Determinants & matrix inversion

M = [[3, 1], [2, 2]]   det(M) = 4

Drag the sliders. The dashed square is the input; the solid shape is where the matrix sends it. Push any slider until the solid shape collapses into a line — that's a singular matrix, and its determinant drops to exactly 0.

For a $2\times2$ matrix, the determinant is:

• $A$ — the $2\times2$ matrix, with entries $a, b, c, d$ (rows top-to-bottom, columns left-to-right).

1. It's a signed area

   $|\det(A)|$ is the area of the parallelogram the unit square maps to. The sign flips exactly when the mapping reverses orientation (flips the plane over).

2. Zero means singular

   $\det(A)=0$ exactly when $A$'s two columns point along the same line — the mapped square collapses to zero area, and $A$ is singular: not invertible.

3. It gates invertibility

   $A^{-1}$ exists if and only if $\det(A)\neq0$, since the inverse formula below divides by it.

The inverse itself, when it exists:

1. Why this undoes A

   $A^{-1}$ is defined precisely so that $A^{-1}A = AA^{-1} = I$ — applying $A$ then $A^{-1}$ (or vice versa) leaves every vector exactly where it started.

2. Singular means no undo

   If $\det(A)=0$, $A$ has already thrown away a dimension (collapsed the plane to a line) — there's no way to recover the lost information, so no inverse can exist.

det(M) = 4   M⁻¹ = [[0.50, -0.25], [-0.50, 0.75]]

Jump between the invertible preset and the singular one. Watch the inverse readout switch from a clean set of numbers to "no inverse" the instant the parallelogram flattens.

For $A = \begin{pmatrix}3&1\\2&2\end{pmatrix}$:

1. Compute the determinant

   $\det(A) = (3)(2) - (1)(2) = 6 - 2 = 4$.

2. Compute the inverse

   $A^{-1} = \frac{1}{4}\begin{pmatrix}2&-1\\-2&3\end{pmatrix} = \begin{pmatrix}0.5&-0.25\\-0.5&0.75\end{pmatrix}$.

3. Check it

   $A A^{-1} = \begin{pmatrix}3&1\\2&2\end{pmatrix}\begin{pmatrix}0.5&-0.25\\-0.5&0.75\end{pmatrix}$. Multiplying entry by entry:

   • Top-left: $3(0.5)+1(-0.5)=1.5-0.5=1$
   • Top-right: $3(-0.25)+1(0.75)=-0.75+0.75=0$
   • Bottom-left: $2(0.5)+2(-0.5)=1-1=0$
   • Bottom-right: $2(-0.25)+2(0.75)=-0.5+1.5=1$

   Exactly $\begin{pmatrix}1&0\\0&1\end{pmatrix}$.