Further Maths/Lessons/Matrices/Inverting a 2×2 Matrix
Core Pure · Matrices · Year 1

Inverting a 2×2 Matrix

Swap, negate, divide — a compact formula undoes any non-singular 2×2 matrix, turning matrix equations into problems you can solve directly.

16 min read 4 worked examples

The inverse of a 2×2 matrix

Definition 21.1 — matrix inverse

The inverse of a square matrix $A$, written $A^{-1}$, is the unique matrix satisfying $AA^{-1}=A^{-1}A=I$. An inverse exists if and only if $\det(A)\neq0$.

Theorem 21.2 — formula for a 2×2 inverse

For $A=\begin{pmatrix}a&b\\c&d\end{pmatrix}$ with $\det(A)=ad-bc\neq0$: $$A^{-1}=\frac{1}{ad-bc}\begin{pmatrix}d&-b\\-c&a\end{pmatrix}$$ Swap the leading-diagonal entries, negate the other two, and divide by the determinant.

ProofVerifying the 2×2 inverse formula
1
Let $A=\begin{pmatrix}a&b\\c&d\end{pmatrix}$ and $B=\dfrac{1}{ad-bc}\begin{pmatrix}d&-b\\-c&a\end{pmatrix}$.
Proposed inverse from the formula.
2
$AB=\dfrac{1}{ad-bc}\begin{pmatrix}a&b\\c&d\end{pmatrix}\begin{pmatrix}d&-b\\-c&a\end{pmatrix}$
Multiply $A$ by the candidate inverse.
3
$=\dfrac{1}{ad-bc}\begin{pmatrix}ad-bc&-ab+ba\\cd-dc&-cb+da\end{pmatrix}=\dfrac{1}{ad-bc}\begin{pmatrix}ad-bc&0\\0&ad-bc\end{pmatrix}$
Direct expansion; off-diagonal entries cancel exactly.
4
$=\begin{pmatrix}1&0\\0&1\end{pmatrix}=I$
Dividing both diagonal entries by $ad-bc$.
∎
Example 1 · Finding a 2×2 inverse
Find $A^{-1}$ for $A=\begin{pmatrix}3&1\\2&4\end{pmatrix}$.

$\det(A)=3(4)-1(2)=10$.

$$A^{-1}=\frac{1}{10}\begin{pmatrix}4&-1\\-2&3\end{pmatrix}$$
Example 2 · Verifying an inverse
Verify the answer to Example 1 by checking $AA^{-1}=I$.

$AA^{-1}=\dfrac{1}{10}\begin{pmatrix}3&1\\2&4\end{pmatrix}\begin{pmatrix}4&-1\\-2&3\end{pmatrix}=\dfrac{1}{10}\begin{pmatrix}12-2&-3+3\\8-8&-2+12\end{pmatrix}=\dfrac{1}{10}\begin{pmatrix}10&0\\0&10\end{pmatrix}=\begin{pmatrix}1&0\\0&1\end{pmatrix}$ ✓

Example 3 · Finding an unknown given an inverse
Given $A=\begin{pmatrix}2&k\\3&5\end{pmatrix}$ and that $\det(A^{-1})=\dfrac{1}{4}$, find the possible value(s) of $k$.

Since $\det(A^{-1})=\dfrac{1}{\det(A)}$, we need $\det(A)=4$.

$\det(A)=2(5)-k(3)=10-3k=4 \Rightarrow 3k=6 \Rightarrow k=2$.

Theorem 21.3 — det(A⁻¹) = 1/det(A)

This follows directly from $\det(AB)=\det(A)\det(B)$ applied to $AA^{-1}=I$: $\det(A)\det(A^{-1})=\det(I)=1$, so $\det(A^{-1})=\dfrac{1}{\det(A)}$.

Example 4 · Using inverses to solve a matrix equation
Given $\begin{pmatrix}5&2\\3&1\end{pmatrix}X=\begin{pmatrix}7\\4\end{pmatrix}$, find the column matrix $X$.

$\det\begin{pmatrix}5&2\\3&1\end{pmatrix}=5(1)-2(3)=-1$, so the inverse is $\dfrac{1}{-1}\begin{pmatrix}1&-2\\-3&5\end{pmatrix}=\begin{pmatrix}-1&2\\3&-5\end{pmatrix}$.

$X=\begin{pmatrix}-1&2\\3&-5\end{pmatrix}\begin{pmatrix}7\\4\end{pmatrix}=\begin{pmatrix}-7+8\\21-20\end{pmatrix}=\begin{pmatrix}1\\1\end{pmatrix}$.

Common mistake

When pre-multiplying both sides of a matrix equation by an inverse, the inverse must be applied on the same side consistently (both left, or both right) — since matrix multiplication isn't commutative, $A^{-1}(AX)=X$ works, but $(AX)A^{-1}$ generally does not simplify usefully.

Exam-style questions

Question 1 5 marks

Find $A^{-1}$ for $A=\begin{pmatrix}5&3\\2&1\end{pmatrix}$.

$\det(A)=5(1)-3(2)=-1$ (M1, A1)

$$A^{-1}=\frac{1}{-1}\begin{pmatrix}1&-3\\-2&5\end{pmatrix}=\begin{pmatrix}-1&3\\2&-5\end{pmatrix}$$

(M1, A1, A1)

Question 2 6 marks

Solve the matrix equation $\begin{pmatrix}2&1\\1&-1\end{pmatrix}X=\begin{pmatrix}5\\1\end{pmatrix}$ using an inverse matrix method.

$\det=2(-1)-1(1)=-3$ (M1, A1)

Inverse: $\dfrac{1}{-3}\begin{pmatrix}-1&-1\\-1&2\end{pmatrix}$ (M1)

$X=\dfrac{1}{-3}\begin{pmatrix}-1&-1\\-1&2\end{pmatrix}\begin{pmatrix}5\\1\end{pmatrix}=\dfrac{1}{-3}\begin{pmatrix}-6\\-3\end{pmatrix}=\begin{pmatrix}2\\1\end{pmatrix}$ (M1, A1, A1)

Question 3 6 marks

Given $A=\begin{pmatrix}4&-1\\6&2\end{pmatrix}$, find $A^{-1}$, and hence find $(A^{-1})^{-1}$ without further calculation, explaining your reasoning.

$\det(A)=4(2)-(-1)(6)=8+6=14$ (M1, A1)

$A^{-1}=\dfrac{1}{14}\begin{pmatrix}2&1\\-6&4\end{pmatrix}$ (M1, A1)

$(A^{-1})^{-1}=A=\begin{pmatrix}4&-1\\6&2\end{pmatrix}$, since taking the inverse of the inverse always returns the original matrix. (B1, B1)

← Previous
Determinants