The Inverse Matrix
Every editing app has an Undo button. Drag a shape, rotate it, stretch it — and if
you don't like the result, one click puts it back exactly where it started. Matrices need an undo
button too. If a matrix A rotates and stretches a picture, its
inverse A^{-1} is the matrix that un-rotates and
un-stretches it, landing every point back on its original spot.
Apply A and then A^{-1} (in either order) and
you return to exactly where you started — the
identity,
the matrix version of "do nothing":
A^{-1}A = AA^{-1} = I.
That's the whole definition. A^{-1} is defined to be whichever
matrix, multiplied by A on either side, gives back
I — nothing more. It plays exactly the role
\tfrac13 plays for the number 3: multiplying
by a number and then by its reciprocal gets you back to 1; applying a
transformation and then its inverse gets you back to "no transformation at all."
The one condition that decides everything
Not every matrix has an undo button. A matrix A is invertible — has an
A^{-1} at all — exactly when its
determinant
is non-zero:
A^{-1} \text{ exists} \iff \det A \neq 0.
Remember what the determinant measures: how much a transformation scales area (in 2D) or volume
(in 3D). If \det A = 0, the transformation squashes the whole plane flat
onto a line — or onto a single point. Area
2 shrinks to area 0. Once two different
starting points have been squeezed onto the very same spot, no matrix on Earth can look at that
spot and tell you which point it used to be — the information is simply gone. An inverse would have
to un-squash a line back into a full plane, stretching one point into two, and no function can do
that. So "singular" (det = 0) and "not invertible" mean exactly the same thing.
Warp, then un-warp
Watch it happen to an entire grid, not just one point. Step through the stages below: first the
grid sits at rest. Apply A and it stretches and skews into a slanted
parallelogram grid. Then apply A^{-1} to that warped grid, and
it snaps perfectly back to the original square grid — every line, every angle, restored. That
round trip, warp then un-warp, is what A^{-1}A = I looks like in motion.
Worked example: checking a claimed inverse
Suppose someone hands you a matrix and claims it's the inverse of another. You don't have to trust
them — just multiply and see if you land on I. Take
A = \begin{bmatrix} 2 & 1 \\ 1 & 1 \end{bmatrix}, \qquad \text{claimed: } A^{-1} = \begin{bmatrix} 1 & -1 \\ -1 & 2 \end{bmatrix}.
Multiply A \cdot A^{-1}, one row-times-column at a time:
A A^{-1} = \begin{bmatrix} 2(1) + 1(-1) & 2(-1) + 1(2) \\ 1(1) + 1(-1) & 1(-1) + 1(2) \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I.
It checks out — but the definition demands both orders agree, so verify
A^{-1}A too:
A^{-1} A = \begin{bmatrix} 1(2) + (-1)(1) & 1(1) + (-1)(1) \\ -1(2) + 2(1) & -1(1) + 2(1) \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I.
Both products come out to I, so the claim is confirmed: this really is
A^{-1}. (You'll meet an actual recipe for finding this inverse,
rather than just checking it, on the
next
page.)
Worked example: solving a system by inverting
Here's why inverses matter beyond checking arithmetic. Two apples and one banana cost
\$5; one apple and one banana cost \$3. What
does each fruit cost? Write it as a matrix equation
A\vec{x} = \vec{b}, with \vec{x} = (a, b) the
unknown prices:
\begin{bmatrix} 2 & 1 \\ 1 & 1 \end{bmatrix}\begin{bmatrix} a \\ b \end{bmatrix} = \begin{bmatrix} 5 \\ 3 \end{bmatrix}.
This is exactly the A from the last example, and we already know its
inverse. Multiply both sides on the left by A^{-1}. On the left,
A^{-1}A collapses to I, leaving just
\vec{x} behind:
\vec{x} = A^{-1}\vec{b} = \begin{bmatrix} 1 & -1 \\ -1 & 2 \end{bmatrix}\begin{bmatrix} 5 \\ 3 \end{bmatrix} = \begin{bmatrix} 1(5) + (-1)(3) \\ -1(5) + 2(3) \end{bmatrix} = \begin{bmatrix} 2 \\ 1 \end{bmatrix}.
Apples cost \$2, bananas cost \$1. Check it:
2(2) + 1(1) = 5 and 1(2) + 1(1) = 3 — both
original equations hold. Notice the whole "solve two equations in two unknowns" problem became one
matrix-vector multiplication, once the inverse was in hand.
Worked example: when there is no undo button
Now try the same trick on a matrix that destroys information. Let
A = \begin{bmatrix} 2 & 4 \\ 1 & 2 \end{bmatrix}, \qquad \det A = 2(2) - 4(1) = 0.
The determinant is zero, so this matrix should not be invertible — and here's the damage in action.
Feed it two different input vectors, (1, 0) and
(3, -1):
A\begin{bmatrix} 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 2 \\ 1 \end{bmatrix}, \qquad A\begin{bmatrix} 3 \\ -1 \end{bmatrix} = \begin{bmatrix} 2(3) + 4(-1) \\ 1(3) + 2(-1) \end{bmatrix} = \begin{bmatrix} 2 \\ 1 \end{bmatrix}.
Two genuinely different starting points, (1,0) and
(3,-1), both land on the exact same output,
(2,1). If somebody only showed you the output (2,1)
and asked "which point did this come from?", there is no way to answer — both are equally valid,
and infinitely many other points map there too. A true inverse would need to send
(2,1) back to one specific place, but it can't decide between
two (or infinitely many) equally good candidates. That's not a computational difficulty to push
through with more effort — it's a hard logical wall, and it's exactly what
\det A = 0 was warning you about.
Two mistakes account for almost every wrong answer involving inverses:
-
If \det A = 0, stop — there is no inverse, and "the inverse
doesn't exist" is the correct, complete answer. It is tempting to feel like you must
have made an arithmetic slip and keep trying to force a formula through, perhaps dividing by that
zero determinant anyway. Don't. A singular matrix genuinely has no undo button; recognising that
and stopping is the right move, not a failure to find one.
-
A^{-1}B^{-1} is not the same as
(AB)^{-1} — the order flips. The correct rule is
(AB)^{-1} = B^{-1}A^{-1}. Check it: (AB)(B^{-1}A^{-1}) =
A(BB^{-1})A^{-1} = AIA^{-1} = AA^{-1} = I — the reversed order is exactly what makes
the middle pair cancel. Think of putting on socks, then shoes: to undo it, you can't peel off
socks first — your shoes are in the way. You must reverse the order too: shoes off, then socks
off. Undoing a sequence of steps always means undoing them in reverse order.
This isn't just textbook tidiness — invertibility is what lets scrambled information come back at
all. An early cipher called the Hill cipher encodes a block of letters by
multiplying them (as numbers) by a secret matrix. Sending the message is easy; the entire scheme
only works, though, if that matrix is invertible — otherwise two different original
messages could scramble into the same coded text, and even the person with the secret key couldn't
tell them apart. Modern error-correcting codes (the ones that let a scratched CD or a weak phone
signal still recover the right data) lean on the very same idea: build the encoding out of an
invertible matrix, and decoding is guaranteed to be possible, in principle, by inverting it.
Even that humble Undo button from the very first paragraph is secretly doing linear algebra: every
rotate, scale or skew you apply to a shape in a drawing program is stored as a matrix, and "Undo"
just multiplies by that matrix's inverse. The whole idea of reversibility — in software, in codes,
in physics — keeps coming back to this one question: does an inverse exist?
See it explained