $[A\,|\,\mathbf b]$. Each operation (swap two rows, scale a row by a non-zero number, add a multiple of one row to another) leaves the solution set unchanged. Forward elimination produces row-echelon form; back substitution finishes the job.
Geometrically each equation is a line (2D) or a plane (3D), and the solution is their common intersection. The rank of a matrix is the number of independent rows (equivalently, the dimension of the space its outputs fill). A unique solution needs $\operatorname{rank}A=n$ ($\det A\neq0$). If $\operatorname{rank}A<\operatorname{rank}[A\,|\,\mathbf b]$ the system is inconsistent (parallel lines). If both ranks are equal but less than $n$ there are infinitely many solutions.
Example 1: where does a vector go?
$A=\begin{bmatrix}1.5&1\\0.5&1.5\end{bmatrix}$ and $\mathbf v=(-1,\,1.5)$: the lab's starting matrix and test vector.
$A\mathbf v=-1\begin{bmatrix}1.5\\0.5\end{bmatrix}+1.5\begin{bmatrix}1\\1.5\end{bmatrix}=\begin{bmatrix}-1.5+1.5\\-0.5+2.25\end{bmatrix}=\begin{bmatrix}0\\1.75\end{bmatrix}$
$\det A=1.5(1.5)-1(0.5)=1.75$: every area grows by 75 %.
Example 2: eigenvalues of a symmetric matrix
$A=\begin{bmatrix}2&1\\1&2\end{bmatrix}$: $\operatorname{tr}A=4$, $\det A=3$, so $\lambda^2-4\lambda+3=(\lambda-3)(\lambda-1)=0$.
$\lambda_1=3$: $(A-3I)\mathbf v=0\Rightarrow -v_1+v_2=0$, so $\mathbf v_1=(1,1)$. $\lambda_2=1$: $v_1+v_2=0$, so $\mathbf v_2=(1,-1)$.
Check: $3+1=4=\operatorname{tr}A$ and $3\times1=3=\det A$. The eigenvectors are perpendicular, as for every symmetric matrix.
Example 3: a rotation has complex eigenvalues
Rotation by 90°: $A=\begin{bmatrix}0&-1\\1&0\end{bmatrix}$, $\operatorname{tr}=0$, $\det=1$, so $\lambda^2+1=0$ and $\lambda=\pm i$. $|\lambda|=1$ (no stretching) and $\arg\lambda=90°$ (the rotation angle). No real vector keeps its direction.
Example 4: AB versus BA
$A=\begin{bmatrix}0&-1\\1&0\end{bmatrix}$ (rotate 90°), $B=\begin{bmatrix}1&1\\0&1\end{bmatrix}$ (shear).
$AB=\begin{bmatrix}0&-1\\1&1\end{bmatrix}$ (shear first, then rotate) but $BA=\begin{bmatrix}1&-1\\1&0\end{bmatrix}$ (rotate first, then shear). Different matrices, different pictures, yet $\det AB=\det BA=1$.
Example 5: homogeneous transform
Rotate the point $(1,0)$ by 90° and then translate by $(3,1)$:
Translating first gives $RT\,(1,0,1)^{\mathsf T}=R\,(4,1,1)^{\mathsf T}=(-1,4,1)^{\mathsf T}$: a different place.
Example 6: Gaussian elimination
Solve $2x+y=5,\ x-y=1$.
- $R_2\to R_2-\tfrac12R_1$: $\left[\begin{smallmatrix}2&1&|&5\\0&-1.5&|&-1.5\end{smallmatrix}\right]$ (forward elimination complete).
- $R_2\to -\tfrac23R_2$: $y=1$.
- $R_1\to R_1-R_2$, then $R_1\to\tfrac12R_1$: $x=2$.
Solution $(x,y)=(2,1)$: the point where the two lines cross. Load the first 2 × 2 example in the lab and press Next step.
Example 7: a market problem
At a market, 2 mangoes, 3 avocados and 1 banana cost KES 190; 1 mango, 2 avocados and 3 bananas cost KES 140; 3 mangoes, 1 avocado and 2 bananas cost KES 150. With unknown prices $(m,a,b)$:
$\det A=2(4-3)-3(2-9)+1(1-6)=2+21-5=18\neq0$, so the answer is unique. Load "Market prices" in the 3 × 3 examples to see every row operation.
Challenges (checked automatically)
Guided experiments
- Determinant through zero. In 2D mode drag ĵ slowly across the line through î. Watch the unit-square area shrink to 0 and the colour change as orientation flips.
- Repeated eigenvalue. Load the shear preset. Only one dashed eigen-line remains: the shear is defective.
- Circle to ellipse. Choose the Circle shape with a symmetric matrix. The ellipse's axes line up with the eigenvectors; their half-lengths are $|\lambda_1|$ and $|\lambda_2|$.
- Rotation commutes with rotation. In Composition set A and B to two different rotations. The two panels end in the same place.
- Flat volume. In 3D load the projection preset: the cube collapses into a flat square and det = 0.
- Inconsistent planes. Load the 3 × 3 "No solution" example and step through. A row reduces to $[0\ 0\ 0\,|\,c\neq0]$, which reads $0=c$.