Module 6 · Lesson 6.3

Inverse Matrices, Column Space & Null Space

Discover when a matrix transformation can be undone — and what happens to the directions it collapses to zero along the way.

Intuition

If a matrix $A$ stretches and shears the plane, can you always find a second matrix that puts everything back where it started? Usually, yes — that undo-matrix is called $A^{-1}$. But in the last lesson you saw that some matrices flatten the plane entirely, collapsing area down to a line or a point. Once information is squashed away like that, there's nothing left to reconstruct from, and no inverse can exist.

Core concept

The column space of $A$ is everywhere the transformation can possibly land — every reachable output. The null space is the opposite: every input vector that $A$ sends straight to the origin, $\vec{0}$. Whenever the null space contains more than just $\vec{0}$ itself, some direction has been squashed flat, the determinant is zero, and no inverse exists. If the null space contains only $\vec{0}$, nothing was lost, and $A^{-1}$ exists.

A A⁻¹
Matrix A[[2, 1], [1, 3]]
det(A)5
Swap diagonal, negate off-diagonal[[3, −1], [−1, 2]]
Divide every entry by det(A)1/5 × [[3, −1], [−1, 2]]
AnswerA⁻¹ = [[0.6, −0.2], [−0.2, 0.4]]

Interactive Lab

The dashed arrow is a fixed test direction, $\vec{v} = (-3, 1)$. The solid arrow is where matrix $A$ sends it, $A\vec{v}$. Slide the matrix entry $b$ and watch the solid arrow shrink toward the origin — right where it vanishes completely, $\vec{v}$ has fallen into the null space.

v Av
matrix A = [[2, 1], [1, 3]] det(A) = 5.00 Av = (−5.00, 0.00) status: invertible

Key Idea

$$ A^{-1} = \frac{1}{\det(A)}\begin{pmatrix} d & -b \\ -c & a \end{pmatrix}, \qquad \det(A) \neq 0 $$

The inverse only exists when the determinant is nonzero — exactly the condition that guarantees the null space contains nothing but the zero vector. Whenever a system like $A\vec{x} = \vec{b}$ shows up in an application, this formula is what lets you solve for $\vec{x}$ directly.

$A^{-1}$The inverse matrix — undoes everything $A$ does
$\det(A)$Must be nonzero for the inverse to exist
column spaceEvery output the transformation can possibly reach
null spaceEvery input that gets sent to the zero vector

Think Further

  1. In circuit analysis, solving $A\vec{x} = \vec{b}$ for unknown currents or voltages requires $A^{-1}$ to exist. What would it mean physically if $A$'s null space contained a nonzero vector?
  2. If $\det(A) = 0$, predict what happens when you try to compute $A^{-1}$ using the formula $\frac{1}{\det(A)}\begin{pmatrix} d & -b \\ -c & a \end{pmatrix}$.
  3. Two matrices could have very different determinants but both have a null space containing only the zero vector. What does that tell you they both share, despite scaling area by different amounts?
Show suggested answers
  1. It would mean the system doesn't pin down a single, unique solution — some combination of currents or voltages could shift without changing the equations, leaving that part of the circuit's behavior undetermined.
  2. The formula requires dividing by $\det(A)$, which is division by zero — undefined. That breakdown confirms directly that no inverse can exist whenever the determinant is exactly zero.
  3. Both are invertible. Having only the zero vector in the null space is exactly the condition for a matrix to be invertible, regardless of how strongly each one individually stretches or shrinks area.