MathsISC Class 12 · Determinants

Adjoint and Inverse of a Matrix: Cofactors, adj A and A⁻¹

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 whenever |A| ≠ 0 the inverse is A⁻¹ = (adj A)/|A|. For a 3 × 3 matrix, |adj A| = |A|². Finding nine cofactors carefully is the heart of every inverse question.

Cofactors, then transpose gives adj A; dividing by |A| gives the inverse, which exists only when |A| is not zero.

Illustration drawn for QuestionBuddy.

Understand Adjoint and Inverse of a Matrix

Minors and cofactors. In a square matrix $A = [a_{ij}]$ , the minor $M_{ij}$ of the entry $a_{ij}$ is the determinant left after deleting row $i$ and column $j$ . The cofactor is the minor with a sign: $C_{ij} = (- 1)^{i + j} M_{ij}$ . The signs follow a chessboard pattern starting with $+$ in the top-left corner:
$+ - + - + - + - +$
A determinant equals the sum of the entries of any row (or column) times their own cofactors: $∣ A ∣ = a_{11} C_{11} + a_{12} C_{12} + a_{13} C_{13}$ . But the entries of one row times the cofactors of a different row always add up to 0. These two facts are what make the adjoint work.

The adjoint. $adj A$ is the transpose of the matrix of cofactors: the cofactors of row 1 go down column 1, and so on. For a $2 \times 2$ matrix this is quick: $adj [a c b d] = [d - c - b a]$ .

The key identity. Multiply $A$ by $adj A$ . The $(i, i)$ entry is row $i$ of $A$ times its own cofactors, which is $∣ A ∣$ . Every off-diagonal entry is a row times another row's cofactors, which is 0. So
$A (adj A) = (adj A) A = ∣ A ∣ I .$
If $∣ A ∣ \neq = 0$ , divide by $∣ A ∣$ : $A (\frac{1}{∣ A ∣} adj A) = I$ , so
$A^{- 1} = \frac{1}{∣ A ∣} adj A .$
This proves that every non-singular matrix is invertible. If $∣ A ∣ = 0$ , then $A (adj A) = O$ and no inverse exists.

Properties of the adjoint (for $n \times n$ matrices, $A$ and $B$ non-singular where needed):

$∣ adj A ∣ = ∣ A ∣^{n - 1}$ , so $∣ A ∣^{2}$ for a $3 \times 3$ matrix. (Take determinants of $A (adj A) = ∣ A ∣ I$ : $∣ A ∣∣ adj A ∣ = ∣ A ∣^{n}$ .)
$adj (A B) = (adj B) (adj A)$ , with the order reversed.
$adj (A^{'}) = (adj A)^{'}$ .
$adj (k A) = k^{n - 1} adj A$ .
$adj (adj A) = ∣ A ∣^{n - 2} A$ , so $∣ adj (adj A) ∣ = ∣ A ∣^{(n - 1)^{2}}$ .
The adjoint of a symmetric matrix is symmetric.

A safe routine for a 3 × 3 inverse. Find $∣ A ∣$ first (expanding along the row or column with most zeros). If it is 0, stop. Otherwise write all nine cofactors in a grid with their signs, transpose the grid to get $adj A$ , and check one diagonal entry of $A (adj A)$ , which must equal $∣ A ∣$ . The ISC 2024 examiners noted computational errors in finding the inverse; this one-line check catches most of them.

Adjoint and Inverse of a Matrix: formulas to remember

Cofactor: $C_{ij} = (- 1)^{i + j} M_{ij}$; Sign pattern + − + / − + − / + − +.
Expansion: $∣ A ∣ = a_{i 1} C_{i 1} + a_{i 2} C_{i 2} + a_{i 3} C_{i 3}$; Any row or column; entries times cofactors of another row give 0.
Adjoint: $adj A = [C_{ij}]^{'}$; Transpose of the cofactor matrix.
2 × 2 adjoint: $adj [a c b d] = [d - c - b a]$; Swap the diagonal, negate the rest.
Key identity: $A (adj A) = (adj A) A = ∣ A ∣ I$; Holds even when |A| = 0.
Inverse: $A^{- 1} = \frac{1}{∣ A ∣} adj A, ∣ A ∣ \neq = 0$; Keep 1/|A| outside the matrix in the answer.
Determinant of adjoint: $∣ adj A ∣ = ∣ A ∣^{n - 1}$; For 3 × 3: |adj A| = |A|².
Adjoint of a product: $adj (A B) = (adj B) (adj A)$; Reverse the order.

How to solve Adjoint and Inverse of a Matrix questions

Use this routine for "find A⁻¹" and "verify A(adj A) = |A|I" questions.

1Expand $∣ A ∣$ along the row or column with the most zeros. If $∣ A ∣ = 0$ , write "A is singular, so A⁻¹ does not exist" and stop.
2Write the sign grid $+ - + / - + - / + - +$ beside the matrix.
3Find each minor by covering its row and column, then attach the sign to get the cofactor. List all nine cofactors.
4Transpose: write the cofactors of row 1 as column 1 of $adj A$ , row 2 as column 2, row 3 as column 3.
5Check: multiply row 1 of $A$ by column 1 of $adj A$ . It must equal $∣ A ∣$ . Row 1 of $A$ by column 2 must give 0.
6Write $A^{- 1} = \frac{1}{∣ A ∣} adj A$ , keeping the fraction outside the matrix.
7For property questions, use $∣ adj A ∣ = ∣ A ∣^{n - 1}$ and $∣ A^{- 1} ∣ = \frac{1}{∣ A ∣}$ instead of computing matrices.

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.

Inverse of a 3 × 3 matrix to undo a transformation

Suppose a photo filter mixes the red, green and blue values of each pixel with $M = 101110011$ : new red = red + green, new green = green + blue, new blue = blue + red. A filtered pixel has values $(150, 130, 120)$ .

Use the adjoint to find $M^{- 1}$ and recover the original pixel.

Try it first, then show the solutionSolution

$∣ M ∣ = 1 (1 - 0) - 1 (0 - 1) + 0 = 2 \neq = 0$ .
Cofactors by rows: $(1, 1, - 1)$ , $(- 1, 1, 1)$ , $(1, - 1, 1)$ . Transposing, $adj M = 11 - 1 - 1 11 1 - 1 1$ .
$M^{- 1} = \frac{1}{2} 11 - 1 - 1 11 1 - 1 1$ .
Original $= \frac{1}{2} 150 - 130 + 120 150 + 130 - 120 - 150 + 130 + 120 = \frac{1}{2} 140160100 = 708050$ .
Check: $70 + 80 = 150$ , $80 + 50 = 130$ , $50 + 70 = 120$ .

Answer: The original pixel was $(70, 80, 50)$ .

Memory hook: A filter can be undone only if its matrix has an inverse: adjoint over determinant.

Inverse of a 2 × 2 matrix by the adjoint

Imagine a juice bar that makes two drinks. Each litre of drink A uses 0.3 L of orange concentrate and 0.1 L of mango concentrate; each litre of drink B uses 0.2 L of each. It has 9 L of orange and 5 L of mango concentrate to use up.

How many litres of each drink should it make? Use $X = \frac{1}{∣ A ∣} (adj A) B$ .

Try it first, then show the solutionSolution

$A = [0.3 0.1 0.2 0.2]$ , $B = [95]$ , $∣ A ∣ = 0.06 - 0.02 = 0.04$ .
$adj A = [0.2 - 0.1 - 0.2 0.3]$ .
$X = \frac{1}{0.04} [1.8 - 1.0 - 0.9 + 1.5] = 25 [0.8 0.6] = [2015]$ .
Check: $0.3 (20) + 0.2 (15) = 9$ and $0.1 (20) + 0.2 (15) = 5$ .

Answer: 20 L of drink A and 15 L of drink B.

Memory hook: Swap, negate, divide: the 2 × 2 adjoint turns a recipe table into quantities.

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.

Chessboard signs
Cofactor signs alternate like a chessboard, starting with + at the top left.
Use it when: Getting all nine cofactor signs right.
Rows become columns
adj A: the cofactors of row 1 are written down column 1.
Use it when: Remembering to transpose.
A adj A = |A| I
Multiplying a matrix by its adjoint gives the determinant on the diagonal and zeros elsewhere.
Use it when: Checking your adjoint and proving properties.
n minus one
|adj A| = |A|ⁿ⁻¹ and adj(kA) = kⁿ⁻¹ adj A: the adjoint always loses one power.
Use it when: One-mark property questions.

Adjoint and Inverse of a Matrix: solved examples

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

Example 1: Determinant of an adjoint · 1 mark · MCQBoard level

If $A$ is a $3 \times 3$ matrix with $∣ A ∣ = 4$ , then $∣ adj A ∣$ is:
(a) 4
(b) 12
(c) 16
(d) 64

Try it first, then show the solutionSolution

$∣ adj A ∣ = ∣ A ∣^{n - 1} = 4^{2} = 16$ .

Answer: (c) 16.

Example 2: Verifying A(adj A) = |A|I for a 2 × 2 matrix · 2 marks · Short answerBoard level

For $A = [24 - 3 5]$ , find $adj A$ and verify that $A (adj A) = ∣ A ∣ I$ .

Try it first, then show the solutionSolution

$adj A = [5 - 4 32]$ and $∣ A ∣ = 10 + 12 = 22$ . (1 mark)
$A (adj A) = [10 + 12 20 - 20 6 - 6 12 + 10] = [220022] = 22 I$ . (1 mark)

Answer: $adj A = [5 - 4 32]$ , and $A (adj A) = 22 I = ∣ A ∣ I$ .

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

Find $A^{- 1}$ for $A = 120231 - 1 12$ .

Try it first, then show the solutionSolution

$∣ A ∣ = 1 (6 - 1) - 2 (4 - 0) + (- 1) (2 - 0) = 5 - 8 - 2 = - 5 \neq = 0$ . (1 mark)
Row 1 cofactors: $C_{11} = 5$ , $C_{12} = - (4 - 0) = - 4$ , $C_{13} = 2$ . (1 mark)
Row 2: $C_{21} = - (4 + 1) = - 5$ , $C_{22} = 2 - 0 = 2$ , $C_{23} = - (1 - 0) = - 1$ . Row 3: $C_{31} = 2 + 3 = 5$ , $C_{32} = - (1 + 2) = - 3$ , $C_{33} = 3 - 4 = - 1$ . (1 mark)
$adj A = 5 - 4 2 - 5 2 - 1 5 - 3 - 1$ . Check: row 1 of $A$ times column 1 $= 5 - 8 - 2 = - 5 = ∣ A ∣$ . (1 mark)
$A^{- 1} = - \frac{1}{5} 5 - 4 2 - 5 2 - 1 5 - 3 - 1$ . (1 mark)

Answer: $A^{- 1} = - \frac{1}{5} 5 - 4 2 - 5 2 - 1 5 - 3 - 1$ .

Example 4: Determinants of adjoints and inverses · 2 marks · Short answerHarder board level

For the matrix $A$ of the previous example ( $∣ A ∣ = - 5$ ), find $∣ adj A ∣$ , $∣ adj (adj A) ∣$ and $∣ A^{- 1} ∣$ without computing any matrix.

Try it first, then show the solutionSolution

$∣ adj A ∣ = ∣ A ∣^{2} = 25$ . (½ mark)
$∣ adj (adj A) ∣ = ∣ adj A ∣^{2} = 625$ . (1 mark)
$∣ A^{- 1} ∣ = \frac{1}{∣ A ∣} = - \frac{1}{5}$ . (½ mark)

Answer: $25$ , $625$ and $- \frac{1}{5}$ .

Example 5: The adjoint of a product · 3 marks · Short answerHarder board level

For $A = [1324]$ and $B = [2103]$ , verify that $adj (A B) = (adj B) (adj A)$ .

Try it first, then show the solutionSolution

$A B = [410612]$ , so $adj (A B) = [12 - 10 - 6 4]$ . (1 mark)
$adj A = [4 - 3 - 2 1]$ , $adj B = [3 - 1 02]$ . (1 mark)
$(adj B) (adj A) = [12 + 0 - 4 - 6 - 6 + 0 2 + 2] = [12 - 10 - 6 4]$ . Verified. (1 mark)

Answer: Both sides equal $[12 - 10 - 6 4]$ .

Example 6: A matrix that is its own inverse · 5 marks · Long answerHOTS / competency

Let $A = 100 0 cos α sin α 0 sin α - cos α$ .
(i) Find $∣ A ∣$ . (1 mark)
(ii) Find $adj A$ . (2 marks)
(iii) Show that $A^{- 1} = A$ and deduce $A^{2}$ . (2 marks)

Try it first, then show the solutionSolution

(i) Expand along $R_{1}$ : $∣ A ∣ = 1 (- cos^{2} α - sin^{2} α) = - 1$ .
(ii) $C_{11} = - 1$ ; $C_{12} = C_{13} = C_{21} = C_{31} = 0$ ; $C_{22} = - cos α$ , $C_{23} = - sin α$ , $C_{32} = - sin α$ , $C_{33} = cos α$ .
The cofactor matrix is symmetric, so $adj A = - 1 00 0 - cos α - sin α 0 - sin α cos α = - A$ .
(iii) $A^{- 1} = \frac{1}{- 1} (- A) = A$ .
Then $A^{2} = AA = A A^{- 1} = I$ .

Answer: $∣ A ∣ = - 1$ , $adj A = - A$ , $A^{- 1} = A$ and $A^{2} = I$ for every $α$ .

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
adj A is the transpose of the cofactor matrix; for non-symmetric matrices the untransposed version gives a wrong inverse.
Do this: Write cofactors of row 1 down column 1, and check A(adj A) = |A|I on one entry.
Usually costs: 2 to 3 marks in a 5-mark question.
Mistake: Sign errors in cofactors
Forgetting the (−1)^{i+j} factor, especially for C₁₂, C₂₁, C₂₃ and C₃₂.
Do this: Draw the chessboard sign grid next to the matrix before you begin.
Usually costs: 1 mark per wrong cofactor, plus the final answer.
Mistake: Computational errors in minors
The ISC 2024 examiners reported computational errors in finding the inverse matrix.
Do this: Write each 2 × 2 minor as "ad − bc" with numbers before simplifying, and check row 1 of A times column 1 of adj A equals |A|.
Usually costs: The final mark and any later part that uses A⁻¹.
Mistake: Writing |adj A| = |A|
For an n × n matrix the adjoint carries |A|ⁿ⁻¹.
Wrong working
$∣ A ∣ = 4 \Rightarrow ∣ adj A ∣ = 4$ (3 × 3).
Right working
$∣ adj A ∣ = 4^{2} = 16$ .
Do this: Derive it from A(adj A) = |A|I whenever unsure.
Usually costs: 1 mark.

Exam tips

Find $∣ A ∣$ before the cofactors. If it is zero, you save several minutes, and if it is not, you already have $C_{11}, C_{12}, C_{13}$ from the expansion.
Present the nine cofactors in a list or grid before the adjoint; method marks are given for them even if one slips.
Keep $\frac{1}{∣ A ∣}$ outside the adjoint in the final answer.
For property questions, quote the formula ( $∣ adj A ∣ = ∣ A ∣^{n - 1}$ ) and substitute; there is no need to build matrices.
If the matrix is symmetric, its adjoint is symmetric: use this to check your cofactors.

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
$adj [p r q s]$ is:
(a) $[s r q p]$
(b) $[s - r - q p]$
(c) $[- s r q - p]$
(d) $[p - q - r s]$
Hint
Swap the diagonal, negate the rest.
Answer
(b).
Q2 · 2 marks · Short answerBoard level
For $A = 21 - 2 - 1 05 341$ , find the cofactors of the elements of the second row and hence $∣ A ∣$ .
Hint
Signs for row 2 are − + −.
Answer
$C_{21} = - (- 1 - 15) = 16$ , $C_{22} = 2 + 6 = 8$ , $C_{23} = - (10 - 2) = - 8$ . $∣ A ∣ = 1 (16) + 0 (8) + 4 (- 8) = - 16$ .
Q3 · 5 marks · Long answerHarder board level
Find the inverse of $210131012$ .
Hint
The matrix is symmetric, so its adjoint is too.
Answer
$∣ A ∣ = 2 (5) - 1 (2) = 8$ . $adj A = 5 - 2 1 - 2 4 - 2 1 - 2 5$ , so $A^{- 1} = \frac{1}{8} 5 - 2 1 - 2 4 - 2 1 - 2 5$ .
Q4 · 2 marks · Short answerHarder board level
Find $k$ for which $123241 k 11$ has no inverse.
Hint
Set $∣ A ∣ = 0$ .
Answer
$∣ A ∣ = 1 (4 - 1) - 2 (2 - 3) + k (2 - 12) = 5 - 10 k = 0$ , so $k = \frac{1}{2}$ .
Q5 · 1 mark · MCQHarder board level
If $A$ is a $3 \times 3$ matrix with $∣ A ∣ = 3$ , then $∣ adj (2 A) ∣$ equals:
(a) 36
(b) 72
(c) 144
(d) 576
Hint
$∣ adj B ∣ = ∣ B ∣^{2}$ with $B = 2 A$ .
Answer
(d) 576: $∣2 A ∣ = 8 \times 3 = 24$ and $2 4^{2} = 576$ .
Q6 · 2 marks · Short answerHOTS / competency
Prove that for a non-singular $3 \times 3$ matrix $A$ , $∣ adj A ∣ = ∣ A ∣^{2}$ .
Hint
Take determinants of $A (adj A) = ∣ A ∣ I$ .
Answer
$∣ A (adj A) ∣ = ∣∣ A ∣ I ∣$ , so $∣ A ∣∣ adj A ∣ = ∣ A ∣^{3} ∣ I ∣ = ∣ A ∣^{3}$ . Dividing by $∣ A ∣ \neq = 0$ gives $∣ adj A ∣ = ∣ A ∣^{2}$ .
Q7 · 4 marks · Case-basedHOTS / competency
Case study: Suppose a coding club sends a block of three numbers $X$ (letters as alphabet positions) as $Y = E X$ , with $E = 101210112$ .
(i) Find $∣ E ∣$ . (1 mark)
(ii) Find $adj E$ . (1 mark)
(iii) Write $E^{- 1}$ . (1 mark)
(iv) Decode $Y = (8, 5, 7)$ and write the letters. (1 mark)
Hint
Decoding is $X = E^{- 1} Y$ .
Answer
(i) $∣ E ∣ = 1 (2) - 2 (- 1) + 1 (- 1) = 3$ . (ii) $adj E = 21 - 1 - 4 12 1 - 1 1$ . (iii) $E^{- 1} = \frac{1}{3} adj E$ . (iv) $X = \frac{1}{3} 16 - 20 + 7 8 + 5 - 7 - 8 + 10 + 7 = 123$ : the letters A, B, C.

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 the cofactor $C_{ij}$ .Show answer
$C_{ij} = (- 1)^{i + j} M_{ij}$ , where $M_{ij}$ is the minor of $a_{ij}$ .
2.What is $adj A$ ?Show answer
The transpose of the matrix of cofactors of $A$ .
3.State the identity linking $A$ , $adj A$ and $∣ A ∣$ .Show answer
$A (adj A) = (adj A) A = ∣ A ∣ I$ .
4.If $∣ A ∣ = - 2$ for a $3 \times 3$ matrix, find $∣ adj A ∣$ .Show answer
4.
5.Write $A^{- 1}$ in terms of the adjoint.Show answer
$A^{- 1} = \frac{1}{∣ A ∣} adj A$ , for $∣ A ∣ \neq = 0$ .
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 a minor and a cofactor?

The minor of an entry is the determinant left after deleting its row and column. The cofactor is the minor multiplied by (−1)^(i+j), so it has the chessboard sign attached.

Why is the adjoint the transpose of the cofactor matrix?

Transposing makes the product A(adj A) multiply each row of A by its own cofactors on the diagonal, giving |A|, and by other rows' cofactors off the diagonal, giving 0. Without the transpose this pattern does not appear.

Can a singular matrix have an adjoint?

Yes. Every square matrix has an adjoint. For a singular matrix A(adj A) = O, so the adjoint exists but there is no inverse.

How can I check my inverse quickly in the exam?

Multiply row 1 of A by column 1 of adj A: you should get |A|. Then multiply row 1 of A by column 2 of adj A: you should get 0. These two checks catch most errors in a few seconds.

Which adjoint properties are in the ISC 2027 syllabus?

The CISCE 2027 syllabus lists the adjoint and inverse of a square matrix, properties of the adjoint, the formula A⁻¹ = (adj A)/|A| for 2 × 2 and 3 × 3 matrices, and properties of the inverse.