Invertible Matrices

The definition

Definition. A square matrix $A$ of order $n$ is invertible if there exists a square matrix $B$ of the same order such that $$AB = BA = I$$ Then $B$ is called the inverse of $A$, written $A^{-1}$.

Only square matrices can be invertible. If $A$ were $m \times n$ with $m \neq n$, the products $AB$ and $BA$ (for any $B$ making them defined) would have different orders — they could never both equal one identity matrix.

Verification is direct multiplication. To confirm $B = A^{-1}$, compute both $AB$ and $BA$ and check each equals $I$: $$A = \begin{pmatrix} 2 & 3 \\ 1 & 2 \end{pmatrix}, \quad B = \begin{pmatrix} 2 & -3 \\ -1 & 2 \end{pmatrix}: \qquad AB = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = BA$$ so $B = A^{-1}$ (and equally $A = B^{-1}$). Not every square matrix qualifies — the zero matrix has no inverse ($OB = O \neq I$ always), and the next chapter's determinant will give the precise test.

Uniqueness

Theorem. If an inverse exists, it is unique.

Proof (a beautiful three-liner): suppose $B$ and $C$ are both inverses of $A$. Then $$B = BI = B(AC) = (BA)C = IC = C \qquad \blacksquare$$ Associativity is the engine — the whole argument is regrouping one triple product two ways.

The two inverse laws

$$\big(A^{-1}\big)^{-1} = A \qquad \qquad (AB)^{-1} = B^{-1}A^{-1} \ \ (\text{socks-shoes: undo in reverse order})$$

Proof of the reversal law: check that $B^{-1}A^{-1}$ undoes $AB$: $$(AB)(B^{-1}A^{-1}) = A(BB^{-1})A^{-1} = AIA^{-1} = AA^{-1} = I$$ and similarly $(B^{-1}A^{-1})(AB) = I$. By uniqueness, $B^{-1}A^{-1}$ is the inverse of $AB$. $\blacksquare$

Two companions worth noting: $\ I^{-1} = I$, and $\big(A'\big)^{-1} = \big(A^{-1}\big)'$ (transpose the equation $AA^{-1} = I$ with the reversal law of transposes).

What invertibility buys back: cancellation

Section 3's bad news was $AB = AC \not\Rightarrow B = C$. With an invertible $A$ the cancellation returns: $$AB = AC \implies A^{-1}(AB) = A^{-1}(AC) \implies IB = IC \implies B = C$$ Multiplying both sides by $A^{-1}$ (on the correct side!) is the matrix world's substitute for division.

Common mistakes to avoid

Mistake 1 — writing $\frac{1}{A}$ or $\frac{B}{A}$. There is no division of matrices — only multiplication by $A^{-1}$, and the side matters: $A^{-1}B \neq BA^{-1}$ in general.

Mistake 2 — $(AB)^{-1} = A^{-1}B^{-1}$. The order must reverse.

Mistake 3 — assuming every nonzero square matrix is invertible. False: $\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}$ has no inverse (next chapter: its determinant is 0).

Mistake 4 — checking only one product. The definition asks for $AB = I$ and $BA = I$. (For square matrices one implies the other — a deep fact — but the safe board answer verifies both.)

Solved Examples

Example 1 — Verifying an inverse pair

Show that $B = \begin{pmatrix} 2 & -3 \\ -1 & 2 \end{pmatrix}$ is the inverse of $A = \begin{pmatrix} 2 & 3 \\ 1 & 2 \end{pmatrix}$.

Step 1 — compute $AB$: $$AB = \begin{pmatrix} 4 - 3 & -6 + 6 \\ 2 - 2 & -3 + 4 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$$

Step 2 — compute $BA$: $$BA = \begin{pmatrix} 4 - 3 & 6 - 6 \\ -2 + 2 & -3 + 4 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$$

Answer: both products equal $I$, so $B = A^{-1}$. (Notice the pattern for $2 \times 2$: swap the diagonal, negate the off-diagonal — it works here because $2 \cdot 2 - 3 \cdot 1 = 1$; the general recipe divides by that number, next chapter.)

Example 2 — A triangular inverse

Find the inverse of $U = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}$ by inspection and verify.

Step 1 — guess the undo: $U$ adds twice the second coordinate to the first; the undo subtracts it: $V = \begin{pmatrix} 1 & -2 \\ 0 & 1 \end{pmatrix}$.

Step 2 — verify: $$UV = \begin{pmatrix} 1 & -2 + 2 \\ 0 & 1 \end{pmatrix} = I, \qquad VU = \begin{pmatrix} 1 & 2 - 2 \\ 0 & 1 \end{pmatrix} = I$$

Answer: $U^{-1} = \begin{pmatrix} 1 & -2 \\ 0 & 1 \end{pmatrix}$ — inverting a unit triangular matrix just negates the off-diagonal entry.

Example 3 — Inverse of a diagonal matrix

Find the inverse of $D = \begin{pmatrix} 2 & 0 \\ 0 & 5 \end{pmatrix}$.

Step 1 — diagonal matrices multiply diagonally, so the inverse must carry the reciprocal entries: $$D^{-1} = \begin{pmatrix} \frac{1}{2} & 0 \\ 0 & \frac{1}{5} \end{pmatrix}$$

Step 2 — verify: $DD^{-1} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = D^{-1}D$ ✓.

Answer: reciprocals down the diagonal — and this fails the moment a diagonal entry is 0, one more glimpse of the invertibility test to come.

Example 4 — Uniqueness of the inverse

Prove that a matrix can have at most one inverse.

Step 1 — suppose two: let $B$ and $C$ both satisfy $AB = BA = I$ and $AC = CA = I$.

Step 2 — evaluate $BAC$ two ways using associativity: $$B = BI = B(AC) = (BA)C = IC = C$$

Answer: $B = C$ — the inverse, when it exists, is unique, which is what entitles us to the notation $A^{-1}$.

Example 5 — The reversal law

If $A$ and $B$ are invertible matrices of the same order, prove that $(AB)^{-1} = B^{-1}A^{-1}$.

Step 1 — multiply $AB$ by the candidate on the right: $$(AB)(B^{-1}A^{-1}) = A(BB^{-1})A^{-1} = AA^{-1} = I$$

Step 2 — and on the left: $$(B^{-1}A^{-1})(AB) = B^{-1}(A^{-1}A)B = B^{-1}B = I$$

Step 3 — invoke uniqueness: the matrix that undoes $AB$ from both sides is its inverse.

Answer: $(AB)^{-1} = B^{-1}A^{-1}$ — undo the last action first.

Example 6 — Cancelling with an inverse

Given that $A$ is invertible and $AB = AC$, prove $B = C$. Show by example that the hypothesis cannot be dropped.

Step 1 — left-multiply by $A^{-1}$: $$A^{-1}(AB) = A^{-1}(AC) \implies (A^{-1}A)B = (A^{-1}A)C \implies B = C$$

Step 2 — the counterexample without invertibility: $A = \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}$, $\ B = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}$, $\ C = \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}$: both products $AB$ and $AC$ equal $\begin{pmatrix} 1 & 0 \\ 1 & 0 \end{pmatrix}$, yet $B \neq C$.

Answer: invertibility restores cancellation; without it, equal products prove nothing.

Example 7 — Transpose meets inverse

Prove that $\big(A'\big)^{-1} = \big(A^{-1}\big)'$ for an invertible matrix $A$.

Step 1 — transpose the equation $AA^{-1} = I$ using the reversal law of transposes: $$\big(AA^{-1}\big)' = I' \implies \big(A^{-1}\big)' A' = I$$

Step 2 — transpose $A^{-1}A = I$ likewise: $A'\big(A^{-1}\big)' = I$.

Step 3 — uniqueness: $\big(A^{-1}\big)'$ undoes $A'$ from both sides.

Answer: the inverse of the transpose is the transpose of the inverse — the two operations commute.