MathsISC Class 12 · Determinants

Properties of Determinants: Row Operations, Factorising and Proofs

Updated 15 October 2026 · Fact-checked

Determinant properties let you simplify before expanding. A determinant is unchanged by transposing or by adding a multiple of one row to another; it changes sign when two rows swap; it is zero if two rows are equal or proportional; and a common factor of a row can be taken outside. The same rules hold for columns.

Understand Properties of Determinants

The value of a $3 \times 3$ determinant can always be found by expansion, but the ISC paper often says "without expanding" or "using properties", and the examiners in 2024 noted that candidates ignored that instruction and lost marks. The properties below turn long expansions into a few lines. Each one holds for rows and equally for columns.

1. Transpose. $∣ A^{'} ∣ = ∣ A ∣$ . So anything true for rows is true for columns.

2. Swapping. Interchanging two rows (or two columns) changes the sign of the determinant.

3. Equal or proportional rows. If two rows are identical, or one is a multiple of the other, the determinant is 0. (Swap them: the value changes sign but the determinant is the same, so it equals its own negative.)

4. Common factor. If every entry of a row is multiplied by $k$ , the determinant is multiplied by $k$ . So a common factor of a row can be taken outside. For a whole $n \times n$ matrix, $∣ k A ∣ = k^{n} ∣ A ∣$ , because each of the $n$ rows gives a factor $k$ .

5. Sum in a row. If each entry of a row is a sum of two terms, the determinant splits into the sum of two determinants, one for each term, with the other rows unchanged.

6. Row operation. Adding a multiple of one row to another, $R_{i} \to R_{i} + k R_{j}$ , does not change the value. This is the workhorse: use it to create zeros or a common factor.

7. Products. $∣ A B ∣ = ∣ A ∣∣ B ∣$ , so $∣ A^{n} ∣ = ∣ A ∣^{n}$ and $∣ A^{- 1} ∣ = \frac{1}{∣ A ∣}$ .

8. Triangular matrices. If all entries on one side of the leading diagonal are zero, the determinant is the product of the diagonal entries.

How these combine in proofs. A typical "prove that" question has a symmetric pattern in $a, b, c$ or $x, y, z$ . The standard moves are:

Add all rows (or columns) into one when every column has the same sum, then take that sum out as a factor. For $a b c b c a c a b$ , $R_{1} \to R_{1} + R_{2} + R_{3}$ gives the factor $a + b + c$ .
Subtract neighbouring columns ( $C_{1} \to C_{1} - C_{2}$ , $C_{2} \to C_{2} - C_{3}$ ) to create factors like $(a - b)$ and $(b - c)$ , as in the ISC 2025 question where $x + y + z = 0$ .
Create two zeros in a row or column, then expand along it: only one $2 \times 2$ determinant is left.

Factor theorem shortcut. If putting $a = b$ makes two columns equal, the determinant is zero then, so $(a - b)$ is a factor. This tells you which factors to aim for.

Area and collinearity. For points $(x_{1}, y_{1}), (x_{2}, y_{2}), (x_{3}, y_{3})$ , the area of the triangle is $\frac{1}{2} x_{1} x_{2} x_{3} y_{1} y_{2} y_{3} 111$ , taking the absolute value. The three points are collinear exactly when this determinant is 0.

Properties of Determinants: formulas to remember

Transpose: $∣ A^{'} ∣ = ∣ A ∣$; Every row rule is also a column rule.
Swap two rows: $R_{i} \leftrightarrow R_{j} \Rightarrow Δ \to - Δ$; The same holds for swapping two columns.
Zero value: $R_{i} = k R_{j} \Rightarrow Δ = 0$; Equal or proportional rows (or columns).
Factor out of a row: $∣ k A ∣ = k^{n} ∣ A ∣ for an n \times n matrix$; One row scaled by k multiplies Δ by k.
Row operation: $R_{i} \to R_{i} + k R_{j} \Rightarrow Δ unchanged$; Use to create zeros or a common factor.
Product rule: $∣ A B ∣ = ∣ A ∣∣ B ∣, ∣ A^{- 1} ∣ = \frac{1}{∣ A ∣}$; A and B square matrices of the same order.
Vandermonde form: $1 a a^{2} 1 b b^{2} 1 c c^{2} = (a - b) (b - c) (c - a)$; Appears in many factorising questions.
Area of a triangle: $Δ = \frac{1}{2} x_{1} x_{2} x_{3} y_{1} y_{2} y_{3} 111$; Determinant zero means the points are collinear.

How to solve Properties of Determinants questions

For "evaluate without expanding" and "prove using properties" questions, follow these steps and write the operation used on every line.

1Look first for equal or proportional rows or columns; if you find them, the value is 0.
2Check whether every row (or column) has the same sum. If so, add all rows into one ( $R_{1} \to R_{1} + R_{2} + R_{3}$ ) and take the sum out as a factor.
3Otherwise subtract neighbouring rows or columns to create differences such as $(a - b)$ , and take them out as factors.
4Aim for two zeros in one row or column. Then expand along it, leaving one $2 \times 2$ determinant.
5Write each operation beside the step, for example " $C_{1} \to C_{1} - C_{2}$ ". Examiners mark the operations as well as the result.
6Factorise the remaining expression and compare with the required answer. If a condition such as $x + y + z = 0$ is given, use it at the end.
7For area or collinearity, set up the determinant with a column of ones, evaluate it, halve it and take the absolute value.

Properties of 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.

Collinearity by a determinant

Suppose an engineer plans to run one straight overhead cable over three electricity poles. On the site plan the poles are at $(2, 3)$ , $(5, 9)$ and $(8, 15)$ , in metres.

Use a determinant to decide whether one straight cable can pass over all three poles.

Try it first, then show the solutionSolution

Form $Δ = 2583915111$ .
Apply $R_{2} \to R_{2} - R_{1}$ and $R_{3} \to R_{3} - R_{1}$ : rows become $(3, 6, 0)$ and $(6, 12, 0)$ .
The new $R_{3}$ is $2 \times$ the new $R_{2}$ , so $Δ = 0$ .
A zero determinant means the triangle formed has zero area: the poles are collinear.

Answer: Yes. The determinant is 0, so the three poles lie on one straight line.

Memory hook: No area, no bend: a zero determinant means the poles are in a straight line.

Product rule |AB| = |A||B|

Imagine a design app that stretches a logo with the matrix $S_{1} = [3001]$ and then slants and stretches it with $S_{2} = [1012]$ . For such a transformation, areas are multiplied by the absolute value of its determinant.

A logo has an area of 5 cm². What is its area after both steps? Use the product rule rather than multiplying the matrices.

Try it first, then show the solutionSolution

$∣ S_{1} ∣ = 3$ and $∣ S_{2} ∣ = 2$ .
The combined transformation is $S_{2} S_{1}$ , and $∣ S_{2} S_{1} ∣ = ∣ S_{2} ∣∣ S_{1} ∣ = 6$ .
New area $= 6 \times 5 = 30$ cm².

Answer: 30 cm².

Memory hook: Scale then scale again: determinants multiply, just like the areas.

Tricks to remember Properties of Determinants

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

SAFE: four zero tests
Same rows, All-zero row, Factor-proportional rows: the determinant is zero. Everything Else needs work.
Use it when: Spotting 1-mark answers instantly.
Same sums? Add them all
If every row sums to the same thing, add all columns into C₁ (or all rows into R₁) and pull out that sum.
Use it when: Circulant determinants like a, b, c / b, c, a / c, a, b.
Put a = b
If a = b makes the determinant zero, then (a − b) is a factor.
Use it when: Predicting the factors before you start.
kⁿ, not k
|kA| = kⁿ|A|: every row gets a k. For a 3 × 3 matrix, |2A| = 8|A|.
Use it when: MCQs on scalar multiples.

Properties of Determinants: solved examples

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

Example 1: Determinant of a scalar multiple · 1 mark · MCQBoard level

If $A$ is a $3 \times 3$ matrix with $∣ A ∣ = 7$ , then $∣2 A ∣$ is:
(a) 14
(b) 28
(c) 56
(d) 49

Try it first, then show the solutionSolution

$∣2 A ∣ = 2^{3} ∣ A ∣ = 8 \times 7 = 56$ .

Answer: (c) 56.

Example 2: Value zero without expanding · 2 marks · Short answerBoard level

Without expanding, show that $a a + 2 x x b b + 2 y y c c + 2 z z = 0$ .

Try it first, then show the solutionSolution

$R_{2} \to R_{2} - R_{1}$ : $a 2 x x b 2 y y c 2 z z$ . (1 mark)
Take 2 out of $R_{2}$ : $2 a x x b y y c z z$ . Now $R_{2} = R_{3}$ , so the value is 0. (1 mark)

Answer: The determinant is 0.

Example 3: Factorising a Vandermonde determinant · 3 marks · Short answerHarder board level

Prove that $1 a a^{2} 1 b b^{2} 1 c c^{2} = (a - b) (b - c) (c - a)$ .

Try it first, then show the solutionSolution

$C_{1} \to C_{1} - C_{2}$ , $C_{2} \to C_{2} - C_{3}$ : $0 a - b a^{2} - b^{2} 0 b - c b^{2} - c^{2} 1 c c^{2}$ . (1 mark)
Take $(a - b)$ from $C_{1}$ and $(b - c)$ from $C_{2}$ : $(a - b) (b - c) 01 a + b 01 b + c 1 c c^{2}$ . (1 mark)
Expand along $R_{1}$ : $1 \times [(b + c) - (a + b)] = c - a$ . So the value is $(a - b) (b - c) (c - a)$ . (1 mark)

Answer: Proved.

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

Prove that $a b c b c a c a b = - (a + b + c) (a^{2} + b^{2} + c^{2} - ab - b c - c a)$ .

Try it first, then show the solutionSolution

$R_{1} \to R_{1} + R_{2} + R_{3}$ and take out $(a + b + c)$ : $(a + b + c) 1 b c 1 c a 1 a b$ . (1 mark)
$C_{2} \to C_{2} - C_{1}$ , $C_{3} \to C_{3} - C_{1}$ : $(a + b + c) 1 b c 0 c - b a - c 0 a - b b - c$ . (1 mark)
Expand along $R_{1}$ : $(c - b) (b - c) - (a - b) (a - c) = - (b - c)^{2} - (a^{2} - ab - a c + b c)$ , which simplifies to $- (a^{2} + b^{2} + c^{2} - ab - b c - c a)$ . (1 mark)

Answer: Proved.

Example 5: Assertion-reason on proportional rows · 1 mark · Assertion-reasonHarder board level

Assertion (A): $2413675109 = 0$ .
Reason (R): If one row of a determinant is a multiple of another row, the determinant is zero.
(a) Both true, R explains A. (b) Both true, R does not explain A. (c) A true, R false. (d) A false, R true.

Try it first, then show the solutionSolution

$R_{2} = 2 R_{1}$ , so by R the determinant is 0.

Answer: (a).

Example 6: A classic cube result · 5 marks · Long answerHOTS / competency

Using properties of determinants, prove that $x + y + 2 z z z x y + z + 2 x x y y z + x + 2 y = 2 (x + y + z)^{3}$ .

Try it first, then show the solutionSolution

$C_{1} \to C_{1} + C_{2} + C_{3}$ : every entry of $C_{1}$ becomes $2 (x + y + z)$ . Take it out: $2 (x + y + z) 111 x y + z + 2 x x y y z + x + 2 y$ . (2 marks)
$R_{2} \to R_{2} - R_{1}$ , $R_{3} \to R_{3} - R_{1}$ : $2 (x + y + z) 100 x x + y + z 0 y 0 x + y + z$ . (2 marks)
This is upper triangular, so the determinant is $2 (x + y + z) \times 1 \times (x + y + z) (x + y + z) = 2 (x + y + z)^{3}$ . (1 mark)

Answer: Proved.

Where students lose marks in Properties of Determinants

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

Mistake: Expanding when told "without expanding"
The ISC 2024 examiners noted that candidates ignored this instruction; an expanded answer gets little or no credit.
Do this: Use row and column operations only, and end with a property such as "two columns are identical".
Usually costs: Most of the 2 marks.
Mistake: Writing |kA| = k|A|
Every row of kA carries the factor k, so an n × n matrix gives kⁿ.
Wrong working
$∣2 A ∣ = 2∣ A ∣$ for a $3 \times 3$ matrix.
Right working
$∣2 A ∣ = 2^{3} ∣ A ∣ = 8∣ A ∣$ .
Do this: Count the rows: one factor of k per row.
Usually costs: 1 mark.
Mistake: Multiplying a row and forgetting to compensate
Replacing R₁ by 2R₁ doubles the determinant; students often write it as an equal value.
Do this: Only use operations of the form Rᵢ → Rᵢ + kRⱼ freely; if you scale a row, divide outside by the same factor.
Usually costs: 1 mark, and the final answer (the 2025 examiners noted candidates "multiplied at several places").
Mistake: Doing two dependent operations at once
Changing R₁ using R₂ and then R₂ using the old R₁ in the same step can change the value.
Do this: When doing several operations in one step, only use rows that are not being changed in that step.
Usually costs: The whole proof if the value changes.

Exam tips

Write each operation ( $R_{1} \to R_{1} + R_{2} + R_{3}$ , $C_{2} \to C_{2} - C_{3}$ ) on the line where you use it.
Before starting a proof, decide what the factors should be by putting $a = b$ , $b = c$ , and so on.
Aim for two zeros in one row or column before you expand; never expand a full $3 \times 3$ with letters if you can avoid it.
If a condition such as $x + y + z = 0$ is given, keep $(x + y + z)$ as a factor and substitute only at the very end.
For area, remember the $\frac{1}{2}$ and the absolute value; for collinearity, you can drop both and just set the determinant to 0.

Practice questions on Properties of 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 two rows of a determinant $Δ$ are interchanged, the new value is:
(a) $Δ$
(b) $- Δ$
(c) $0$
(d) $2Δ$
Hint
Swapping changes the sign.
Answer
(b) $- Δ$ .
Q2 · 2 marks · Short answerBoard level
Without expanding, find $10211718333646$ .
Hint
Compare $R_{1}$ with $R_{3}$ .
Answer
$R_{1} = 6 R_{3}$ (since $102 = 6 \times 17$ , $18 = 6 \times 3$ , $36 = 6 \times 6$ ), so the value is 0.
Q3 · 3 marks · Short answerHarder board level
Prove that $111 a b c b c c a ab = (a - b) (b - c) (c - a)$ .
Hint
$R_{1} \to R_{1} - R_{2}$ , $R_{2} \to R_{2} - R_{3}$ , then take out factors.
Answer
Apply $R_{1} \to R_{1} - R_{2}$ and $R_{2} \to R_{2} - R_{3}$ : $R_{1} = (0, a - b, c (b - a))$ and $R_{2} = (0, b - c, a (c - b))$ . Take out $(a - b)$ and $(b - c)$ : $(a - b) (b - c) 001 11 c - c - a ab$ . Expanding along $C_{1}$ : $1 \times (- a + c) = c - a$ . So the value is $(a - b) (b - c) (c - a)$ .
Q4 · 2 marks · Short answerHarder board level
If $A$ is a $3 \times 3$ matrix with $∣ A ∣ = 3$ , find $∣3 A^{'} ∣$ and $∣ A^{- 1} ∣$ .
Hint
$∣ k A^{'} ∣ = k^{3} ∣ A ∣$ .
Answer
$∣3 A^{'} ∣ = 27 \times 3 = 81$ ; $∣ A^{- 1} ∣ = \frac{1}{3}$ .
Q5 · 2 marks · Short answerHarder board level
Using a determinant, find $k$ if the points $(2, - 3)$ , $(k, - 1)$ and $(0, 4)$ are collinear.
Hint
Set the area determinant to zero.
Answer
$2 k 0 - 3 - 1 4 111 = 2 (- 1 - 4) + 3 (k - 0) + (4 k - 0) = 7 k - 10 = 0$ , so $k = \frac{10}{7}$ .
Q6 · 3 marks · Short answerHOTS / competency
Prove that $b + c b c a c + a c a b a + b = 4 ab c$ .
Hint
$R_{1} \to R_{1} - R_{2} - R_{3}$ creates a row $(0, - 2 c, - 2 b)$ .
Answer
$R_{1} \to R_{1} - R_{2} - R_{3}$ gives $R_{1} = (0, - 2 c, - 2 b)$ . Expanding along $R_{1}$ : $2 c [b (a + b) - b c] - 2 b [b c - c (c + a)] = 2 b c (a + b - c) + 2 b c (c + a - b) = 2 b c \cdot 2 a = 4 ab c$ .
Q7 · 4 marks · Case-basedHOTS / competency
Case study: Suppose a farmer's triangular plot has corners at $O (0, 0)$ , $P (60, 20)$ and $Q (20, 80)$ on a plan in metres.
(i) Find the area of the plot using a determinant. (2 marks)
(ii) A well is at $W (40, 50)$ . Show that it lies on the boundary $P Q$ . (1 mark)
(iii) If $Q$ is moved to $(20, k)$ with $k > 0$ so that the area becomes 1600 m², find $k$ . (1 mark)
Hint
Area $= \frac{1}{2} ∣60 y_{Q} - 20 x_{Q} ∣$ when one corner is the origin.
Answer
(i) $\frac{1}{2} ∣60 \times 80 - 20 \times 20∣ = \frac{1}{2} (4800 - 400) = 2200$ m². (ii) The determinant for $P$ , $Q$ , $W$ is $60 (80 - 50) - 20 (20 - 40) + (20 \times 50 - 40 \times 80) = 1800 + 400 - 2200 = 0$ , so they are collinear. (iii) $\frac{1}{2} ∣60 k - 400∣ = 1600 \Rightarrow 60 k - 400 = 3200 \Rightarrow k = 60$ (the other sign gives a negative $k$ ).

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 happens to a determinant when two columns are interchanged?Show answer
It changes sign.
2.What is $∣ k A ∣$ for an $n \times n$ matrix?Show answer
$k^{n} ∣ A ∣$ .
3.Which row operation leaves a determinant unchanged?Show answer
$R_{i} \to R_{i} + k R_{j}$ (adding a multiple of another row).
4.Find $∣ A B ∣$ if $∣ A ∣ = 2$ and $∣ B ∣ = - 5$ .Show answer
$- 10$ .
5.Evaluate $125247369$ without expanding.Show answer
0, since $R_{2} = 2 R_{1}$ .
6.What is the condition for three points to be collinear using determinants?Show answer
The determinant with rows $(x_{i}, y_{i}, 1)$ equals 0.

Properties of Determinants: frequently asked questions

What does "without expanding" mean in a determinant question?

You must find the value using properties such as row and column operations, common factors and identical rows, rather than writing out the full expansion. The final step is usually a property, for example two identical columns giving zero.

Do determinant properties work for columns as well as rows?

Yes. Since the determinant of a matrix equals the determinant of its transpose, every property stated for rows also holds for columns.

How do I choose which row or column operation to use?

Look for a common sum (then add all rows or columns), or for entries that differ by simple expressions (then subtract neighbours). The aim is to create a common factor or two zeros in one line.

Why is the determinant zero when two rows are equal?

Swapping the two equal rows changes the sign of the determinant but leaves the matrix the same. So the determinant equals its own negative, which is only possible if it is zero.

Is the area of a triangle using determinants in the ISC 2027 syllabus?

Yes. The CISCE 2027 syllabus lists applications of determinants in finding the area of a triangle and checking collinearity, along with problems based on properties of determinants.