Inverse Relation - Definition, Formula & Examples

What Is An Inverse Relation?

An inverse relation is the relation obtained by reversing the order of every ordered pair in a given relation. If $R$ is a relation, its inverse is written $R^{-1}$, and a pair $(a, b)$ belongs to $R$ exactly when $(b, a)$ belongs to $R^{-1}$.

$$R^{-1} = {(b, a) : (a, b) \in R}$$

If $R$ links set $A$ to set $B$ (so $R \subseteq A \times B$), then $R^{-1}$ links $B$ back to $A$ (so $R^{-1} \subseteq B \times A$). The swap is total: every pair flips, no pair is added or dropped.

How Do You Find the Inverse of a Relation?

For a relation listed as ordered pairs, swap each pair's coordinates. For a relation given by an equation, swap $x$ and $y$ and (where useful) solve for $y$.

A frequent reader question — "how do you find the inverse of $y = 3x + 2$?" — runs like this:

1. Swap the variables: $x = 3y + 2$.

2. Solve for $y$: $x - 2 = 3y$, so $y = \dfrac{x - 2}{3}$.

The original relation's domain becomes the inverse's range, and the original range becomes the inverse's domain. That swap is automatic — it falls straight out of reversing the pairs.

What Is the Inverse Relation Theorem?

The inverse relation theorem states that inverting twice returns the original relation:

$$(R^{-1})^{-1} = R$$

The proof is short. A pair $(a, b) \in R$ becomes $(b, a) \in R^{-1}$ by definition. Inverting again sends $(b, a)$ back to $(a, b)$, which lands in $(R^{-1})^{-1}$.

Since every pair returns to where it started, the two relations are identical. The flip is its own undo — which is exactly why the graph is a clean mirror image.

Examples of Inverse Relation

Six examples, building from a discrete list to algebraic equations and a graph reflection.

Example 1

Find the inverse of $R = {(1, 4), (2, 5), (3, 6)}$.

Swap each pair.

$R^{-1} = {(4, 1), (5, 2), (6, 3)}$

Final answer: $R^{-1} = {(4, 1), (5, 2), (6, 3)}$.

Example 2

State the domain and range of $R = {(a, 2), (b, 4), (c, 1)}$ and of $R^{-1}$.

Original domain $= {a, b, c}$; original range $= {1, 2, 4}$.

Inverse: $R^{-1} = {(2, a), (4, b), (1, c)}$.

Inverse domain $= {1, 2, 4}$; inverse range $= {a, b, c}$.

Final answer: the domain and range trade places — the inverse's domain is the original's range, and vice versa.

Example 3

A student inverts $R = {(1, 5), (2, 5), (3, 7)}$ and concludes the inverse is a function. Is it?

Wrong attempt. The student swaps to get $R^{-1} = {(5, 1), (5, 2), (7, 3)}$ and reasons, "the original was a function, so its inverse must be too."

Why it breaks. Look at the inverse's first coordinates: $5$ appears twice, paired with both $1$ and $2$. A function cannot send one input to two outputs. So the inverse fails the function test even though it is a perfectly valid inverse relation.

Correct. $R^{-1} = {(5, 1), (5, 2), (7, 3)}$ is an inverse relation but not an inverse function, because the original relation was not one-to-one.

Final answer: the inverse relation exists; it is a function only when the original is one-to-one.

Example 4

Find the inverse of the algebraic relation $y = x^2$.

Swap the variables.

$x = y^2$

Solve for $y$: $y = \pm\sqrt{x}$.

Final answer: $y = \pm\sqrt{x}$, a valid inverse relation, but the $\pm$ shows it is not a function (each positive $x$ gives two $y$-values).

Example 5

Find the inverse of $y = \dfrac{2x + 1}{5}$.

Swap the variables.

$x = \dfrac{2y + 1}{5}$

Multiply both sides by $5$: $5x = 2y + 1$.

Subtract $1$: $5x - 1 = 2y$.

Divide by $2$: $y = \dfrac{5x - 1}{2}$.

Final answer: $y = \dfrac{5x - 1}{2}$.

Example 6

A relation passes through $(0, 3)$, $(2, 7)$, and $(-1, 1)$. Where does its inverse pass through, and how does the graph relate?

Swap each pair: the inverse passes through $(3, 0)$, $(7, 2)$, and $(1, -1)$.

Each original point and its inverse are mirror images across $y = x$.

Final answer: the inverse passes through $(3, 0)$, $(7, 2)$, $(1, -1)$, and its graph is the original reflected over the line $y = x$.

Where Inverse Relations Show Up

"Reversing a mapping is how we ask a question backwards."

The inverse relation exists because we constantly need to run a correspondence in reverse — and the swap formalises that.

• Encoding and decoding. A cipher maps letters to symbols; reading a message back uses the inverse relation. The decode step is the inverse.

• Unit conversion. Celsius-to-Fahrenheit and its reverse are inverse relations of each other. The reflection across $y = x$ is literally the "convert the other way" rule.

• Databases and lookups. Any "who has this?" table can be flipped to "what does this person have?" — the same pairs, read backward.

The destination is the inverse function: once a relation is one-to-one, its inverse relation graduates into an inverse function, the form you will use across algebra, trigonometry, and calculus.

Common Errors With Inverse Relations

Mistake 1: Assuming the inverse of a function is always a function

Where it slips in: Any relation that is not one-to-one, as in Examples 3 and 4.

Don't do this: Conclude that because $R$ is a function, $R^{-1}$ must be too.

The correct way: Check whether the original is one-to-one. Every relation has an inverse relation; the inverse is a function only when the original passes the horizontal line test.

The first instinct is to treat "inverse" as a property that preserves functionhood — but the inverse of $y = x^2$ being $\pm\sqrt{x}$ is exactly where that assumption drops the second branch and goes wrong.

Mistake 2: Forgetting to swap domain and range

Where it slips in: Stating the domain and range of an inverse, as in Example 2.

Don't do this: Keep the original domain as the inverse's domain.

The correct way: The inverse's domain is the original's range, and the inverse's range is the original's domain. They always trade.

Mistake 3: Solving an equation before swapping the variables

Where it slips in: Finding the inverse of $y = mx + b$ style relations.

Don't do this: Solve the original for $x$ in terms of $y$ and call that the inverse.

The correct way: Swap $x$ and $y$ first, then solve for $y$. The rusher who reorders the steps gets an expression that looks similar but represents the wrong relation.

Practice Questions

Try these, then check the answers below.

1. Find the inverse of $R = {(2, 8), (3, 8), (5, 1)}$ and decide whether the inverse is a function.

2. Find the inverse of $y = 4x - 7$.

3. Reflect the points $(1, 2)$ and $(4, 0)$ across $y = x$.

4. State the domain and range of $R = {(p, 3), (q, 5)}$ and of $R^{-1}$.

5. Apply the inverse relation theorem: what is $(R^{-1})^{-1}$ for $R = {(1, 9)}$?

Answers

Answer to Question 1: $R^{-1} = {(8, 2), (8, 3), (1, 5)}$. It is not a function — the input $8$ maps to both $2$ and $3$, because the original relation was not one-to-one.

Answer to Question 2: Swap first: $x = 4y - 7$, so $4y = x + 7$ and $y = \dfrac{x + 7}{4}$.

Answer to Question 3: $(1, 2) \to (2, 1)$ and $(4, 0) \to (0, 4)$. Swap each pair's coordinates.

Answer to Question 4: Original domain $= {p, q}$, range $= {3, 5}$; inverse domain $= {3, 5}$, range $= {p, q}$. The two trade places.

Answer to Question 5: $(R^{-1})^{-1} = R = {(1, 9)}$. Inverting twice returns the original.

Conclusion

• An inverse relation $R^{-1}$ is formed by swapping every ordered pair $(a, b)$ in $R$ into $(b, a)$.

• The graph of an inverse relation is the original reflected across the line $y = x$, and the domain and range trade places.

• The inverse relation theorem says $(R^{-1})^{-1} = R$, so inverting twice returns the original.

• An inverse relation is always a relation, but it is a function only when the original is one-to-one.

• Always swap the variables before solving for $y$ when inverting an equation.

Your Next Move

Work through the practice questions above to solidify your understanding. If you get stuck on the function-or-not question, return to Example 3 and check the inverse's first coordinates for repeats. At Bhanzu, trainers anchor the inverse relation on the $y = x$ mirror first, so the algebra never feels like a memorised swap. Want a live Bhanzu trainer to walk through more inverse relation problems? Book a free demo class.