MHT-CET Maths · Determinants and Matrices

Inverse of a Matrix — Adjoint Formula, Products and Verification

A⁻¹ = adj(A)/|A| — for a 2 × 2 that is swap-the-diagonal, negate-the-off-diagonal, divide by the determinant — and (AB)⁻¹ = B⁻¹A⁻¹ handles anything built from products.

All Determinants and Matrices notes MHT-CET Maths strategy

Why this matters

44 PYQs at 25% HARD — the chapter's steady marks, and now its largest page: 2026 alone added 29. A plain 2 × 2 inverse is set most years, and the HARD variants only add one step before it: compute A² − 5A or A + B first, or notice that a matrix with tan x or cot(θ/2) entries has a trigonometric determinant and an adjoint equal to its transpose. The two-matrix stems — (AB)⁻¹ given, find B⁻¹ — and the unknown-entries stems are the same fact, AA⁻¹ = I, read in two directions.

Concept 1 of 5: Inverse of a 2 × 2 Matrix

Since

A\,\operatorname{adj}A = |A|\,I

, dividing the adjoint by the determinant gives a matrix that multiplies

A

to the identity. For

2 \times 2

the adjoint is a swap and two sign changes, so the inverse takes ten seconds.

Definition

$A^{-1} = \dfrac{1}{|A|}\operatorname{adj}A$ , defined only when $|A| \ne 0$ .
$\begin{pmatrix} a & b \\ c & d \end{pmatrix}^{-1} = \dfrac{1}{ad - bc}\begin{pmatrix} d & -b \\ -c & a \end{pmatrix}$ .
The options are usually written with the scalar outside: $\dfrac{1}{14}\begin{pmatrix} 3 & 2 \\ -4 & 2 \end{pmatrix}$ ; match the sign of the scalar AND the signs inside — $-\dfrac{1}{14}\begin{pmatrix} 3 & 2 \\ -4 & 2 \end{pmatrix}$ is a different matrix.
A matrix with $|A| = 1$ (like $\begin{pmatrix} 0.8 & -0.6 \\ 0.6 & 0.8 \end{pmatrix}$ ) has $A^{-1} = \operatorname{adj}A$ ; a rotation-type matrix has $A^{-1} = A^T$ .
Verify when in doubt: $AA^{-1}$ must be $I$ .

2 × 2 inverse

\begin{pmatrix} a & b \\ c & d \end{pmatrix}^{-1} = \frac{1}{ad - bc}\begin{pmatrix} d & -b \\ -c & a \end{pmatrix}, \quad ad - bc \ne 0

Worked example

Find the inverse of

A = \begin{pmatrix} 4 & 3 \\ 2 & 5 \end{pmatrix}

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 2nd May Shift 1 · Q150Easy

Example 1 · Determinants and Matrices · Inverse of a Matrix — Adjoint Formula, Products and Verification

A = \begin{pmatrix} 3 & -1 \\ -4 & 2 \end{pmatrix}

, then

A^{-1}

Swapping signs on the diagonal instead of the off-diagonal

The adjoint of

\begin{pmatrix} a & b \\ c & d \end{pmatrix}

\begin{pmatrix} d & -b \\ -c & a \end{pmatrix}

: the diagonal entries SWAP places and keep their signs; the off-diagonal entries stay put and change sign. Every option list contains the other three sign patterns.

Drill 2 more on inverse of a 2 × 2 matrix

Concept 2 of 5: Invert an Expression: Compute A² − 5A or A + B First, Then Invert

(A^2 - 5A)^{-1}

is not

(A^2)^{-1} - (5A)^{-1}

. Multiply out

A^2

, subtract

5A

, and invert the single

2 \times 2

matrix that results — the difficulty is arithmetic discipline, not theory.

Definition

Compute the expression entry by entry first: $A^2 = A\cdot A$ , then combine, then invert the result.
For $A = \begin{pmatrix} 2 & 1 \\ 7 & 4 \end{pmatrix}$ : $A^2 - 5A = \begin{pmatrix} 1 & 1 \\ 7 & 3 \end{pmatrix}$ , determinant $-4$ , inverse $\dfrac14\begin{pmatrix} -3 & 1 \\ 7 & -1 \end{pmatrix}$ .
$(A + B)^{-1}$ : add first — $(A + B)^{-1} \ne A^{-1} + B^{-1}$ .
A power with a parameter: $A = \begin{pmatrix} x & 1 \\ 1 & 0 \end{pmatrix}$ gives $A^4 = \begin{pmatrix} x^4 + 3x^2 + 1 & x^3 + 2x \\ x^3 + 2x & x^2 + 1 \end{pmatrix}$ ; $a_{11} = 109$ forces $x = 3$ , $A^4 = \begin{pmatrix} 109 & 33 \\ 33 & 10 \end{pmatrix}$ with determinant $1$ , so its inverse is $\begin{pmatrix} 10 & -33 \\ -33 & 109 \end{pmatrix}$ .
Factor out a scalar at the end: $\dfrac{1}{475}\begin{pmatrix} 35 & 15 \\ 15 & 20 \end{pmatrix} = \dfrac{1}{95}\begin{pmatrix} 7 & 3 \\ 3 & 4 \end{pmatrix}$ — the options are reduced.

Order of operations

(A^2 - 5A)^{-1} = \left(\text{the matrix } A^2 - 5A\right)^{-1} \ne (A^2)^{-1} - (5A)^{-1}, \qquad (A + B)^{-1} \ne A^{-1} + B^{-1}

Worked example

For

A = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}

, find

(A^2 + A)^{-1}

Practise this conceptself-check

The same idea in a real exam question:

MHT-CET · 2023 · 11th May Shift 2 · Q141Hard

Example 2 · Determinants and Matrices · Inverse of a Matrix — Adjoint Formula, Products and Verification

A = \begin{pmatrix}2 & 1\\7 & 4\end{pmatrix}

then

(A^2-5A)^{-1}

Distributing the inverse over a sum

There is no rule for

(A + B)^{-1}

(A^2 - 5A)^{-1}

except to form the matrix and invert it. The distractors are exactly what distributing produces.

Drill 3 more on invert an expression: compute a² − 5a or a + b first, then invert

Concept 3 of 5: Matrices with tan x Entries: |A| = sec²x and adj A = Aᵀ

A = \begin{pmatrix} 1 & \tan x \\ -\tan x & 1 \end{pmatrix}

has determinant

1 + \tan^2x = \sec^2x

, and its adjoint is exactly its transpose. So

A^{-1} = \cos^2x\,A^T

, and

A^TA^{-1}

turns out to be a rotation by

2x

Definition

For $A = \begin{pmatrix} 1 & t \\ -t & 1 \end{pmatrix}$ : $|A| = 1 + t^2$ , $\operatorname{adj}A = \begin{pmatrix} 1 & -t \\ t & 1 \end{pmatrix} = A^T$ , so $A^{-1} = \dfrac{A^T}{1 + t^2}$ .
With $t = \tan x$ : $A^TA^{-1} = \dfrac{(A^T)^2}{1 + \tan^2x} = \dfrac{1}{1 + \tan^2x}\begin{pmatrix} 1 - \tan^2x & -2\tan x \\ 2\tan x & 1 - \tan^2x \end{pmatrix} = \begin{pmatrix} \cos 2x & -\sin 2x \\ \sin 2x & \cos 2x \end{pmatrix}$ .
With $t = \cot\dfrac{\theta}{2}$ : $|A| = \csc^2\dfrac{\theta}{2}$ , so $A^{-1} = \sin^2\dfrac{\theta}{2}\,A^T = \dfrac{1 - \cos\theta}{2}A^T$ .
The identities used: $\dfrac{1 - \tan^2x}{1 + \tan^2x} = \cos 2x$ , $\dfrac{2\tan x}{1 + \tan^2x} = \sin 2x$ , $\sin^2\dfrac{\theta}{2} = \dfrac{1 - \cos\theta}{2}$ .

The tan-entry matrix

A = \begin{pmatrix} 1 & \tan x \\ -\tan x & 1 \end{pmatrix}:\ |A| = \sec^2x,\ \operatorname{adj}A = A^T,\ A^{-1} = \cos^2x\,A^T,\ A^TA^{-1} = \begin{pmatrix} \cos 2x & -\sin 2x \\ \sin 2x & \cos 2x \end{pmatrix}

Worked example

For

A = \begin{pmatrix} 1 & \tan x \\ -\tan x & 1 \end{pmatrix}

, find

A^{-1}

in terms of

A^T

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 20 April Shift I · Q148Moderate

Example 3 · Determinants and Matrices · Inverse of a Matrix — Adjoint Formula, Products and Verification

A =\begin{bmatrix} 1 & \tan x \\ - \tan x & 1 \end{bmatrix}

, then

A^{T}A^{- 1}=

Sign of the sin 2x entries

A^TA^{-1}

has

-\sin 2x

in the top-right and

+\sin 2x

in the bottom-left. The four sign arrangements are all offered; compute the

(1,2)

entry explicitly —

\dfrac{-\tan x - \tan x}{1 + \tan^2x} = -\sin 2x

Drill 2 more on matrices with tan x entries: |a| = sec²x and adj a = aᵀ

Concept 4 of 5: (AB)⁻¹ = B⁻¹A⁻¹: Inverting Products and Recovering a Factor

To undo 'first

B

, then

A

' you undo

A

first, then

B

— so the inverse of a product reverses the order. Given

(AB)^{-1}

and

A^{-1}

, multiplying by

A

on the right peels

A^{-1}

off and leaves

B^{-1}

Definition

$(AB)^{-1} = B^{-1}A^{-1}$ ; hence $B^{-1} = (AB)^{-1}A$ and $A^{-1} = B\,(AB)^{-1}$ .
$(A^{-1})^{-1} = A$ : to recover $A$ from $A^{-1} = \dfrac13\begin{pmatrix} 4 & 3 \\ -1 & 0 \end{pmatrix}$ , invert it — determinant $\dfrac13$ , so $A = \begin{pmatrix} 0 & -3 \\ 1 & 4 \end{pmatrix}$ .
$(A^T)^{-1} = (A^{-1})^T$ , $(kA)^{-1} = \dfrac1k A^{-1}$ , $(A^n)^{-1} = (A^{-1})^n$ .
For a non-square product like $A_{2\times3}B_{3\times2}$ : compute $AB$ (a $2 \times 2$ ) and invert it directly — the factors themselves have no inverse.

Inverse of a product

(AB)^{-1} = B^{-1}A^{-1} \qquad B^{-1} = (AB)^{-1}A \qquad (A^{-1})^{-1} = A \qquad (A^T)^{-1} = (A^{-1})^T

Worked example

(AB)^{-1} = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}

and

A = \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix}

, find

B^{-1}

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 23 April Shift I · Q110Moderate

Example 4 · Determinants and Matrices · Inverse of a Matrix — Adjoint Formula, Products and Verification

If A and B are non-singular matrices of order 2 such that

(AB)^{- 1}=\frac{1}{6}\begin{bmatrix} - 7 & - 3 \\ 2 & 3 \end{bmatrix}

and

A^{- 1}=\frac{1}{3}\begin{bmatrix} 4 & 3 \\ - 1 & 0 \end{bmatrix}

then

B^{- 1}=

Keeping the order

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

is false unless the matrices commute. Reversing is the whole rule; with it,

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

and NOT

A\,(AB)^{-1}

Drill 1 more on (ab)⁻¹ = b⁻¹a⁻¹: inverting products and recovering a factor

Concept 5 of 5: Unknown Entries and A⁻¹ = A³: Use AA⁻¹ = I

B

is claimed to be

A^{-1}

, then

AB = I

, and each entry of that product is an equation. Three well-chosen entries give the three unknowns. The same fact read backwards —

A^{-1} = A^3

means

A^4 = I

— settles 'which power is the inverse' questions.

Definition

Write out only the entries of $AB$ that contain the unknowns; each must equal the corresponding entry of $I$ ( $1$ on the diagonal, $0$ off it).
Pick entries where ONE unknown appears: $15b = 0$ gives $b$ at once; then use $b$ in the next.
$A^{-1} = A^k \iff A^{k+1} = I$ . Test $A^2$ , then $A^4 = (A^2)^2$ : if $A^4 = I$ then $A^{-1} = A^3$ .
A stem with a printed inconsistency (two entries forcing different values of $a$ ) has happened; the official key follows the entry the setter used. Prefer the entry involving the most unknowns' pairing that the answer options confirm.

The defining property of the inverse

AB = I \iff B = A^{-1} \qquad A^{-1} = A^{3} \iff A^{4} = I

Worked example

A = \begin{pmatrix} 2 & 1 \\ 3 & 2 \end{pmatrix}

and

A^{-1} = \begin{pmatrix} p & -1 \\ q & 2 \end{pmatrix}

, find

p

and

q

Practise this conceptself-check

The same idea in a real exam question:

MHT-CET · 2025 · 22 April Shift II · Q109Moderate

Example 5 · Determinants and Matrices · Inverse of a Matrix — Adjoint Formula, Products and Verification

If matrix

A =\frac{1}{11}\begin{bmatrix} - 1 & 7 & - 24 \\ 2 & a & 4 \\ 2 & - 3 & 15 \end{bmatrix}

and

A^{- 1}=\begin{bmatrix} 3 & 3 & 4 \\ 2 & - 3 & 4 \\ b & - 1 & c \end{bmatrix}

, then the values of

a,b,c

respectively are

\_\_\_\_

Solving all nine entries

Three unknowns need three equations; the other six entries are checks. Choosing the entries that isolate one unknown each keeps the arithmetic to a few lines — solving the whole product is where time goes.

Drill 2 more on unknown entries and a⁻¹ = a³: use aa⁻¹ = i

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (5)

Inverse of a 2 × 2 Matrix
2 × 2 inverse
$\begin{pmatrix} a & b \\ c & d \end{pmatrix}^{-1} = \frac{1}{ad - bc}\begin{pmatrix} d & -b \\ -c & a \end{pmatrix}, \quad ad - bc \ne 0$
Invert an Expression: Compute A² − 5A or A + B First, Then Invert
Order of operations
$(A^2 - 5A)^{-1} = \left(\text{the matrix } A^2 - 5A\right)^{-1} \ne (A^2)^{-1} - (5A)^{-1}, \qquad (A + B)^{-1} \ne A^{-1} + B^{-1}$
Matrices with tan x Entries: |A| = sec²x and adj A = Aᵀ
The tan-entry matrix
$A = \begin{pmatrix} 1 & \tan x \\ -\tan x & 1 \end{pmatrix}:\ |A| = \sec^2x,\ \operatorname{adj}A = A^T,\ A^{-1} = \cos^2x\,A^T,\ A^TA^{-1} = \begin{pmatrix} \cos 2x & -\sin 2x \\ \sin 2x & \cos 2x \end{pmatrix}$
(AB)⁻¹ = B⁻¹A⁻¹: Inverting Products and Recovering a Factor
Inverse of a product
$(AB)^{-1} = B^{-1}A^{-1} \qquad B^{-1} = (AB)^{-1}A \qquad (A^{-1})^{-1} = A \qquad (A^T)^{-1} = (A^{-1})^T$
Unknown Entries and A⁻¹ = A³: Use AA⁻¹ = I
The defining property of the inverse
$AB = I \iff B = A^{-1} \qquad A^{-1} = A^{3} \iff A^{4} = I$

Watch out for (5)

Swapping signs on the diagonal instead of the off-diagonal
→ Inverse of a 2 × 2 Matrix
Distributing the inverse over a sum
→ Invert an Expression: Compute A² − 5A or A + B First, Then Invert
Sign of the sin 2x entries
→ Matrices with tan x Entries: |A| = sec²x and adj A = Aᵀ
Keeping the order
→ (AB)⁻¹ = B⁻¹A⁻¹: Inverting Products and Recovering a Factor
Solving all nine entries
→ Unknown Entries and A⁻¹ = A³: Use AA⁻¹ = I

Test yourself on Determinants and Matrices

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