Inverse Matrices

The inverse of a square matrix $A$A, written $A^{-1}$A−1, is the matrix that "undoes" $A$A: it satisfies $AA^{-1} = A^{-1}A = I$AA−1=A−1A=I. It is the matrix analogue of a reciprocal — the thing you multiply by to get back to the identity, just as $x \cdot x^{-1} = 1$x⋅x−1=1 for a non-zero number. But there is a crucial catch: only non-singular matrices (those with $\det A \neq 0$detA=0) have an inverse. A singular matrix has collapsed space irreversibly, so there is nothing to undo it. This lesson gives the slick $2\times2$2×2 formula, the full $3\times3$3×3 adjugate method, the algebraic properties (especially the order-reversing $(AB)^{-1} = B^{-1}A^{-1}$(AB)−1=B−1A−1), and the verification habits that protect your marks. The inverse is where "matrix division" finally makes sense — not as an entrywise operation, but as multiplication by $A^{-1}$A−1 — and it is the key that unlocks solving $A\mathbf{x} = \mathbf{b}$Ax=b in a single stroke. Because the whole edifice rests on the determinant, the very first thing you compute for any inverse problem is $\det A$detA: it tells you immediately whether an inverse even exists, and it appears as the dividing factor in the answer.

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1. Where this sits in AQA 7367

Inverses are compulsory pure content for Papers 1 and 2. The $2\times2$2×2 and $3\times3$3×3 computations are AO1; choosing and justifying the method, and using $A^{-1}$A−1 to solve equations or prove identities, are AO2/AO3. The inverse is the direct prerequisite for solving systems (lesson 5: $\mathbf{x} = A^{-1}\mathbf{b}$x=A−1b) and underlies transformations (the inverse transformation has matrix $M^{-1}$M−1) and diagonalisation ($A = PDP^{-1}$A=PDP−1). Everything depends on the determinant from the previous lesson: $A^{-1}$A−1 exists if and only if $\det A \neq 0$detA=0.

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2. The 2×2 inverse

For $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$A=(ac​bd​) with $\det(A) = ad - bc \neq 0$det(A)=ad−bc=0,

$$A^{-1} = \frac{1}{ad - bc}\begin{pmatrix} d & -b \\ -c & a \end{pmatrix}.$$A−1=ad−bc1​(d−c​−ba​).

The recipe in words: swap the main-diagonal entries, negate the off-diagonal entries, and divide by the determinant. You can derive it from $\operatorname{adj}A / \det A$adjA/detA (below) or simply verify that the product is $I$I:

$$\frac{1}{ad-bc}\begin{pmatrix} a & b \\ c & d \end{pmatrix}\begin{pmatrix} d & -b \\ -c & a \end{pmatrix} = \frac{1}{ad-bc}\begin{pmatrix} ad-bc & 0 \\ 0 & ad-bc \end{pmatrix} = I.$$ad−bc1​(ac​bd​)(d−c​−ba​)=ad−bc1​(ad−bc0​0ad−bc​)=I.

Worked Example 1 — a 2×2 inverse with verification (with mark scheme)

Find the inverse of $A = \begin{pmatrix} 3 & 1 \\ 5 & 2 \end{pmatrix}$A=(35​12​).

$\det(A) = 3\cdot2 - 1\cdot5 = 1$det(A)=3⋅2−1⋅5=1 (non-zero, so the inverse exists). Apply the recipe:

$$A^{-1} = \frac{1}{1}\begin{pmatrix} 2 & -1 \\ -5 & 3 \end{pmatrix} = \begin{pmatrix} 2 & -1 \\ -5 & 3 \end{pmatrix}.$$A−1=11​(2−5​−13​)=(2−5​−13​).

Check: $AA^{-1} = \begin{pmatrix} 3 & 1 \\ 5 & 2 \end{pmatrix}\begin{pmatrix} 2 & -1 \\ -5 & 3 \end{pmatrix} = \begin{pmatrix} 6-5 & -3+3 \\ 10-10 & -5+6 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = I$AA−1=(35​12​)(2−5​−13​)=(6−510−10​−3+3−5+6​)=(10​01​)=I ✓.

Mark scheme: M1 for $\det A = 1$detA=1; M1 for applying the swap/negate recipe; A1 for the correct $A^{-1}$A−1. The verification is not required but is the single best way to catch a sign slip — examiners credit a clear check if the question says "verify".

Worked Example 2 — when the determinant is not 1 (with mark scheme)

Find the inverse of $B = \begin{pmatrix} 4 & 6 \\ 1 & 2 \end{pmatrix}$B=(41​62​).

$\det(B) = 4\cdot2 - 6\cdot1 = 2$det(B)=4⋅2−6⋅1=2. Then

$$B^{-1} = \frac{1}{2}\begin{pmatrix} 2 & -6 \\ -1 & 4 \end{pmatrix} = \begin{pmatrix} 1 & -3 \\ -\tfrac{1}{2} & 2 \end{pmatrix}.$$B−1=21​(2−1​−64​)=(1−21​​−32​).

Mark scheme: M1 for $\det B = 2$detB=2; M1 for the adjugate $\begin{pmatrix} 2 & -6 \\ -1 & 4 \end{pmatrix}$(2−1​−64​); A1 for dividing every entry by $2$2 (the $-\tfrac12$−21​ must be exact). Forgetting to divide the whole matrix by the determinant — a very common error — loses the A1.

A quick sanity check that costs nothing: the product of the diagonal entries of $B^{-1}$B−1 here is $1 \cdot 2 = 2$1⋅2=2 and the off-diagonal product is $(-3)(-\tfrac12) = \tfrac32$(−3)(−21​)=23​, so $\det(B^{-1}) = 2 - \tfrac32 = \tfrac12 = 1/\det B$det(B−1)=2−23​=21​=1/detB — exactly as the property table predicts. Verifying $\det(B^{-1}) = 1/\det B$det(B−1)=1/detB is a fast way to catch a dropped factor.

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3. The 3×3 inverse via the adjugate

For a $3\times3$3×3 (or larger) matrix, use

$A^{-1} = \frac{1}{\det(A)}\operatorname{adj}(A),$A−1=det(A)1​adj(A),

where the adjugate $\operatorname{adj}(A)$adj(A) is the transpose of the matrix of cofactors. It is worth understanding why this works rather than treating it as a ritual. The cofactors are exactly the numbers that appear when you expand a determinant, and the adjugate arranges them so that multiplying by $A$A reconstructs $\det A$detA on the diagonal and zeros off it. The transpose in step 4 is therefore not a cosmetic flourish: it is what aligns each cofactor with the right entry of $A$A so the products telescope correctly. Skip it and you compute $A \cdot (\text{cofactor matrix})$A⋅(cofactor matrix), which is not $(\det A)I$(detA)I in general — the single most common reason a $3\times3$3×3 inverse comes out wrong.

The full procedure:

1. Find $\det(A)$det(A). If it is $0$0, stop — no inverse exists.
2. Matrix of minors: replace each entry by the $2\times2$2×2 determinant left after deleting its row and column.
3. Apply the checkerboard signs $\begin{pmatrix} + & - & + \\ - & + & - \\ + & - & + \end{pmatrix}$​+−+​−+−​+−+​​ to get the matrix of cofactors.
4. Transpose the cofactor matrix to get $\operatorname{adj}(A)$adj(A).
5. Divide every entry by $\det(A)$det(A).

Worked Example 3 — a full 3×3 inverse (with mark scheme)

Find the inverse of $A = \begin{pmatrix} 1 & 2 & 0 \\ 0 & 1 & 1 \\ 2 & 0 & 1 \end{pmatrix}$A=​102​210​011​​.

Step 1 — determinant (expand along row 1): $\det A = 1(1\cdot1 - 1\cdot0) - 2(0\cdot1 - 1\cdot2) + 0 = 1 + 4 = 5 \neq 0$detA=1(1⋅1−1⋅0)−2(0⋅1−1⋅2)+0=1+4=5=0.

Step 2 — matrix of minors:

$$\begin{pmatrix} 1 & -2 & -2 \\ 2 & 1 & -4 \\ 2 & 1 & 1 \end{pmatrix}.$$​122​−211​−2−41​​.

Step 3 — apply signs (cofactors):

$$\begin{pmatrix} 1 & 2 & -2 \\ -2 & 1 & 4 \\ 2 & -1 & 1 \end{pmatrix}.$$​1−22​21−1​−241​​.

Step 4 — transpose to the adjugate:

$$\operatorname{adj}(A) = \begin{pmatrix} 1 & -2 & 2 \\ 2 & 1 & -1 \\ -2 & 4 & 1 \end{pmatrix}.$$adj(A)=​12−2​−214​2−11​​.

Step 5 — divide by the determinant:

$$A^{-1} = \frac{1}{5}\begin{pmatrix} 1 & -2 & 2 \\ 2 & 1 & -1 \\ -2 & 4 & 1 \end{pmatrix}.$$A−1=51​​12−2​−214​2−11​​.

Check (one column is enough): $A \cdot \tfrac15(1, 2, -2)^{T} = \tfrac15(1\cdot1+2\cdot2+0, \, 0\cdot1+1\cdot2+1\cdot(-2), \, 2\cdot1+0+1\cdot(-2))^{T} = \tfrac15(5, 0, 0)^{T} = (1,0,0)^{T}$A⋅51​(1,2,−2)T=51​(1⋅1+2⋅2+0,0⋅1+1⋅2+1⋅(−2),2⋅1+0+1⋅(−2))T=51​(5,0,0)T=(1,0,0)T ✓ — the first column of $I$I.

Mark scheme: M1 $\det A = 5$detA=5; M1 matrix of minors; M1 apply signs; A1 correct adjugate (transpose); A1 final $A^{-1}$A−1. Up to 5 marks; the transpose step (cofactors $\to$→ adjugate) is the most common omission and loses an A mark.

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4. Properties of the inverse

Property	Statement
$(A^{-1})^{-1} = A$(A−1)−1=A	the inverse of the inverse is the original
$(AB)^{-1} = B^{-1}A^{-1}$(AB)−1=B−1A−1	inverses reverse the order
$(A^{T})^{-1} = (A^{-1})^{T}$(AT)−1=(A−1)T	inverse and transpose commute
$\det(A^{-1}) = \dfrac{1}{\det(A)}$det(A−1)=det(A)1​	from $\det(A)\det(A^{-1}) = 1$det(A)det(A−1)=1
$(kA)^{-1} = \dfrac{1}{k}A^{-1}$(kA)−1=k1​A−1	for a scalar $k \neq 0$k=0

Key point: $(AB)^{-1} = B^{-1}A^{-1}$(AB)−1=B−1A−1 — the order reverses, like taking off shoes and socks: last on, first off. The proof is one line: $(AB)(B^{-1}A^{-1}) = A(BB^{-1})A^{-1} = AIA^{-1} = AA^{-1} = I$(AB)(B−1A−1)=A(BB−1)A−1=AIA−1=AA−1=I.

The order-reversal generalises: $(ABC)^{-1} = C^{-1}B^{-1}A^{-1}$(ABC)−1=C−1B−1A−1, and $(A^n)^{-1} = (A^{-1})^n$(An)−1=(A−1)n. These are the matrix versions of index laws, with the all-important caveat that order must be respected throughout.

The property $(A^{T})^{-1} = (A^{-1})^{T}$(AT)−1=(A−1)T is worth a moment too. It says transposing and inverting can be done in either order. The proof is the same telescoping trick: $(A^{-1})^{T}A^{T} = (AA^{-1})^{T} = I^{T} = I$(A−1)TAT=(AA−1)T=IT=I, using $(XY)^{T} = Y^{T}X^{T}$(XY)T=YTXT from the first lesson. A neat corollary is that the inverse of a symmetric matrix is symmetric: if $A = A^{T}$A=AT then $(A^{-1})^{T} = (A^{T})^{-1} = A^{-1}$(A−1)T=(AT)−1=A−1. Facts like these are exactly what a "show that" question in this strand will ask you to prove, and they all reduce to one or two applications of the order-reversal laws — no computation needed.

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5. When the inverse fails to exist

$A$A has no inverse exactly when $\det(A) = 0$det(A)=0 — i.e. when $A$A is singular. For instance $\begin{pmatrix} 2 & 4 \\ 1 & 2 \end{pmatrix}$(21​42​) has $\det = 4 - 4 = 0$det=4−4=0, so it is singular: its rows are proportional, the transformation squashes the plane onto a line, and no matrix can recover the lost information.

The "no recovery" picture is the most intuitive way to see why a singular matrix cannot be inverted. If a transformation flattens a whole plane of points down onto a single line, then infinitely many starting points land on the same image point — and an inverse would have to send that image point back to all of them at once, which no function can do. The determinant being zero is precisely the algebra of "the image is lower-dimensional than the input". Contrast a non-singular matrix, whose determinant measures a genuine (non-zero) area or volume scaling: nothing is lost, so the process can be reversed. This is why $\det A = 0$detA=0 and "non-invertible" are not two separate facts but one fact viewed two ways.

A common exam task is to find the parameter that makes a matrix singular (hence non-invertible).

Worked Example 4 — the non-invertible parameter (with mark scheme)

For what value of $k$k does $\begin{pmatrix} 1 & k \\ 3 & 6 \end{pmatrix}$(13​k6​) have no inverse?

No inverse $\iff \det = 0$⟺det=0: $\det = 1\cdot6 - k\cdot3 = 6 - 3k = 0$det=1⋅6−k⋅3=6−3k=0, so $k = 2$k=2.

Mark scheme: M1 set $\det = 0$det=0; A1 $k = 2$k=2. Stating "singular when $\det = 0$det=0" earns the method mark even before solving.

Worked Example 5 — a second route via Cayley–Hamilton (with mark scheme)

For a $2\times2$2×2 matrix the inverse can be found without the swap/negate recipe, using the identity $A^{-1} = (\operatorname{tr}(A)\,I - A)/\det A$A−1=(tr(A)I−A)/detA (a preview of lesson 10). Take $A = \begin{pmatrix} 3 & 1 \\ 5 & 2 \end{pmatrix}$A=(35​12​): $\operatorname{tr}(A) = 5$tr(A)=5, $\det A = 1$detA=1, so

$$A^{-1} = \frac{1}{1}\left(5\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} - \begin{pmatrix} 3 & 1 \\ 5 & 2 \end{pmatrix}\right) = \begin{pmatrix} 5-3 & 0-1 \\ 0-5 & 5-2 \end{pmatrix} = \begin{pmatrix} 2 & -1 \\ -5 & 3 \end{pmatrix},$$A−1=11​(5(10​01​)−(35​12​))=(5−30−5​0−15−2​)=(2−5​−13​),

agreeing exactly with Worked Example 1. The two methods must coincide — they are different ways of writing the same adjugate-over-determinant formula.

Mark scheme: M1 for $\operatorname{tr}A = 5$trA=5, $\det A = 1$detA=1; M1 for forming $\operatorname{tr}(A)I - A$tr(A)I−A; A1 for the correct $A^{-1}$A−1. Either route earns full marks; choosing the most reliable one for the matrix in front of you is good exam strategy.

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6. Specimen-style exam question

(Specimen-style — not a real past paper.) The matrix $A = \begin{pmatrix} 2 & -1 \\ 1 & 3 \end{pmatrix}$A=(21​−13​). (a) Find $A^{-1}$A−1. (b) Hence solve the matrix equation $A\mathbf{x} = \begin{pmatrix} 4 \\ 5 \end{pmatrix}$Ax=(45​). (5 marks)

Model solution. (a) $\det A = 2\cdot3 - (-1)\cdot1 = 7$detA=2⋅3−(−1)⋅1=7, so

$$A^{-1} = \frac{1}{7}\begin{pmatrix} 3 & 1 \\ -1 & 2 \end{pmatrix}.$$A−1=71​(3−1​12​).