NDA Maths · Matrices & Determinants

Determinants: Evaluation & Properties

A determinant collapses a square matrix to one number; a handful of properties (multiplicativity, row operations, the factor theorem) evaluate almost any exam determinant without brute force.

All Matrices & Determinants notes NDA Maths strategy

Why this matters

Fifty-nine PYQs — the largest and hardest area in the chapter (46% HARD). This is where the marks and the traps both live: determinant of products and scalars (det(kA) = kⁿ det A is the #1 trap), the row/column properties, the factor-theorem and Vandermonde determinants, cyclic determinants, and telescoping sums of determinants. The eight concepts below cover the lot.

Concept 1 of 8

Evaluating 2×2 and 3×3 determinants

Intuition

2\times2

determinant is

ad - bc

; geometrically it's the signed area of the parallelogram spanned by the columns. A

3\times3

is found by cofactor expansion along any row/column, or by Sarrus' diagonal rule.

Definition

$\begin{vmatrix}a&b\\c&d\end{vmatrix} = ad - bc$ . For $3\times3$ , expand along a row: $\sum_j a_{1j}(-1)^{1+j}M_{1j}$ , or use Sarrus — add the three down-right diagonal products, subtract the three down-left ones. Expanding along the row/column with the most zeros is fastest.

2×2 determinant

\begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc

Worked example

Evaluate

\begin{vmatrix}1 & 2 & 3\\0 & 4 & 5\\1 & 0 & 6\end{vmatrix}

by expanding along the first column.

Down column 1 the entries are $1, 0, 1$ with cofactor signs $+, -, +$ — the middle (zero) term drops out.
$= 1\cdot\begin{vmatrix}4 & 5\\0 & 6\end{vmatrix} + 1\cdot\begin{vmatrix}2 & 3\\4 & 5\end{vmatrix}$ .
$\begin{vmatrix}4 & 5\\0 & 6\end{vmatrix} = 24$ ; $\begin{vmatrix}2 & 3\\4 & 5\end{vmatrix} = 10 - 12 = -2$ .
Sum: $1(24) + 1(-2) = 22$ .

Answer:

22

Practice this conceptself-check · 4 quick reps

Try it yourself

Evaluate

\begin{vmatrix}2 & 0 & 1\\3 & 1 & 2\\1 & 0 & 4\end{vmatrix}

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
$\begin{vmatrix}3&1\\2&4\end{vmatrix}$ ?
2.
$\begin{vmatrix}5&2\\10&4\end{vmatrix}$ ?
3.
Geometric meaning of a 2×2 determinant?
4.
Best row/column to expand a 3×3 along?

From the bank · past-year question

Example 1Matrices & DeterminantsMODERATE

What is the value of the determinant

\begin{vmatrix}1&1&1\\1&1+xyz&1\\1&1&1+xyz\end{vmatrix}

[Q33 · Apr · 2017]

Drill 2 more on evaluating 2×2 and 3×3 determinants

Concept 2 of 8

Determinant of products, scalars, and powers

Intuition

Determinant is multiplicative:

\det(AB) = \det A \cdot \det B

. The single most-tested trap is scaling: pulling a scalar out of the WHOLE matrix multiplies the determinant by

k^n

(not

k

), because it scales all

n

rows.

Definition

For $n\times n$ matrices: $\det(AB) = \det A\,\det B$ ; $\det(A^T) = \det A$ ; $\det(kA) = k^n \det A$ ; $\det(A^m) = (\det A)^m$ ; $\det(A^{-1}) = 1/\det A$ ; $\det(B^{-1}AB) = \det A$ ; $\det(AA^T) = (\det A)^2$ .

Multiplicativity and scaling

\det(AB) = \det A\,\det B, \qquad \det(kA) = k^{\,n}\det A

Worked example

A

is a square matrix with

|A| = -2

, find

|AA^T|

$|AA^T| = |A|\cdot|A^T|$ .
$|A^T| = |A| = -2$ .
$|AA^T| = (-2)(-2) = 4 = |A|^2$ .

Answer:

4

Practice this conceptself-check · 4 quick reps

Try it yourself

A

3\times3

with

\det A = 4

, find

\det(2A)

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
$\det(kA)$ for $n\times n$ ?
2.
$\det(A^{-1})$ ?
3.
$\det(B^{-1}AB)$ ?
4.
$\det(A^T)$ vs $\det A$ ?

From the bank · past-year question

Example 2Matrices & DeterminantsEASY

If A is a square matrix such that

|A|=-2

, then

|AA^T|

, where

A^T

is the transpose of A, is equal to

[Q17 · Apr · 2026]

$\det(kA) = k^n \det A$ , NOT $k\det A$

Pulling a scalar out of a determinant works one ROW at a time. Factoring

k

from the whole

n\times n

matrix factors it from each of the

n

rows →

k^n

. The single most common determinant mistake in this bank.

Drill 17 more on determinant of products, scalars, and powers

Concept 3 of 8

Core row and column properties

Intuition

A short list of properties handles most symbolic determinants: swapping two rows flips the sign, two equal (or proportional) rows make it 0, a common factor pulls out of a row, and adding a multiple of one row to another leaves it unchanged — the workhorse for simplifying.

Definition

Transpose: $\det A^T = \det A$ (rows and columns play identical roles).
Swap: swapping two rows/columns multiplies the determinant by $-1$ .
Equal/proportional: two identical or proportional rows/columns $\Rightarrow \det = 0$ .
Common factor: a factor common to a row/column pulls outside.
Row operation: $R_i \to R_i + \lambda R_j$ leaves the determinant unchanged (the key simplification move).

Worked example

\Delta = \begin{vmatrix}a&b&c\\d&e&f\\g&h&i\end{vmatrix}

, what happens to

\Delta

if you multiply row 1 by 3 and swap rows 2 and 3?

Multiplying one row by 3 multiplies the determinant by 3.
Swapping two rows multiplies by $-1$ .
Net effect: $3 \times (-1) = -3$ , so the new determinant is $-3\Delta$ .

Answer:

-3\Delta

Practice this conceptself-check · 4 quick reps

Try it yourself

Evaluate

\begin{vmatrix}a & b & c\\ l & m & n\\ p & q & r\end{vmatrix}

compared with

\begin{vmatrix}l & m & n\\ a & b & c\\ p & q & r\end{vmatrix}

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
Two identical rows ⇒ determinant?
2.
Swapping two columns multiplies det by?
3.
$R_2 \to R_2 + 5R_1$ changes det by?
4.
$\det A^T$ vs $\det A$ ?

From the bank · past-year question

Example 3Matrices & DeterminantsHARD

\Delta=\begin{vmatrix}a&b&c\\d&e&f\\g&h&i\end{vmatrix}

, then what is

\begin{vmatrix}3d+5g&4a+7g&6g\\3e+5h&4b+7h&6h\\3f+5i&4c+7i&6i\end{vmatrix}

equal to?

[Q9 · Sep · 2021]

Drill 5 more on core row and column properties

Concept 4 of 8

Singular matrices and determinant equations

Intuition

A matrix is singular exactly when its determinant is 0 — that's also when it has no inverse. Many questions set a determinant equal to 0 (or to a value) and ask for the unknown: expand, then solve the resulting polynomial in

x

Definition

$A$ is singular $\iff \det A = 0 \iff A^{-1}$ does not exist. Determinant equations $\det(\cdot) = 0$ become polynomial equations once expanded; simplify first with row operations to lower the degree of work.

Worked example

Find

x

\begin{vmatrix}1 & 2 & 3\\2 & x & 6\\0 & 0 & 1\end{vmatrix} = 0

Expand along row 3 (it has two zeros): only the $(3,3)$ entry $= 1$ contributes.
Cofactor of $(3,3)$ : $(+1)\begin{vmatrix}1&2\\2&x\end{vmatrix} = x - 4$ .
So the determinant $= 1 \cdot (x - 4) = 0$ .
$x = 4$ .

Answer:

x = 4

Practice this conceptself-check · 4 quick reps

Try it yourself

For what value of

x

does

\begin{pmatrix}2 & 4\\-8 & x\end{pmatrix}

fail to have an inverse?

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
Singular means determinant equals?
2.
A singular matrix has an inverse?
3.
$\begin{vmatrix}x&2\\2&x\end{vmatrix}=0$ ⇒ x?
4.
Best first step on a determinant equation?

From the bank · past-year question

Example 4Matrices & DeterminantsMODERATE

If the determinant

\begin{vmatrix}x & 1 & 3\\0 & 0 & 1\\1 & x & 4\end{vmatrix} = 0

, then what is

x

equal to?

[Q18 · Apr · 2021]

Drill 7 more on singular matrices and determinant equations

Concept 5 of 8

Factor-theorem and Vandermonde determinants

Intuition

When a determinant has symbolic entries, treat it as a polynomial: if setting

a = b

makes two rows equal (determinant 0), then

(a - b)

is a factor. Chaining this gives the famous Vandermonde factorisation

(a-b)(b-c)(c-a)

Definition

If substituting $x = c$ makes two rows/columns identical, $(x - c)$ divides the determinant (factor theorem). The Vandermonde determinant $\begin{vmatrix}1&1&1\\a&b&c\\a^2&b^2&c^2\end{vmatrix} = (a-b)(b-c)(c-a)$ . Use known factors plus a degree/leading-coefficient check to pin the constant.

Vandermonde (3×3)

\begin{vmatrix}1&1&1\\a&b&c\\a^2&b^2&c^2\end{vmatrix} = (a-b)(b-c)(c-a)

Worked example

Show

(a-b)

is a factor of

D = \begin{vmatrix}1&1&1\\a&b&c\\a^2&b^2&c^2\end{vmatrix}

, and state

D

Put $a = b$ : columns 1 and 2 become identical $\Rightarrow D = 0$ . So $(a-b)$ divides $D$ .
By symmetry $(b-c)$ and $(c-a)$ also divide $D$ ; the product has the right degree (3).
A leading-term check fixes the constant as $+1$ .

Answer:

D = (a-b)(b-c)(c-a)

Practice this conceptself-check · 4 quick reps

Try it yourself

Which factor does

\begin{vmatrix}x & y & 3\\ x^2 & y^2 & 9\\ x^3 & y^3 & 27\end{vmatrix}

contain —

(x-y)

(x-3)

, or

(y-3)

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
If $a=b$ makes two columns equal, a factor is?
2.
Vandermonde $\begin{vmatrix}1&1&1\\a&b&c\\a^2&b^2&c^2\end{vmatrix}$ ?
3.
Two equal rows give determinant?
4.
How to fix the leftover constant after finding factors?

From the bank · past-year question

Example 5Matrices & DeterminantsHARD

Which one of the following factors does the expansion of the determinant

\begin{vmatrix} x & y & 3 \\ x^2 & 5y^3 & 9 \\ x^3 & 10y^5 & 27 \end{vmatrix}

contain?

[Q13 · Sep · 2018]

Drill 6 more on factor-theorem and vandermonde determinants

Concept 6 of 8

Cyclic determinants

Intuition

The cyclic determinant

\begin{vmatrix}a&b&c\\b&c&a\\c&a&b\end{vmatrix}

factors as

-(a^3 + b^3 + c^3 - 3abc) = -(a+b+c)(a^2+b^2+c^2-ab-bc-ca)

. So it vanishes exactly when

a+b+c = 0

a = b = c

Definition

$\begin{vmatrix}a&b&c\\b&c&a\\c&a&b\end{vmatrix} = -(a^3+b^3+c^3-3abc) = -(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$ . It equals 0 iff $a+b+c = 0$ (real case) or $a=b=c$ . Recognising the cyclic pattern saves a full expansion.

Cyclic determinant

\begin{vmatrix}a&b&c\\b&c&a\\c&a&b\end{vmatrix} = -(a^3+b^3+c^3-3abc)

Worked example

Under what condition does

\begin{vmatrix}a&b&c\\b&c&a\\c&a&b\end{vmatrix} = 0

(for real

a,b,c

It factors as $-(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$ .
The quadratic factor $= \tfrac12[(a-b)^2+(b-c)^2+(c-a)^2] = 0$ only when $a=b=c$ .
So the determinant is 0 iff $a+b+c = 0$ or $a = b = c$ .

Answer:When

a + b + c = 0

a = b = c

Practice this conceptself-check · 4 quick reps

Try it yourself

a + b + c = 4

and

ab + bc + ca = 0

, find

\begin{vmatrix}a&b&c\\b&c&a\\c&a&b\end{vmatrix}

given also

abc = 0

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
$\begin{vmatrix}a&b&c\\b&c&a\\c&a&b\end{vmatrix}$ in terms of cubes?
2.
It vanishes (real) when?
3.
Factor of $a^3+b^3+c^3-3abc$ ?
4.
If $a=b=c=2$ , the cyclic determinant is?

From the bank · past-year question

Example 6Matrices & DeterminantsHARD

Under which of the following conditions does the determinant

\begin{vmatrix}a & b & c\\b & c & a\\c & a & b\end{vmatrix}

vanish? 1.

a+b+c=0

a^3+b^3+c^3=3abc

a^2+b^2+c^2-ab-bc-ca=0

Select the correct answer using the code given below:

[Q3 · Apr · 2022]

Drill 4 more on cyclic determinants

Concept 7 of 8

Sums and sequences of determinants

Intuition

When asked for

\sum_k \det(A_k)

with

A_k

depending on

k

, first get a clean formula for

\det(A_k)

(often linear or telescoping in

k

), then apply the standard series sum — don't compute 100 determinants.

Definition

Evaluate $\det(A_k)$ symbolically in $k$ ; it is frequently a constant, linear, or telescoping expression. Then sum with $\sum_{k=1}^{n} k = \tfrac{n(n+1)}{2}$ , $\sum k^2 = \tfrac{n(n+1)(2n+1)}{6}$ , or telescoping cancellation.

Worked example

A_k = \begin{pmatrix}k & 1\\2 & 3\end{pmatrix}

, find

\sum_{k=1}^{100}\det(A_k)

$\det(A_k) = (k)(3) - (1)(2) = 3k - 2$ .
$\sum_{k=1}^{100}(3k - 2) = 3\sum k - 2\cdot100 = 3\cdot\tfrac{100\cdot101}{2} - 200 = 15150 - 200$ .
$= 14950$ .

Answer:

14950

Practice this conceptself-check · 4 quick reps

Try it yourself

M_k = \begin{pmatrix}k & k-1\\k-1 & k\end{pmatrix}

, find

\det(M_1) + \det(M_2) + \dots + \det(M_n)

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
$\det\begin{pmatrix}k&k-1\\k-1&k\end{pmatrix}$ ?
2.
$\sum_{k=1}^{n}(2k-1)$ ?
3.
First step on $\sum \det(A_k)$ ?
4.
$\sum_{k=1}^{10} k$ ?

From the bank · past-year question

Example 7Matrices & DeterminantsMODERATE

A_k = \begin{pmatrix}k-1&k\\k-2&k+1\end{pmatrix}

, then what is

\det(A_1)+\det(A_2)+\cdots+\det(A_{100})

equal to?

[Q20 · Sep · 2023]

Drill 3 more on sums and sequences of determinants

Concept 8 of 8

Structured and bounded determinants

Intuition

Some determinants are decided by structure, not arithmetic: if every row is a fixed multiple pattern (rank 1) the determinant is 0; if entries are bounded (all

\pm 1

), the determinant is bounded too. Spot the structure instead of expanding.

Definition

Rank-1 patterns: if $a_{ij}$ factors as $f(i)g(j)$ (e.g. $a_{ij} = 2(i+j)$ is a sum of two rank-1 pieces), the $3\times3$ determinant collapses to 0.
Bounded entries: a third-order determinant with entries all $\pm1$ lies in a small range; the maximum magnitude is 4.
Counting determinants from a fixed set of numbers uses permutations of the placements.

Worked example

The element in row

i

, column

j

of a 3rd-order determinant is

2(i+j)

. Find its value.

Entries: $a_{ij} = 2i + 2j$ — a sum of a row-only term and a column-only term.
Such a determinant is rank ≤ 2, so for a $3\times3$ it must be 0 (rows are linear combinations).
Concretely $R_3 - R_2 = R_2 - R_1$ (constant row differences) → rows dependent.

Answer:

0

Practice this concept4 quick reps

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
If $a_{ij} = f(i)g(j)$ , a 3×3 determinant is?
2.
Rows in arithmetic progression ⇒ determinant?
3.
Max magnitude of a 3×3 determinant with entries all $\pm1$ ?
4.
$a_{ij} = i + j$ for 3×3 — determinant?

From the bank · past-year question

Example 8Matrices & DeterminantsMODERATE

The element in the

i

th row and the

j

th column of a determinant of third order is equal to

2(i+j)

. What is the value of the determinant?

[Q54 · Apr · 2021]

Drill 6 more on structured and bounded determinants

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 (4)

Evaluating 2×2 and 3×3 determinants
2×2 determinant
$\begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc$
Determinant of products, scalars, and powers
Multiplicativity and scaling
$\det(AB) = \det A\,\det B, \qquad \det(kA) = k^{\,n}\det A$
Factor-theorem and Vandermonde determinants
Vandermonde (3×3)
$\begin{vmatrix}1&1&1\\a&b&c\\a^2&b^2&c^2\end{vmatrix} = (a-b)(b-c)(c-a)$
Cyclic determinants
Cyclic determinant
$\begin{vmatrix}a&b&c\\b&c&a\\c&a&b\end{vmatrix} = -(a^3+b^3+c^3-3abc)$

Watch out for (1)

$\det(kA) = k^n \det A$ , NOT $k\det A$
→ Determinant of products, scalars, and powers

Mastery check — 5 interleaved questions

Try each one before clicking. Questions are interleaved across the concepts above, not grouped — interleaving sharpens transfer.

Example 1Matrices & DeterminantsMODERATE

The value of the determinant

\begin{vmatrix}a&b&c\\l&m&n\\p&q&r\end{vmatrix}

is equal to

[Q16 · Sep · 2025]

Example 2Matrices & DeterminantsMODERATE

\begin{vmatrix}a-b & p-q & x-y\\b-c & q-r & y-z\\c-a & r-p & z-x\end{vmatrix}=k\begin{vmatrix}a & b & c\\p & q & r\\x & y & z\end{vmatrix}

, then what is the value of

k

[Q15 · Apr · 2026]

Example 3Matrices & DeterminantsMODERATE

\Delta_1 = \begin{vmatrix}1 & p & q\\1 & q & r\\1 & r & p\end{vmatrix}

and

\Delta_2 = \begin{vmatrix}1 & 1 & 1\\q & r & p\\r & p & q\end{vmatrix}

where

p \neq q \neq r

, then

\Delta_1 + \Delta_2

[Q1 · Apr · 2022]

Example 4Matrices & DeterminantsMODERATE

a \neq b \neq c

, then one value of

x

which satisfies the equation

\begin{vmatrix} 0 & x-a & x-b \\ x+a & 0 & x-c \\ x+b & x+c & 0 \end{vmatrix} = 0

is given by

[Q29 · Apr · 2017]

Example 5Matrices & DeterminantsHARD

a + b + c = 0

, then one of the solutions of

\begin{vmatrix} a-x & c & b \\ c & b-x & a \\ b & a & c-x \end{vmatrix} = 0

[Q19 · Sep · 2018]

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