MathsCBSE Class 12 · Determinants

Evaluating Determinants of 2 × 2 and 3 × 3 Matrices (Class 12)

Updated 15 October 2026 · Fact-checked

For a 2 × 2 matrix, |A| = ad − bc. For a 3 × 3 matrix, expand along any row or column: multiply each entry by its cofactor (its minor with sign (−1)^(i+j)) and add. Pick the row or column with the most zeros. Useful results: |kA| = kⁿ|A|, |A′| = |A| and |AB| = |A||B|.

Expand along the row with a zero: that term vanishes, and the chessboard signs (− + − for row 2) tell you how to sign the rest.

Illustration drawn for QuestionBuddy.

Understand Evaluating Determinants

Every square matrix $A$ has a number attached to it, its determinant, written $∣ A ∣$ or $det A$ . It tells you whether $A$ has an inverse ( $∣ A ∣ \neq = 0$ ) and, for systems of equations, whether there is a unique solution. The 2026-27 syllabus covers determinants up to order 3.

Order 1 and 2. For $A = [a]$ , $∣ A ∣ = a$ . For a $2 \times 2$ matrix,
$a c b d = a d - b c$ :
product of the leading diagonal minus product of the other diagonal.

Order 3: expansion by a row or column. The minor $M_{ij}$ of an entry $a_{ij}$ is the $2 \times 2$ determinant left after deleting row $i$ and column $j$ . The cofactor is $A_{ij} = (- 1)^{i + j} M_{ij}$ , so the signs follow the chessboard pattern
$+ - + - + - + - +$ .
Then, along row 1,
$∣ A ∣ = a_{11} A_{11} + a_{12} A_{12} + a_{13} A_{13} = a_{11} M_{11} - a_{12} M_{12} + a_{13} M_{13}$ .
You may expand along any row or column and get the same value, so choose the one with the most zeros; each zero wipes out a whole $2 \times 2$ calculation. For example, along row 2 of $201 - 1 45 3 - 2 0$ only two terms survive.

Results used in board questions. For square matrices of order $n$ :

$∣ k A ∣ = k^{n} ∣ A ∣$ , because each of the $n$ rows is multiplied by $k$ . For a $3 \times 3$ matrix, $∣2 A ∣ = 8∣ A ∣$ , not $2∣ A ∣$ .
$∣ A^{'} ∣ = ∣ A ∣$ : transposing does not change the determinant.
$∣ A B ∣ = ∣ A ∣∣ B ∣$ , so $∣ A B ∣ = ∣ B A ∣$ even when $A B \neq = B A$ .
$∣ A^{- 1} ∣ = \frac{1}{∣ A ∣}$ (from $A A^{- 1} = I$ and $∣ I ∣ = 1$ ), and $∣ A^{m} ∣ = ∣ A ∣^{m}$ .
$∣ - A ∣ = (- 1)^{n} ∣ A ∣$ , so for odd $n$ it is $- ∣ A ∣$ .

Combining the last results: a skew-symmetric matrix of odd order has $∣ A ∣ = ∣ A^{'} ∣ = ∣ - A ∣ = - ∣ A ∣$ , so $∣ A ∣ = 0$ .

Determinant equations. Questions like 'find $x$ if $2 x 8 5 x = 6853$ ' reduce to an ordinary equation after expansion. For $3 \times 3$ determinants with $x$ in several places, expand fully, collect the polynomial, and factor it, often by spotting an easy root such as $x = 1$ or $x = 2$ .

The NCERT section on properties of determinants (row operations used to simplify) was dropped in the rationalisation, so evaluate by direct expansion. This topic is NCERT Section 4.2 (with minors and cofactors in 4.4), and 1-mark determinant MCQs appear in every recent board paper.

Evaluating Determinants: formulas to remember

2 × 2 determinant: $a c b d = a d - b c$; Leading diagonal minus the other diagonal.
Cofactor: $A_{ij} = (- 1)^{i + j} M_{ij}$; Signs follow $+ - +$ / $- + -$ / $+ - +$ .
Expansion along row 1: $∣ A ∣ = a_{11} M_{11} - a_{12} M_{12} + a_{13} M_{13}$; Any row or column gives the same value.
Scalar multiple: $∣ k A ∣ = k^{n} ∣ A ∣$; n is the order of A.
Transpose and product: $∣ A^{'} ∣ = ∣ A ∣, ∣ A B ∣ = ∣ A ∣∣ B ∣$; Hence $∣ A B ∣ = ∣ B A ∣$ .
Inverse and powers: $∣ A^{- 1} ∣ = \frac{1}{∣ A ∣}, ∣ A^{m} ∣ = ∣ A ∣^{m}$; $∣ A^{- 1} ∣$ exists only if $∣ A ∣ \neq = 0$ .

How to solve Evaluating Determinants questions

Use these steps for evaluating a determinant, using the scalar and product results, and solving determinant equations.

1For $2 \times 2$ , write $a d - b c$ with brackets around negative entries.
2For $3 \times 3$ , choose the row or column with the most zeros, and write its sign pattern ( $+ - +$ or $- + -$ ).
3For each non-zero entry, delete its row and column, write the $2 \times 2$ minor, and attach the sign.
4Evaluate each minor, multiply by its entry and add; double-check the sign of every term.
5For results questions, identify the order $n$ first, then apply $∣ k A ∣ = k^{n} ∣ A ∣$ , $∣ A^{'} ∣ = ∣ A ∣$ , $∣ A B ∣ = ∣ A ∣∣ B ∣$ , $∣ A^{- 1} ∣ = \frac{1}{∣ A ∣}$ one at a time.
6For an equation in $x$ , expand both sides, simplify to a polynomial equation, and solve; test small integers to factor a cubic.
7Check the answer by substituting back into the determinant or by expanding along a different row.

Evaluating Determinants 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.

Determinant gives the area of a triangle

Suppose a farmer's triangular field has corners at $(1, 2)$ , $(7, 4)$ and $(3, 8)$ on a map where one unit is 10 m.

Using $Area = \frac{1}{2} x_{1} x_{2} x_{3} y_{1} y_{2} y_{3} 111$ , find the area of the field in square metres.

Try it first, then show the solutionSolution

$Δ = 173248111 = 1 (4 - 8) - 2 (7 - 3) + 1 (56 - 12)$ .
$= - 4 - 8 + 44 = 32$ .
Area $= \frac{1}{2} ∣32∣ = 16$ square units.
One square unit is $10 \times 10 = 100$ m², so the area is $16 \times 100 = 1600$ m².

Answer: 1600 m²

Memory hook: Three corners in a determinant, halve it, drop the sign: that is the field's area.

A zero determinant means no unique answer

Imagine two friends describe their stationery bills. One bought 2 pens and 3 notepads for Rs 70; the other bought 4 pens and 6 notepads for Rs 140. You try to work out the price of one pen and one notepad.

Write the coefficient matrix and evaluate its determinant. Can the prices be found uniquely?

Try it first, then show the solutionSolution

Equations: $2 p + 3 n = 70$ and $4 p + 6 n = 140$ , so the coefficient matrix is $[2436]$ .
Its determinant is $2 (6) - 3 (4) = 12 - 12 = 0$ .
A zero determinant means the coefficient matrix has no inverse, so there is no unique solution.
Indeed, the second bill is just double the first; for example $p = 20, n = 10$ and $p = 5, n = 20$ both fit.

Answer: The determinant is 0, so the prices cannot be found uniquely; the second bill gives no new information.

Memory hook: Determinant zero: the information repeats itself.

Tricks to remember Evaluating Determinants

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 + in the top-left corner.
Use it when: Expanding along any row or column.
Hunt the zeros
Expand along the row or column with the most zeros: every zero is one less $2 \times 2$ to compute.
Use it when: Saving time and avoiding slips.
k to the power of the order
Multiplying a $3 \times 3$ matrix by $k$ multiplies each of its 3 rows by $k$ , so the determinant gets $k^{3}$ .
Use it when: MCQs on $∣ k A ∣$ , which appear almost every year.
Product in, product out
$∣ A B ∣ = ∣ A ∣∣ B ∣$ : determinants turn matrix products into number products, so order no longer matters.
Use it when: Questions on $∣ A B ∣$ , $∣ A^{2} ∣$ , $∣ A^{- 1} ∣$ .

Evaluating Determinants: solved examples

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

Example 1: Expanding along the row with a zero · 2 marks · Very short answerBoard level

Evaluate $201 - 1 45 3 - 2 0$ .

Try it first, then show the solutionSolution

Expand along row 2 (signs $- + -$ ); the first entry is 0. (½ mark)
$+ 42130 = 4 (0 - 3) = - 12$ . (½ mark)
$- (- 2) 21 - 1 5 = 2 (10 + 1) = 22$ . (½ mark)
Determinant $= - 12 + 22 = 10$ . (Expanding along row 1 gives $2 (0 + 10) + 1 (0 + 2) + 3 (0 - 4) = 10$ too.) (½ mark)

Answer: 10

Example 2: A 2 × 2 determinant equation · 2 marks · Very short answerBoard level

Find $x$ if $2 x 8 5 x = 6853$ .

Try it first, then show the solutionSolution

Left side: $2 x \cdot x - 5 \cdot 8 = 2 x^{2} - 40$ . Right side: $18 - 40 = - 22$ . (1 mark)
$2 x^{2} - 40 = - 22 \Rightarrow 2 x^{2} = 18 \Rightarrow x^{2} = 9 \Rightarrow x = \pm 3$ . (1 mark)

Answer: $x = 3$ or $x = - 3$

Example 3: Using the results for kA, A′ and A⁻¹ · 2 marks · Very short answerHarder board level

A is a square matrix of order 3 with $∣ A ∣ = 4$ . Find (i) $∣2 A ∣$ , (ii) $∣ - A ∣$ , (iii) $∣ A^{- 1} ∣$ , (iv) $∣3 A^{'} A ∣$ .

Try it first, then show the solutionSolution

(i) $∣2 A ∣ = 2^{3} ∣ A ∣ = 32$ . (½ mark)
(ii) $∣ - A ∣ = (- 1)^{3} ∣ A ∣ = - 4$ . (½ mark)
(iii) $∣ A^{- 1} ∣ = \frac{1}{4}$ . (½ mark)
(iv) $∣3 A^{'} A ∣ = 3^{3} ∣ A^{'} ∣∣ A ∣ = 27 \times 4 \times 4 = 432$ . (½ mark)

Answer: (i) 32 (ii) $- 4$ (iii) $\frac{1}{4}$ (iv) 432

Example 4: A 3 × 3 determinant equation · 3 marks · Short answerHarder board level

Solve $x 11 1 x 1 11 x = 0$ .

Try it first, then show the solutionSolution

Expand along row 1: $x (x^{2} - 1) - 1 (x - 1) + 1 (1 - x)$ . (1 mark)
$= x^{3} - x - x + 1 + 1 - x = x^{3} - 3 x + 2$ . (1 mark)
$x = 1$ is a root ( $1 - 3 + 2 = 0$ ), so $x^{3} - 3 x + 2 = (x - 1) (x^{2} + x - 2) = (x - 1)^{2} (x + 2)$ . (½ mark)
Hence $x = 1$ or $x = - 2$ . (½ mark)

Answer: $x = 1$ (repeated) or $x = - 2$

Example 5: Why an odd-order skew-symmetric determinant vanishes · 3 marks · Short answerHOTS / competency

Show that the determinant of every $3 \times 3$ skew-symmetric matrix is 0, and check it for $0 - 3 1 30 - 4 - 1 40$ .

Try it first, then show the solutionSolution

For skew-symmetric A, $A^{'} = - A$ . Then $∣ A ∣ = ∣ A^{'} ∣ = ∣ - A ∣ = (- 1)^{3} ∣ A ∣ = - ∣ A ∣$ . (1 mark)
So $2∣ A ∣ = 0$ and $∣ A ∣ = 0$ . (½ mark)
Check by expanding along row 1: $0 - 3 - 3 1 40 + (- 1) - 3 1 0 - 4$ . (½ mark)
$= - 3 (0 - 4) - (12 - 0) = 12 - 12 = 0$ . (1 mark)

Answer: $∣ A ∣ = - ∣ A ∣$ , so $∣ A ∣ = 0$ ; the example gives $12 - 12 = 0$ .

Example 6: Case study: rotating and stretching a design · 4 marks · Case-basedHOTS / competency

Suppose a graphics app first stretches a design by $S = [2003]$ and then rotates it by $R = [cos θ sin θ - sin θ cos θ]$ . The combined effect is $R S$ , and (given) areas get multiplied by $∣ det (R S) ∣$ .
(i) Find $∣ R ∣$ . (1 mark)
(ii) Find $∣ R S ∣$ . (1 mark)
(iii) A logo of area 5 cm² is transformed. Find its new area. (1 mark)
(iv) Find $∣ (R S)^{- 1} ∣$ and interpret it. (1 mark)

Try it first, then show the solutionSolution

(i) $∣ R ∣ = cos^{2} θ + sin^{2} θ = 1$ . (1 mark)
(ii) $∣ R S ∣ = ∣ R ∣∣ S ∣ = 1 \times 6 = 6$ . (1 mark)
(iii) New area $= 6 \times 5 = 30$ cm². (1 mark)
(iv) $∣ (R S)^{- 1} ∣ = \frac{1}{6}$ : undoing the transformation shrinks areas by a factor of 6. (1 mark)

Answer: (i) 1 (ii) 6 (iii) 30 cm² (iv) $\frac{1}{6}$

Where students lose marks in Evaluating Determinants

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

Mistake: Using |kA| = k|A|
Every row is multiplied by k, so k comes out n times.
Wrong working
$∣ A ∣ = 4$ (order 3): $∣2 A ∣ = 8$ .
Right working
$∣2 A ∣ = 2^{3} \times 4 = 32$ .
Do this: Write the order first: 'A is of order 3, so $∣ k A ∣ = k^{3} ∣ A ∣$ '.
Usually costs: The MCQ.
Mistake: Wrong sign on the middle term
The cofactor sign pattern is forgotten when expanding along row 2 or a column.
Wrong working
Along row 2: $+ 2 M_{21} + (- 1) M_{22} + 0$ .
Right working
Along row 2 the signs are $- + -$ : $- 2 M_{21} + (- 1) M_{22} - 0 \cdot M_{23}$ .
Do this: Write the sign pattern above the chosen row before expanding.
Usually costs: The value, and everything that depends on it.
Mistake: Deleting the wrong row or column for a minor
Students delete only the row of the entry.
Wrong working
Minor of $a_{12}$ in $123 2 - 1 1 304$ taken as $23 - 1 1$ .
Right working
Delete row 1 and column 2: $M_{12} = 2304 = 8$ .
Do this: Cover the whole row and the whole column of the entry with your fingers.
Usually costs: 1 mark or more.
Mistake: Losing a root when factoring
Students divide by $(x - 1)$ and stop, missing the repeated or remaining roots.
Wrong working
$x^{3} - 3 x + 2 = 0$ gives $x = 1$ .
Right working
$(x - 1)^{2} (x + 2) = 0$ gives $x = 1$ or $x = - 2$ .
Do this: Factor the quadratic left after division and list all roots.
Usually costs: The final answer mark.

Exam tips

Every recent board paper has 1-mark MCQs on $∣ k A ∣$ , $∣ A^{'} ∣$ , $∣ A B ∣$ or $∣ adj A ∣$ ; write the order of the matrix before applying any result.
Show the expansion line with all three terms, even if one is zero; it earns the method mark.
The properties of determinants (row and column operations) are not in the 2026-27 syllabus; direct expansion is always enough.
For trigonometric determinants, simplify with an identity at the end ( $cos^{2} θ + sin^{2} θ = 1$ , compound angles).
Re-check a $3 \times 3$ answer by expanding along a different row if time allows.

Practice questions on Evaluating Determinants

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 $∣ A ∣ = 5$ , then $∣3 A ∣$ is:
(a) 15
(b) 45
(c) 135
(d) 125
Hint
$∣ k A ∣ = k^{n} ∣ A ∣$ with $n = 3$ .
Answer
(c) $27 \times 5 = 135$ .
Q2 · 1 mark · MCQBoard level
The value of $cos 15° sin 75° sin 15° cos 75°$ is:
(a) 1
(b) 0
(c) $\frac{1}{2}$
(d) $- 1$
Hint
$cos A cos B - sin A sin B = cos (A + B)$ .
Answer
(b) $cos 15° cos 75° - sin 15° sin 75° = cos 90° = 0$ .
Q3 · 2 marks · Very short answerBoard level
Evaluate $123 2 - 1 1 304$ .
Hint
Row 2 or column 3 contains a zero.
Answer
Along row 2 (signs $- + -$ ): $- 22134 + (- 1) 1334 - 0 = - 2 (5) - (4 - 9) = - 10 + 5 = - 5$ .
Q4 · 3 marks · Short answerHarder board level
Solve $x 22 2 x 3 33 x = 0$ .
Hint
Expand along row 1 and test $x = 2$ in the cubic.
Answer
$x (x^{2} - 9) - 2 (2 x - 6) + 3 (6 - 2 x) = x^{3} - 19 x + 30$ . $x = 2$ is a root, so $x^{3} - 19 x + 30 = (x - 2) (x^{2} + 2 x - 15) = (x - 2) (x - 3) (x + 5)$ . Hence $x = 2, 3, - 5$ .
Q5 · 1 mark · Assertion-reasonHarder board level
Assertion (A): For square matrices A and B of the same order, $∣ A B ∣ = ∣ B A ∣$ even if $A B \neq = B A$ .
Reason (R): $∣ A B ∣ = ∣ A ∣∣ B ∣$ .
(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
Numbers commute even when matrices do not.
Answer
(a). $∣ A B ∣ = ∣ A ∣∣ B ∣ = ∣ B ∣∣ A ∣ = ∣ B A ∣$ .
Q6 · 2 marks · Very short answerHarder board level
For $A = [1324]$ and $B = [2103]$ , verify that $∣ A B ∣ = ∣ A ∣∣ B ∣$ .
Hint
Compute AB first.
Answer
$∣ A ∣ = 4 - 6 = - 2$ , $∣ B ∣ = 6$ , so $∣ A ∣∣ B ∣ = - 12$ . $A B = [410612]$ , $∣ A B ∣ = 48 - 60 = - 12$ . Verified.
Q7 · 3 marks · Short answerHOTS / competency
A and B are $3 \times 3$ matrices with $∣ A ∣ = - 2$ and $∣ B ∣ = 3$ . Find (i) $∣ A^{- 1} (2 A)^{'} ∣$ and (ii) $∣ A^{2} B ∣$ .
Hint
Use the product rule, then $∣ k A^{'} ∣ = k^{3} ∣ A ∣$ .
Answer
(i) $∣ A^{- 1} ∣∣ (2 A)^{'} ∣ = \frac{1}{∣ A ∣} \times 2^{3} ∣ A ∣ = 8$ . (ii) $∣ A ∣^{2} ∣ B ∣ = 4 \times 3 = 12$ .

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.What is $a c b d$ ?Show answer
$a d - b c$ .
2.What is the cofactor $A_{ij}$ ?Show answer
$(- 1)^{i + j} M_{ij}$ , where $M_{ij}$ is the minor.
3.If A is $2 \times 2$ with $∣ A ∣ = 3$ , what is $∣5 A ∣$ ?Show answer
$25 \times 3 = 75$ .
4.What is $∣ A^{- 1} ∣$ in terms of $∣ A ∣$ ?Show answer
$\frac{1}{∣ A ∣}$ .
5.What is the determinant of a $3 \times 3$ skew-symmetric matrix?Show answer
0.
6.Which row should you expand along?Show answer
The row or column with the most zeros.

Evaluating Determinants: frequently asked questions

Can I expand a determinant along any row or column?

Yes. Expanding along any row or any column gives the same value, as long as you use the correct cofactor signs. Choose the one with the most zeros.

Are properties of determinants in the 2026-27 syllabus?

No. The NCERT section on properties of determinants (using row and column operations to simplify) was dropped. You evaluate determinants by expansion, and use results such as |AB| = |A||B| and |kA| = kⁿ|A|.

Why is |2A| = 8|A| for a 3 × 3 matrix?

Multiplying A by 2 multiplies every entry, so each of the three rows gets a factor of 2. Each row contributes one factor to the determinant, giving 2 × 2 × 2 = 8.

Which NCERT sections cover evaluating determinants?

Section 4.2 (Determinant) and Section 4.4 (Minors and Cofactors) of Chapter 4, Determinants, in NCERT Mathematics Part I for Class 12.

Do I need determinants of order 4?

No. The 2026-27 syllabus limits determinants to square matrices up to order 3.