Matrix Inverses

The inverse as the undo machine

In the Sets stratum you proved that a function $f$ has an inverse $f^{-1}$ if and only if $f$ is bijective (injective and surjective). A matrix $A$ is a linear function $\mathbb{R}^n \to \mathbb{R}^n$. So $A$ has an inverse when the corresponding function is bijective.

The matrix inverse $A^{-1}$ is the matrix such that:

$AA^{-1} = I \quad \text{and} \quad A^{-1}A = I$

In words: applying $A$ then $A^{-1}$ puts you back where you started — undoes all the damage $A$ did. Both conditions are required — for square matrices they turn out to be equivalent (a theorem from the Vector Spaces stratum), but stating both is the honest definition.

The identity matrix $I$ plays the role of "1": it's the do-nothing transformation, so $AA^{-1} = I$ says "undo of $A$ composes with $A$ to give the identity," exactly the way $f^{-1} \circ f = \text{id}$ in the Sets stratum. Same fucking move, matrix clothes.

Invertible iff det ≠ 0: three ways to say one thing

From the geometric picture:

• If $\det(A) \ne 0$: $A$ is a bijective transformation (it neither collapses the plane nor misses any point). An inverse exists.
• If $\det(A) = 0$: $A$ collapses the plane to a line (or point). Multiple inputs map to the same output — not injective; many outputs are unreachable — not surjective. No inverse exists.

These three statements are equivalent and should be remembered as one fact — tattoo them on your brain:

$A \text{ is invertible} \iff \det(A) \ne 0 \iff A\text{'s columns are linearly independent}$

A singular matrix (zero determinant) is not invertible. The word "singular" is historically ominous — something went badly wrong with this transformation, and no amount of wishful thinking will fix it.

The $2\times 2$ inverse formula

For $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ with $\det(A) = ad - bc \ne 0$:

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

The recipe: swap the diagonal entries ($a$ and $d$ exchange places); negate the off-diagonal entries ($b \to -b$, $c \to -c$); divide every entry by $\det(A)$.

Verify it directly. Let $\delta = ad-bc$. Then: $AA^{-1} = \frac{1}{\delta}\begin{pmatrix}a&b\\c&d\end{pmatrix}\begin{pmatrix}d&-b\\-c&a\end{pmatrix} = \frac{1}{\delta}\begin{pmatrix}ad-bc & -ab+ba\\cd-dc & -bc+da\end{pmatrix} = \frac{1}{\delta}\begin{pmatrix}\delta & 0\\0&\delta\end{pmatrix} = I$

The $\frac{1}{\det(A)}$ out front is what forces the diagonal entries to be $1$ rather than $\det(A)$. This is why zero determinant kills the formula — you'd be dividing by zero.

Solving Ax = b

One of the most important uses of the inverse: solving $A\vec{x} = \vec{b}$.

If $A$ is invertible, multiply both sides on the LEFT by $A^{-1}$:

$A^{-1}(A\vec{x}) = A^{-1}\vec{b} \quad \Rightarrow \quad I\vec{x} = A^{-1}\vec{b} \quad \Rightarrow \quad \vec{x} = A^{-1}\vec{b}$

Order matters: multiply on the LEFT. $\vec{b}A^{-1}$ is often not even defined (or is a completely different operation). This is the non-commutativity of matrix multiplication showing up where it matters.

When would you actually do this? The formula $\vec{x} = A^{-1}\vec{b}$ is clean and satisfying, but in practice — say, when solving a $100 \times 100$ system — computing $A^{-1}$ and then multiplying is slower than running Gaussian elimination directly on $[A | \vec{b}]$ (next lesson). Don't be the person who inverts a hundred-by-hundred matrix to solve one system. The inverse formula shines for $2\times 2$ systems and for theoretical understanding. For large-scale computation, elimination is smarter.

The socks-and-shoes formula

For invertible matrices $A$ and $B$:

$(AB)^{-1} = B^{-1}A^{-1}$

Socks and shoes: if you put on socks ($B$) then shoes ($A$), to reverse the process you take off shoes ($A^{-1}$) then socks ($B^{-1}$). The order reverses.

Proof: $(AB)(B^{-1}A^{-1}) = A(BB^{-1})A^{-1} = A \cdot I \cdot A^{-1} = AA^{-1} = I$

Similarly $(B^{-1}A^{-1})(AB) = B^{-1}(A^{-1}A)B = B^{-1}IB = I$. Both identities verified.

The full generalization: $(ABC)^{-1} = C^{-1}B^{-1}A^{-1}$. The entire product reverses.