Matrices and Determinants Formula Sheet — JEE Main Mathematics

Every key Matrices and Determinants formula, definition and theorem for JEE Main Mathematics in one place — with common examiner traps and worked examples. Free to read; blurt from memory, then check your gaps.

Syllabus — topics coveredNTA · 9 sub-topics

Matrices
Algebra of matrices
Types of matrices
Determinants and matrices of order two and three
Properties of determinants
Evaluation of determinants
Area of triangles using determinants
Adjoint and evaluation of inverse of a square matrix using determinants
Test of consistency and solution of simultaneous linear equations in two or three variables using determinants and matrices

Matrices — Order, Types & Algebra

Matrix: A rectangular array

A = [a_{ij}]_{m \times n}

of m rows and n columns (mn entries). Two matrices are equal only if they have the

same order

and every corresponding entry matches.

Algebra of matrices

Operation	Rule
$Add / subtract$	only for the SAME order; entrywise
$Scalar k A$	multiply every entry by $k$
$Product A B$	needs cols $(A) =$ rows(B): $(m \times n) (n \times p) = (m \times p)$
$A B \neq = B A$	NOT commutative (but associative)
$Identity$	$A I = I A = A$ ; $A B = O$ does NOT force $A = O$ or $B = O$

Types of matrices

▸ $Row$ ( $1 \times n$ ), $column$ ( $m \times 1$ ), $square$ ( $n \times n$ ).
▸ $Diagonal$ : off-diagonal entries 0; $scalar$ : equal diagonal entries.
▸ $Identity$ I: diagonal $1$ , rest 0; $zero$ O: all entries $0$ .
▸ $Triangular$ : all zeros below (upper) or above (lower) the diagonal.

🎯 Exam · When is the product AB defined?

AB exists only if

cols (A) = rows (B)

(m \times n) (n \times p) = (m \times p)

. Addition needs the

same order

; scalar kA multiplies every entry by k.

⚠️ Watch out · Matrix multiplication is not commutative

In general

A B \neq = B A

(and one may exist while the other doesn't). Also

A B = O

does

not

force

A = O

B = O

, and there is no cancellation:

A B = A C

does not give

B = C

★ Remember · Identity and powers

A I = I A = A

. Powers

A^{2}, A^{3}, \dots

make sense only for

square

matrices, and

(A^{m}) (A^{n}) = A^{m + n}

🚫 Examiner Trap · Examiner traps

(1) Matrix multiplication is NOT commutative:

A B \neq = B A

in general (and one may not even be defined). (2) AB needs cols

(A) =

rows(B); the product is

(m \times p)

. (3)

A B = O

does NOT force

A = O

B = O

(zero divisors exist). (4) Add/subtract only for the SAME order; scalar kA multiplies EVERY entry by k.

Transpose & Special Square Matrices

Named square matrices

Name	Condition
$Symmetric$	$A^{T} = A$ ( $a_{ij} = a_{ji}$ )
$Skew-symmetric$	$A^{T} = - A$ (diagonal all $0$ )
$Orthogonal$	$A A^{T} = A^{T} A = I$ (so $A^{- 1} = A^{T}$ )
$Idempotent$	$A^{2} = A$
$Involutory$	$A^{2} = I$ (so $A^{- 1} = A$ )
$Nilpotent$	$A^{k} = O$ for some k

Transpose Aᵀ

▸ $(A^{T})^{T} = A$ and $(k A)^{T} = k A^{T}$ .
▸ $(A + B)^{T} = A^{T} + B^{T}$ .
▸ $(A B)^{T} = B^{T} A^{T}$ — $reverse$ the order.
▸ $(A^{T})^{- 1} = (A^{- 1})^{T}$ .

★ Remember · Symmetric + skew decomposition

Symmetric

A^{T} = A

Skew-symmetric

A^{T} = - A

(so the diagonal is all

0

). Every square matrix splits uniquely as

A = \frac{1}{2} (A + A^{T}) + \frac{1}{2} (A - A^{T})

(symmetric

+

skew).

🎯 Exam · Named matrices to recognise

Orthogonal

A A^{T} = I

(so

A^{T} = A^{- 1}

);

idempotent

A^{2} = A

;

involutory

A^{2} = I

;

nilpotent

A^{k} = O

. A skew-symmetric matrix of

odd

order has determinant

0

★ Remember · Trace

tr (A)

= sum of the diagonal entries. It is linear,

tr (A + B) = tr (A) + tr (B)

, and

tr (A B) = tr (B A)

even though

A B \neq = B A

🚫 Examiner Trap · Examiner traps

(1) Transpose reverses a product:

(A B)^{T} = B^{T} A^{T}

(NOT

A^{T} B^{T}

). (2) A skew-symmetric matrix has ALL diagonal entries

0

(since

a_{ii} = - a_{ii}

). (3) Every square matrix

= \frac{1}{2} (A + A^{T}) + \frac{1}{2} (A - A^{T})

(symmetric

+

skew). (4) Orthogonal means

A A^{T} = I

(so

A^{- 1} = A^{T}

); don't confuse with symmetric.

Determinants — Evaluation, Minors & Cofactors

Determinant: A single number

∣ A ∣

attached to a

square

matrix. For

2 \times 2

a c b d = a d - b c

; for

3 \times 3

(and larger) expand along any row or column using cofactors.

Minors & cofactors

▸ $Minor$ $M_{ij}$ : determinant left after deleting row i, column j.
▸ $Cofactor$ $C_{ij} = (- 1)^{i + j} M_{ij}$ (the $+ - +$ sign-board).
▸ $∣ A ∣ = \sum_{j} a_{ij} C_{ij}$ along any row i (or any column).
▸Expand along the row/column with the $most zeros$ — least work.

🎯 Exam · The 'wrong-row' sum is zero

Entries of one row against the cofactors of a

different

row sum to zero:

\sum_{j} a_{ij} C_{kj} = 0

when

i \neq = k

. This is what makes

A (adj A) = ∣ A ∣ I

work.

★ Remember · Sarrus for 3×3

For a

3 \times 3

you may use Sarrus' diagonal rule (sum of the three

↘

products minus the three

↙

products) — but it works for

3 \times 3

only

, not larger.

⚠️ Watch out · Don't forget the cofactor sign

The minor and the cofactor differ by

(- 1)^{i + j}

. Forgetting the alternating sign — especially on the middle entry of a row — is the most common evaluation error.

🚫 Examiner Trap · Examiner traps

(1) Cofactor carries the sign-board:

C_{ij} = (- 1)^{i + j} M_{ij}

— forgetting

(- 1)^{i + j}

is the classic slip. (2) Expand along the row/column with the MOST zeros (less work, same answer). (3) Mixing rows gives ZERO:

\sum_{j} a_{ij} C_{kj} = 0

when

i \neq = k

. (4) Determinant is defined ONLY for square matrices.

Properties of Determinants

Properties of determinants

Property	Effect
Transpose	$∣ A^{T} ∣ = ∣ A ∣$ — rows & columns play the same role
Swap two rows / cols	the determinant changes SIGN
Two identical / proportional rows	$∣ A ∣ = 0$
Common factor $k$ in a row	pull it out: that row scales $∣ A ∣$ by $k$
Row op $R_{i} \to R_{i} + k R_{j}$	leaves $∣ A ∣$ UNCHANGED (the workhorse)
Scalar multiple	$∣ k A ∣ = k^{n} ∣ A ∣$ for $n \times n$
Product	$∣ A B ∣ = ∣ A ∣ ∣ B ∣$ (so $∣ A^{- 1} ∣ = 1/∣ A ∣$ )
Triangular / diagonal	$∣ A ∣ =$ product of the diagonal entries
Singular	$∣ A ∣ = 0 \Leftrightarrow A$ has no inverse

The properties you use most

▸ $∣ A^{T} ∣ = ∣ A ∣$ — rows and columns are interchangeable.
▸Swapping two rows/columns $\Rightarrow$ sign changes; identical/proportional rows $\Rightarrow ∣ A ∣ = 0$ .
▸ $R_{i} \to R_{i} + k R_{j}$ leaves $∣ A ∣$ $unchanged$ (use it to make zeros).
▸A common factor of a row comes out front.

🎯 Exam · Scalar and product rules

∣ k A ∣ = k^{n} ∣ A ∣

for an

n \times n

matrix (the factor comes out of

every

row).

∣ A B ∣ = ∣ A ∣∣ B ∣

, hence

∣ A^{- 1} ∣ = \frac{1}{∣ A ∣}

and

∣ A^{n} ∣ = ∣ A ∣^{n}

★ Remember · Triangular = product of the diagonal

If a matrix is triangular (or diagonal),

∣ A ∣

is just the product of its diagonal entries. Row operations that create such a form (without scaling rows) give a quick evaluation.

⚠️ Watch out · |kA| is not k|A|

Scaling the whole matrix scales

each

of the n rows, so

∣ k A ∣ = k^{n} ∣ A ∣

— not

k ∣ A ∣

. For a

3 \times 3

∣2 A ∣ = 8∣ A ∣

, not

2∣ A ∣

🚫 Examiner Trap · Examiner traps

(1) SWAP of two rows/cols flips the SIGN of

∣ A ∣

(a frequent miss). (2) Two identical OR proportional rows

\Rightarrow ∣ A ∣ = 0

. (3)

∣ k A ∣ = k^{n} ∣ A ∣

for

n \times n

(the factor is

k^{n}

, not k). (4) The safe row operation

R_{i} \to R_{i} + k R_{j}

leaves

∣ A ∣

UNCHANGED;

∣ A B ∣ = ∣ A ∣∣ B ∣

but

∣ A + B ∣ \neq = ∣ A ∣ + ∣ B ∣

Adjoint & Inverse of a Matrix

Adjoint:

adj A

is the

transpose of the cofactor matrix

[C_{ij}]^{T}

. Its defining property is

A (adj A) = (adj A) A = ∣ A ∣ I

, which is exactly what turns it into the inverse.

★ Remember · The inverse formula

A^{- 1} = \frac{1}{∣ A ∣} adj A

, which exists

\Leftrightarrow

∣ A ∣ \neq = 0

non-singular

matrix). For a

2 \times 2

: swap the diagonal entries and negate the off-diagonal, then divide by

∣ A ∣

Adjoint & inverse properties

▸ $∣ adj A ∣ = ∣ A ∣^{n - 1}$ for an $n \times n$ matrix.
▸ $adj (A B) = adj B \cdot adj A$ (reverse order).
▸ $(A B)^{- 1} = B^{- 1} A^{- 1}$ , $(A^{- 1})^{- 1} = A$ , $(A^{T})^{- 1} = (A^{- 1})^{T}$ .
▸ $∣ A^{- 1} ∣ = 1/∣ A ∣$ and $(k A)^{- 1} = \frac{1}{k} A^{- 1}$ .

🎯 Exam · Singular ⇒ no inverse

∣ A ∣ = 0

the matrix is

singular

and has no inverse — so before inverting, always check the determinant. A product AB is invertible iff both A and B are.

★ Remember · Row-reduction alternative

You can also invert by elementary row operations: reduce

[A ∣ I]

[I ∣ A^{- 1}]

. If a full row of zeros appears on the left, A is singular (no inverse).

🚫 Examiner Trap · Examiner traps

(1)

A^{- 1} = \frac{1}{∣ A ∣} adj A

exists ONLY if

∣ A ∣ \neq = 0

(non-singular). (2)

adj A = [C_{ij}]^{T}

— it's the TRANSPOSE of the cofactor matrix (don't forget to transpose). (3)

∣ adj A ∣ = ∣ A ∣^{n - 1}

adj (k A) = k^{n - 1} adj A

. (4)

(A B)^{- 1} = B^{- 1} A^{- 1}

(reverse order). For

2 \times 2

: swap the diagonal, negate the off-diagonal, divide by

∣ A ∣

System of Linear Equations — Cramer & Consistency

★ Remember · Two ways to solve AX = B

Matrix method

X = A^{- 1} B

(needs

∣ A ∣ \neq = 0

Cramer’s rule

x = \frac{Δ _{1}}{Δ}

y = \frac{Δ _{2}}{Δ}

z = \frac{Δ _{3}}{Δ}

, where

Δ = ∣ A ∣

and

Δ_{i}

replaces column i of A with B.

Consistency of AX = B

▸ $Δ \neq = 0$ : a $unique$ solution (consistent).
▸ $Δ = 0$ and some $Δ_{i} \neq = 0$ : $no$ solution (inconsistent).
▸ $Δ = 0$ and all $Δ_{i} = 0$ : infinitely many solutions (or none — check).
▸Rank test: consistent $\Leftrightarrow$ rank $(A) =$ rank of the augmented matrix.

🎯 Exam · Homogeneous systems AX = O

Always consistent (

X = O

is the trivial solution). If

Δ \neq = 0

the trivial solution is the

only

one; if

Δ = 0

there are

infinitely many non-trivial

solutions.

⚠️ Watch out · Δ = 0 doesn't always mean 'infinite'

When

Δ = 0

you must check the

Δ_{i}

: if any is non-zero the system is

inconsistent

(no solution); only when all

Δ_{i} = 0

can it have infinitely many. Don't jump straight to 'infinite'.

🚫 Examiner Trap · Examiner traps

(1) Cramer:

Δ_{i}

replaces the i-th column of A with B — and Cramer needs

Δ \neq = 0

. (2)

Δ = 0

with some

Δ_{i} \neq = 0 \Rightarrow

NO solution (inconsistent);

Δ = 0

with all

Δ_{i} = 0 \Rightarrow

infinitely many OR none (check further). (3) Homogeneous

A X = O

is always consistent; it has NON-trivial solutions IFF

Δ = 0

. (4) Matrix method

X = A^{- 1} B

needs

∣ A ∣ \neq = 0

Area of a Triangle & Special Results

★ Remember · Area of a triangle

Area

= \frac{1}{2} det x_{1} x_{2} x_{3} y_{1} y_{2} y_{3} 111

— always take the

absolute value

. The determinant is

0

\Leftrightarrow

the three points are

collinear

🎯 Exam · Cayley-Hamilton theorem

Every square matrix satisfies its own characteristic equation

∣ A - λ I ∣ = 0

. Rearranging that equation gives

A^{- 1}

(or high powers

A^{n}

) in terms of lower powers — a big shortcut.

★ Remember · Eigenvalues

The eigenvalues

λ

solve

∣ A - λ I ∣ = 0

. Their

sum = tr (A)

and their

product = ∣ A ∣

— a quick check on your characteristic equation.

★ Remember · Line through two points

The line through

(x_{1}, y_{1})

and

(x_{2}, y_{2})

is found by setting the area determinant with a general point (x,y) equal to

0

— the collinearity condition in reverse.

🚫 Examiner Trap · Examiner traps

(1) Triangle area

= \frac{1}{2} ∣ det ∣

— take the ABSOLUTE value (area

\geq 0

). (2) Determinant

= 0 \Rightarrow

the three points are COLLINEAR (zero area), which is also how you write the line through two points. (3) Cayley–Hamilton: a matrix satisfies its OWN characteristic equation — use it to get

A^{- 1}

or high powers. (4) Eigenvalues solve

∣ A - λ I ∣ = 0

;

\sum λ = tr (A)

\prod λ = ∣ A ∣

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