MathsCBSE Class 12 · Determinants

Adjoint and Inverse of a Matrix by the Cofactor Method (Class 12)

Updated 15 October 2026 · Fact-checked

The adjoint of a square matrix A is the transpose of its matrix of cofactors. It satisfies A(adj A) = (adj A)A = |A|I, so when |A| ≠ 0 the inverse is A⁻¹ = (1/|A|) adj A. For a 3 × 3 matrix, |adj A| = |A|². A matrix with |A| = 0 is singular and has no inverse.

Each row of cofactors becomes a column of adj A (same colour), then divide by |A| to get the inverse unless |A| = 0.

Illustration drawn for QuestionBuddy.

Understand Adjoint and Inverse of a Matrix

Building the adjoint. For a square matrix $A$ , replace every entry by its cofactor $A_{ij} = (- 1)^{i + j} M_{ij}$ to get the cofactor matrix, then transpose it. The result is the adjoint (or adjugate), written $adj A$ :
$adj A = [A_{ij}]^{'}$ .
So the cofactor of row 1 entries goes down column 1 of $adj A$ . Forgetting the transpose is the commonest mistake.

**The $2 \times 2$ shortcut.** For $A = [a c b d]$ , $adj A = [d - c - b a]$ : swap the diagonal entries and change the signs of the other two.

The key identity. For every square matrix,
$A (adj A) = (adj A) A = ∣ A ∣ I$ .
Why: multiplying row $i$ of $A$ by column $i$ of $adj A$ is expanding $∣ A ∣$ along row $i$ ; multiplying row $i$ by a different column gives the expansion of a determinant with two equal rows, which is 0.

Singular and non-singular. $A$ is singular if $∣ A ∣ = 0$ and non-singular if $∣ A ∣ \neq = 0$ . A square matrix is invertible exactly when it is non-singular, and then dividing the identity by $∣ A ∣$ gives
$A^{- 1} = \frac{1}{∣ A ∣} adj A$ .
If $∣ A ∣ = 0$ , there is no inverse; say so and stop.

Results for the adjoint (order $n$ ):

$∣ adj A ∣ = ∣ A ∣^{n - 1}$ ; for $3 \times 3$ , $∣ adj A ∣ = ∣ A ∣^{2}$ . (Take determinants of $A (adj A) = ∣ A ∣ I$ : $∣ A ∣∣ adj A ∣ = ∣ A ∣^{n}$ .)
$∣ A (adj A) ∣ = ∣ A ∣^{n}$ .
$adj (A B) = (adj B) (adj A)$ and $(A B)^{- 1} = B^{- 1} A^{- 1}$ .
$adj (A^{'}) = (adj A)^{'}$ .

A 3 × 3 inverse in five steps: find $∣ A ∣$ (stop if 0); find all nine cofactors; arrange them and transpose to get $adj A$ ; divide by $∣ A ∣$ ; check one or two entries of $A A^{- 1} = I$ . Write the cofactors in a list such as $A_{11} = - 2, A_{12} = - 5, \dots$ before arranging, so the examiner can award marks even if one sign slips.

This is NCERT Section 4.5 (Adjoint and Inverse of a Matrix), and the adjoint method is the route to solving equations by the matrix method in the next topic.

Adjoint and Inverse of a Matrix: formulas to remember

Adjoint: $adj A = [A_{ij}]^{'}, A_{ij} = (- 1)^{i + j} M_{ij}$; Transpose of the cofactor matrix.
2 × 2 adjoint: $adj [a c b d] = [d - c - b a]$; Swap the diagonal, negate the rest.
Fundamental identity: $A (adj A) = (adj A) A = ∣ A ∣ I$; Holds for every square matrix.
Inverse: $A^{- 1} = \frac{1}{∣ A ∣} adj A, ∣ A ∣ \neq = 0$; No inverse if $∣ A ∣ = 0$ .
Determinant of the adjoint: $∣ adj A ∣ = ∣ A ∣^{n - 1}$; For order 3: $∣ A ∣^{2}$ ; for order 2: $∣ A ∣$ .
Adjoint of a product: $adj (A B) = (adj B) (adj A)$; Reverse order, like inverses and transposes.

How to solve Adjoint and Inverse of a Matrix questions

Use this routine for 'find adj A', 'find A⁻¹' and adjoint MCQs.

1Compute $∣ A ∣$ . If $∣ A ∣ = 0$ , write 'A is singular, so $A^{- 1}$ does not exist'.
2List all cofactors: for each entry, delete its row and column, find the $2 \times 2$ minor, and attach the sign $(- 1)^{i + j}$ .
3Write the cofactor matrix with $A_{ij}$ in position $(i, j)$ .
4Transpose it to get $adj A$ (rows of cofactors become columns).
5Write $A^{- 1} = \frac{1}{∣ A ∣} adj A$ , keeping the fraction outside the matrix.
6Check: multiply row 1 of A by column 1 of $adj A$ ; you should get $∣ A ∣$ . Multiply row 1 by column 2; you should get 0.
7For MCQs, use $∣ adj A ∣ = ∣ A ∣^{n - 1}$ and $∣ A (adj A) ∣ = ∣ A ∣^{n}$ with the correct order n.

Adjoint and Inverse of a Matrix in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Using the inverse to recover quantities

Suppose a juice stall makes two blends. Each batch of blend P uses 3 L of orange pulp and 1 L of mango pulp; each batch of blend Q uses 2 L of orange and 4 L of mango. In one week the stall used 26 L of orange and 22 L of mango.

Write the information as $M X = C$ , find $M^{- 1}$ by the adjoint method, and find the number of batches of each blend.

Try it first, then show the solutionSolution

$M = [3124]$ , $X = [p q]$ , $C = [2622]$ .
$∣ M ∣ = 12 - 2 = 10 \neq = 0$ ; $adj M = [4 - 1 - 2 3]$ , so $M^{- 1} = \frac{1}{10} [4 - 1 - 2 3]$ .
$X = M^{- 1} C = \frac{1}{10} [104 - 44 - 26 + 66] = [64]$ .
Check: orange $3 (6) + 2 (4) = 26$ , mango $6 + 4 (4) = 22$ .

Answer: 6 batches of P and 4 batches of Q.

Memory hook: Swap, negate, divide by the determinant: the 2 × 2 inverse in one breath.

Singular matrices lose information

Imagine a friend designs a code that multiplies pairs of numbers by the key $K = [2142]$ .

Find $∣ K ∣$ . Show that the messages $(3, 0)$ and $(1, 1)$ get the same code, and explain why the code cannot be decoded.

Try it first, then show the solutionSolution

$∣ K ∣ = 4 - 4 = 0$ , so K is singular.
$K [30] = [63]$ and $K [11] = [2 + 4 1 + 2] = [63]$ .
Two different messages give the same code, so the receiver cannot tell which was sent.
Decoding would need $K^{- 1} = \frac{1}{∣ K ∣} adj K$ , which does not exist because $∣ K ∣ = 0$ .

Answer: $∣ K ∣ = 0$ ; both messages encode to $(6, 3)$ , so decoding is impossible.

Memory hook: No determinant, no inverse, no decoding.

Tricks to remember Adjoint and Inverse of a Matrix

Short memory aids. Say each one aloud once, then use it on a question the same day.

Cofactors, then flip
Adjoint = cofactor matrix flipped over the diagonal. Say 'cofactor, transpose' every time.
Use it when: Avoiding the most common slip in 3 × 3 adjoints.
2 × 2: swap and negate
Swap $a$ and $d$ ; put minus signs on $b$ and $c$ .
Use it when: Quick 2 × 2 inverses.
The adjoint power is one less
$∣ adj A ∣ = ∣ A ∣^{n - 1}$ : for a 3 × 3 matrix, square $∣ A ∣$ .
Use it when: 1-mark adjoint MCQs.
Row times own column gives |A|
Row $i$ of $A$ times column $i$ of $adj A$ is $∣ A ∣$ ; times any other column is 0.
Use it when: Checking an adjoint in seconds.

Adjoint and Inverse of a Matrix: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: 2 × 2 adjoint and inverse · 2 marks · Very short answerBoard level

Find the adjoint and the inverse of $A = [32 - 4 5]$ .

Try it first, then show the solutionSolution

$∣ A ∣ = 15 - (- 8) = 23 \neq = 0$ . (½ mark)
$adj A = [5 - 2 43]$ (swap 3 and 5, negate $- 4$ and 2). (1 mark)
$A^{- 1} = \frac{1}{23} [5 - 2 43]$ . (½ mark)

Answer: $adj A = [5 - 2 43]$ , $A^{- 1} = \frac{1}{23} [5 - 2 43]$

Example 2: Adjoint results for a 3 × 3 matrix · 2 marks · Very short answerBoard level

A is a square matrix of order 3 with $∣ A ∣ = 5$ . Find (i) $∣ adj A ∣$ , (ii) $∣ A (adj A) ∣$ , (iii) $A (adj A)$ .

Try it first, then show the solutionSolution

(i) $∣ adj A ∣ = ∣ A ∣^{3 - 1} = 25$ . (½ mark)
(ii) $∣ A (adj A) ∣ = ∣ A ∣ \cdot ∣ adj A ∣ = 5 \times 25 = 125$ (also $= ∣5 I ∣ = 5^{3}$ ). (1 mark)
(iii) $A (adj A) = ∣ A ∣ I = 5 I = 500050005$ . (½ mark)

Answer: (i) 25 (ii) 125 (iii) $5 I$

Example 3: Inverse of a 3 × 3 matrix · 5 marks · Long answerHarder board level

Find $A^{- 1}$ for $A = 13 - 1 201 - 1 21$ .

Try it first, then show the solutionSolution

$∣ A ∣ = 1 (0 - 2) - 2 (3 + 2) + (- 1) (3 - 0) = - 2 - 10 - 3 = - 15 \neq = 0$ , so $A^{- 1}$ exists. (1 mark)
Row 1 cofactors: $A_{11} = (0 - 2) = - 2$ , $A_{12} = - (3 + 2) = - 5$ , $A_{13} = (3 - 0) = 3$ . (1 mark)
Row 2: $A_{21} = - (2 + 1) = - 3$ , $A_{22} = (1 - 1) = 0$ , $A_{23} = - (1 + 2) = - 3$ . Row 3: $A_{31} = (4 - 0) = 4$ , $A_{32} = - (2 + 3) = - 5$ , $A_{33} = (0 - 6) = - 6$ . (1 mark)
$adj A = - 2 - 5 3 - 3 0 - 3 4 - 5 - 6$ (cofactors of row $i$ placed in column $i$ ). (1 mark)
$A^{- 1} = - \frac{1}{15} - 2 - 5 3 - 3 0 - 3 4 - 5 - 6 = \frac{1}{15} 25 - 3 303 - 4 56$ . Check: row 1 of A times column 1 of adj A: $- 2 - 10 - 3 = - 15 = ∣ A ∣$ . (1 mark)

Answer: $A^{- 1} = \frac{1}{15} 25 - 3 303 - 4 56$

Example 4: When the inverse is a multiple of A · 2 marks · Very short answerHarder board level

For $A = [21 3 - 2]$ , show that $A^{- 1} = k A$ and find $k$ .

Try it first, then show the solutionSolution

$∣ A ∣ = - 4 - 3 = - 7$ ; $adj A = [- 2 - 1 - 3 2] = - A$ . (1 mark)
$A^{- 1} = \frac{1}{- 7} (- A) = \frac{1}{7} A$ , so $k = \frac{1}{7}$ . (1 mark)

Answer: $k = \frac{1}{7}$

Example 5: Working back from the adjoint · 3 marks · Short answerHOTS / competency

A is a $3 \times 3$ matrix with $∣ adj A ∣ = 64$ . Find (i) the possible values of $∣ A ∣$ , (ii) $∣ A^{- 1} ∣$ , (iii) $∣ adj (adj A) ∣$ .

Try it first, then show the solutionSolution

(i) $∣ adj A ∣ = ∣ A ∣^{2} = 64$ , so $∣ A ∣ = 8$ or $∣ A ∣ = - 8$ . (1 mark)
(ii) $∣ A^{- 1} ∣ = \frac{1}{∣ A ∣} = \pm \frac{1}{8}$ (same sign as $∣ A ∣$ ). (1 mark)
(iii) Apply the rule to $B = adj A$ : $∣ adj B ∣ = ∣ B ∣^{2} = 6 4^{2} = 4096$ . (1 mark)

Answer: (i) $\pm 8$ (ii) $\pm \frac{1}{8}$ (iii) 4096

Example 6: Case study: a three-letter code · 4 marks · Case-basedHOTS / competency

Suppose a club encodes three-letter words. Letters become numbers (A = 1, ..., Z = 26) and each word, written as a column, is multiplied by $K = 110011101$ .
(i) Find $∣ K ∣$ . (1 mark)
(ii) Find $adj K$ and $K^{- 1}$ . (2 marks)
(iii) Decode the received column $(23, 4, 21)$ . (1 mark)

Try it first, then show the solutionSolution

(i) $∣ K ∣ = 1 (1 - 0) - 0 + 1 (1 - 0) = 2$ . (1 mark)
(ii) Cofactors: row 1: $1, - 1, 1$ ; row 2: $1, 1, - 1$ ; row 3: $- 1, 1, 1$ . Transposing, $adj K = 1 - 1 1 11 - 1 - 1 11$ and $K^{- 1} = \frac{1}{2} adj K$ . (2 marks)
(iii) $K^{- 1} 23421 = \frac{1}{2} 23 + 4 - 21 - 23 + 4 + 21 23 - 4 + 21 = \frac{1}{2} 6240 = 3120$ , which spells C, A, T. (1 mark)

Answer: (i) 2 (ii) $adj K = 1 - 1 1 11 - 1 - 1 11$ , $K^{- 1} = \frac{1}{2} adj K$ (iii) 'CAT'

Where students lose marks in Adjoint and Inverse of a Matrix

The slips examiners see most often, and what to write instead.

Mistake: Not transposing the cofactor matrix
The adjoint is the transpose of the cofactor matrix.
Wrong working
$adj A = A_{11} A_{21} A_{31} A_{12} A_{22} A_{32} A_{13} A_{23} A_{33}$ .
Right working
$adj A = A_{11} A_{12} A_{13} A_{21} A_{22} A_{23} A_{31} A_{32} A_{33}$ .
Do this: Place the cofactors of row $i$ down column $i$ .
Usually costs: The adjoint and inverse marks (2 to 3).
Mistake: Sign errors in cofactors
Students forget $(- 1)^{i + j}$ for entries like $a_{12}$ and $a_{21}$ .
Wrong working
$A_{12} = 3 - 1 21 = 5$ .
Right working
$A_{12} = - 3 - 1 21 = - 5$ .
Do this: Write the chessboard sign before each minor.
Usually costs: Half a mark per wrong cofactor, plus the final answer.
Mistake: Using |adj A| = |A|ⁿ
The exponent is one less than the order.
Wrong working
Order 3, $∣ A ∣ = 5$ : $∣ adj A ∣ = 125$ .
Right working
$∣ adj A ∣ = 5^{2} = 25$ ; it is $∣ A (adj A) ∣$ that equals 125.
Do this: Remember $∣ A ∣∣ adj A ∣ = ∣ A ∣^{n}$ .
Usually costs: The MCQ.
Mistake: Finding an inverse without checking |A| first
If $∣ A ∣ = 0$ the inverse does not exist and all the cofactor work is wasted.
Wrong working
Cofactors computed, then $A^{- 1} = \frac{1}{0} adj A$ .
Right working
$∣ A ∣ = 0$ , so A is singular and $A^{- 1}$ does not exist.
Do this: Always compute $∣ A ∣$ first and state that it is non-zero.
Usually costs: Time, and the existence mark.

Exam tips

Adjoint MCQs ( $∣ adj A ∣$ , $A (adj A)$ ) appeared in the 2023, 2024, 2025 and 2026 board papers; learn $∣ adj A ∣ = ∣ A ∣^{n - 1}$ well.
In a 5-mark inverse, list all nine cofactors with signs before writing adj A; each correct cofactor row earns credit.
Leave $A^{- 1}$ as $\frac{1}{∣ A ∣}$ times a matrix; do not divide every entry unless asked.
Check your adjoint with one row-column product; it takes 20 seconds and catches most sign slips.
Elementary row operations are not in the 2026-27 syllabus; use the adjoint method for inverses.

Practice questions on Adjoint and Inverse of a Matrix

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
If A is a $3 \times 3$ matrix with $∣ A ∣ = - 3$ , then $∣ adj A ∣$ is:
(a) $- 3$
(b) 9
(c) $- 27$
(d) 27
Hint
$∣ adj A ∣ = ∣ A ∣^{n - 1}$ .
Answer
(b) $(- 3)^{2} = 9$ .
Q2 · 1 mark · MCQBoard level
If $A (adj A) = [5005]$ , then $∣ A ∣$ is:
(a) 25
(b) 5
(c) 10
(d) 0
Hint
$A (adj A) = ∣ A ∣ I$ .
Answer
(b) 5.
Q3 · 2 marks · Very short answerBoard level
Find the inverse of $[23 - 1 4]$ .
Hint
Swap, negate, divide by the determinant.
Answer
$∣ A ∣ = 8 + 3 = 11$ ; $A^{- 1} = \frac{1}{11} [4 - 3 12]$ .
Q4 · 3 marks · Short answerHarder board level
Let $A = 101 x 12 235$ . (i) For which value of $x$ is A singular? (ii) For $x = 2$ , find $∣ adj A ∣$ .
Hint
Expand along row 1 and keep $x$ as a letter.
Answer
(i) $∣ A ∣ = 1 (5 - 6) - x (0 - 3) + 2 (0 - 1) = 3 x - 3$ , which is 0 when $x = 1$ . (ii) For $x = 2$ , $∣ A ∣ = 3$ , so $∣ adj A ∣ = 3^{2} = 9$ .
Q5 · 1 mark · Assertion-reasonHarder board level
Assertion (A): If $∣ A ∣ = 0$ , then $A^{- 1}$ does not exist.
Reason (R): If $A^{- 1}$ exists, then $∣ A ∣ ∣ A^{- 1} ∣ = ∣ I ∣ = 1$ .
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
What would R give if |A| were 0?
Answer
(a). If $∣ A ∣ = 0$ , the product $∣ A ∣ ∣ A^{- 1} ∣$ would be 0, not 1, so no inverse can exist; R is exactly the reason for A.
Q6 · 5 marks · Long answerHarder board level
Find the inverse of $A = 201110 - 1 32$ .
Hint
$∣ A ∣ = 8$ . List all nine cofactors before arranging them.
Answer
$∣ A ∣ = 2 (2 - 0) - 1 (0 - 3) + (- 1) (0 - 1) = 4 + 3 + 1 = 8$ . Cofactors: row 1: $2, 3, - 1$ ; row 2: $- 2, 5, 1$ ; row 3: $4, - 6, 2$ . $adj A = 23 - 1 - 2 51 4 - 6 2$ , so $A^{- 1} = \frac{1}{8} 23 - 1 - 2 51 4 - 6 2$ .
Q7 · 3 marks · Short answerHOTS / competency
For $A = [cos α - sin α sin α cos α]$ , find $adj A$ and $A^{- 1}$ , and show that $A^{- 1} = A^{'}$ .
Hint
$∣ A ∣ = cos^{2} α + sin^{2} α$ .
Answer
$∣ A ∣ = 1$ . $adj A = [cos α sin α - sin α cos α]$ , so $A^{- 1} = adj A$ , which is exactly $A^{'}$ .

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Define adj A.Show answer
The transpose of the matrix of cofactors of A.
2.What is $A (adj A)$ ?Show answer
$∣ A ∣ I$ .
3.Write $adj [a c b d]$ .Show answer
$[d - c - b a]$ .
4.When does $A^{- 1}$ exist?Show answer
Exactly when $∣ A ∣ \neq = 0$ .
5.For a $3 \times 3$ matrix with $∣ A ∣ = 2$ , what is $∣ adj A ∣$ ?Show answer
4.
6.What is $adj (A B)$ ?Show answer
$(adj B) (adj A)$ .

Adjoint and Inverse of a Matrix: frequently asked questions

What is the difference between the cofactor matrix and the adjoint?

The cofactor matrix has the cofactor of each entry in that entry's position. The adjoint is its transpose, so the cofactors of the first row become the first column of adj A.

Is there a shortcut for the inverse of a 2 × 2 matrix?

Yes. Swap the two diagonal entries, change the signs of the other two entries, and divide by the determinant ad − bc.

Why is |adj A| = |A|² for a 3 × 3 matrix?

Take determinants of A(adj A) = |A|I. The left side gives |A| × |adj A| and the right side gives |A|³. Dividing by |A| (when it is non-zero) gives |adj A| = |A|².

Which NCERT section covers the adjoint and inverse?

Section 4.5 (Adjoint and Inverse of a Matrix) of Chapter 4, Determinants, in NCERT Mathematics Part I for Class 12.

What is a singular matrix?

A square matrix whose determinant is zero. It has no inverse.