Matrices and Determinants Formulas for JEE Main and Advanced

MathsClass 12, Matrices and DeterminantsFor JEE Main and Advanced

Matrices and determinants formulas for JEE Main and Advanced on one page: matrix algebra and the transpose, special matrices and powers, the properties of determinants, the adjoint and inverse, and the conditions for a system of linear equations to have one, many or no solutions, each with the condition under which it holds. The table shows which formulas decided the most questions in recent JEE papers, and section 7 solves ten of them step by step. These are the Class 12 chapters Matrices and Determinants.

Which formulas JEE uses most

We read all 84 questions on matrices and determinants in our 48 past papers (JEE Main 2025-2026, JEE Advanced 2022-2026) and noted the formulas each solution needs; one question can need several. The formulas used most:

Formula	Questions
$A X = B$ : unique solution iff $∣ A ∣ \neq = 0$ ; Cramer's $D, D_{1}, D_{2}, D_{3}$ for none or infinitely many; $A X = O$ has a non-trivial solution iff $∣ A ∣ = 0$	2833%
Adjoint: $A (adj A) = ∣ A ∣ I$ , $∣ adj A ∣ = ∣ A ∣^{n - 1}$ , $adj (adj A) = ∣ A ∣^{n - 2} A$ , $adj (k A) = k^{n - 1} adj A$	2024%
Evaluating a determinant: expansion along a row or column, row and column operations	2024%
Powers: spot the pattern; $A = I + N$ with $N^{2} = O$ gives $A^{n} = I + n N$ ; Cayley-Hamilton ( $2 \times 2$ ) $A^{2} - (tr A) A + ∣ A ∣ I = O$	1518%
$∣ A B ∣ = ∣ A ∣∣ B ∣$ , $∣ A^{T} ∣ = ∣ A ∣$ , $∣ k A ∣ = k^{n} ∣ A ∣$ , $∣ A^{m} ∣ = ∣ A ∣^{m}$	1417%
Product row by column: $(A B)_{ij} = \sum_{k} a_{ik} b_{k j}$ ; $A e_{j}$ is column $j$ of $A$	911%
$A^{- 1} = \frac{adj A}{∣ A ∣}$ , $(A B)^{- 1} = B^{- 1} A^{- 1}$ ; $A$ is invertible iff $∣ A ∣ \neq = 0$	78%
$(A B)^{T} = B^{T} A^{T}$ ; symmetric $A^{T} = A$ , skew-symmetric $A^{T} = - A$	78%

Show the other 2

Formula	Questions
Trace: $tr (A + B)$ , $tr (k A)$ , $tr (A B) = tr (B A)$ , $tr (A^{T} A) = \sum a_{ij}^{2}$	56%
Orthogonal $A A^{T} = I$ , idempotent $A^{2} = A$ , involutory $A^{2} = I$ , nilpotent $A^{k} = O$	56%

1. Matrix algebra

Expression	Equals	Note
$(A B)_{ij}$	$\sum_{k} a_{ik} b_{k j}$	$A$ is $m \times n$ and $B$ is $n \times p$ : row $i$ of $A$ times column $j$ of $B$
$(A + B)^{2}$	$A^{2} + A B + B A + B^{2}$	It is $A^{2} + 2 A B + B^{2}$ only when $A B = B A$
$(A^{T})^{T}$	$A$
$(A + B)^{T}$	$A^{T} + B^{T}$
$(k A)^{T}$	$k A^{T}$
$(A B)^{T}$	$B^{T} A^{T}$	The order reverses
$tr (A + B)$	$tr A + tr B$	Trace: the sum of the diagonal entries
$tr (k A)$	$k tr A$
$tr (A B)$	$tr (B A)$	Even when $A B \neq = B A$
$tr (A^{T} A)$	$\sum_{i, j} a_{ij}^{2}$	The sum of the squares of all the entries
$A e_{j}$	column $j$ of $A$	$e_{j}$ has $1$ in place $j$ and $0$ elsewhere; $e_{i}^{T} A$ is row $i$ of $A$

Symmetric and skew-symmetric parts

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

Every square matrix splits this way, uniquely. Symmetric: $A^{T} = A$ . Skew-symmetric: $A^{T} = - A$ , so every diagonal entry is $0$ .

2. Special matrices and powers

Matrix	Definition	What follows
Orthogonal	$A A^{T} = A^{T} A = I$	$A^{- 1} = A^{T}$ and $∣ A ∣ = \pm 1$ . Rotation matrices are orthogonal.
Idempotent	$A^{2} = A$	$A^{n} = A$ , and $(I + A)^{n} = I + (2^{n} - 1) A$
Involutory	$A^{2} = I$	$A^{- 1} = A$
Nilpotent	$N^{k} = O$	If $N^{2} = O$ : $(I + N)^{n} = I + n N$ . If $N^{3} = O$ : $(I + N)^{n} = I + n N +^{n} C_{2} N^{2}$

Cayley-Hamilton for a

2 \times 2

matrix

A^{2} - (tr A) A + ∣ A ∣ I = O

Use it to reduce $A^{3}$ , $A^{5}$ , ... to $p A + q I$ , or to read $tr A$ and $∣ A ∣$ from a given relation. For $3 \times 3$ : $A^{3} - (tr A) A^{2} + c A - ∣ A ∣ I = O$ , where $c$ is the sum of the three $2 \times 2$ principal minors.

For a high power, first look for one of these shortcuts: write $A = I + N$ and check whether $N^{2} = O$ , check $A^{2} = A$ or $A^{2} = I$ , or compute $A^{2}$ , $A^{3}$ and spot the pattern (then prove it by induction or by the binomial theorem).

3. Determinants

Expanding a

3 \times 3

determinant

a_{1} a_{2} a_{3} b_{1} b_{2} b_{3} c_{1} c_{2} c_{3} = a_{1} (b_{2} c_{3} - b_{3} c_{2}) - b_{1} (a_{2} c_{3} - a_{3} c_{2}) + c_{1} (a_{2} b_{3} - a_{3} b_{2})

Expand along the row or column with the most zeros; make zeros first with row or column operations.

Property	Effect on the determinant	Note
Transpose	$∣ A^{T} ∣ = ∣ A ∣$	Every row rule also holds for columns
Swap two rows	The sign changes
Two equal (or proportional) rows	$∣ A ∣ = 0$
Multiply one row by $k$	$∣ A ∣$ is multiplied by $k$	So a common factor of a row can be taken out, and $∣ k A ∣ = k^{n} ∣ A ∣$ for an $n \times n$ matrix
Add a multiple of one row to another	No change	$R_{i} \to R_{i} + k R_{j}$ ; the main tool for simplifying
A row written as a sum	Splits into two determinants	The determinant is linear in each row
Product and powers	$∣ A B ∣ = ∣ A ∣ ∣ B ∣$ , $∣ A^{m} ∣ = ∣ A ∣^{m}$
Skew-symmetric of odd order	$∣ A ∣ = 0$	Because $∣ A ∣ = ∣ A^{T} ∣ = ∣ - A ∣ = (- 1)^{n} ∣ A ∣$
Row times other row's cofactors	$\sum_{j} a_{ij} C_{k j} = 0$ for $i \neq = k$	For $i = k$ the sum is $∣ A ∣$

4. Adjoint and inverse

$A$ is a square matrix of order $n$ and $adj A$ is the transpose of its cofactor matrix.

Expression	Equals	Valid for
$A (adj A)$	$(adj A) A = ∣ A ∣ I_{n}$	all square $A$
$∣ adj A ∣$	$∣ A ∣^{n - 1}$	$n \geq 2$
$adj (adj A)$	$∣ A ∣^{n - 2} A$	$n \geq 2$
$∣ adj (adj A) ∣$	$∣ A ∣^{(n - 1)^{2}}$	$n \geq 2$
$adj (k A)$	$k^{n - 1} adj A$	any number $k$
$adj (A B)$	$adj B adj A$	square $A$ , $B$ of the same order (note the reversed order)
$adj (A^{T})$	$(adj A)^{T}$	all square $A$
$A^{- 1}$	$\frac{adj A}{∣ A ∣}$	$∣ A ∣ \neq = 0$
$(A B)^{- 1}$	$B^{- 1} A^{- 1}$	$A$ , $B$ invertible
$(A^{- 1})^{T}$	$(A^{T})^{- 1}$	$∣ A ∣ \neq = 0$
$∣ A^{- 1} ∣$	$\frac{1}{∣ A ∣}$	$∣ A ∣ \neq = 0$
$(adj A)^{- 1}$	$adj (A^{- 1}) = \frac{A}{∣ A ∣}$	$∣ A ∣ \neq = 0$

Inverse of a

2 \times 2

matrix

[a c b d]^{- 1} = \frac{1}{a d - b c} [d - c - b a]

Swap the diagonal entries and change the signs of the other two: that matrix is $adj A$ . For $2 \times 2$ , $∣ adj A ∣ = ∣ A ∣$ .

5. Systems of linear equations

Matrix form

A X = B, D = ∣ A ∣

$D_{1}, D_{2}, D_{3}$ : replace the first, second, third column of $A$ by $B$ and take the determinant.

Case	Result
$D \neq = 0$	Unique solution: $X = A^{- 1} B$ , or by Cramer's rule $x = \frac{D _{1}}{D}$ , $y = \frac{D _{2}}{D}$ , $z = \frac{D _{3}}{D}$
$D = 0$ and at least one $D_{i} \neq = 0$	No solution (inconsistent)
$D = 0$ and $D_{1} = D_{2} = D_{3} = 0$	Infinitely many solutions, or none. If some two left-hand sides are not proportional: infinitely many. If all three left-hand sides are proportional (parallel planes): infinitely many when every equation is a multiple of one of them (one plane), none otherwise.
Matrix test: $D = 0$ and $(adj A) B \neq = O$	No solution
$A X = O$ with $D \neq = 0$	Only the trivial solution $X = O$
$A X = O$ with $D = 0$	Infinitely many solutions, so non-trivial ones exist

6. Exam traps

Watch out

$A B \neq = B A$ in general. $(A + B)^{2} = A^{2} + 2 A B + B^{2}$ and the binomial expansion need $A B = B A$ ; $(I + N)^{n}$ works because $I$ commutes with everything.
$∣ k A ∣ = k^{n} ∣ A ∣$ , not $k ∣ A ∣$ ; $n$ is the order. For $n = 3$ , $∣2 A ∣ = 8∣ A ∣$ .
$adj (k A) = k^{n - 1} adj A$ : for $n = 3$ the factor is $k^{2}$ .
$(A B)^{- 1} = B^{- 1} A^{- 1}$ and $(A B)^{T} = B^{T} A^{T}$ : the order reverses.
$D = 0$ with at least one $D_{i} \neq = 0$ means no solution. $D = 0$ alone decides nothing, and $D_{1} = D_{2} = D_{3} = 0$ does not promise a solution (next point).
$D = D_{1} = D_{2} = D_{3} = 0$ can still give no solution, for example $x + y + z = 1$ , $x + y + z = 2$ , $x + y + z = 3$ .
A homogeneous system $A X = O$ always has $X = O$ ; it has other solutions only when $D = 0$ .
For a $2 \times 2$ matrix $∣ adj A ∣ = ∣ A ∣$ ; for $3 \times 3$ , $∣ adj A ∣ = ∣ A ∣^{2}$ .
Only odd-order skew-symmetric determinants are always $0$ .

7. Past-paper questions, solved

Ten questions from our past papers, chosen to cover the formulas used most. The question and the answer key are as set in the paper; the working is ours.

PYQ 1: No solution:

D = 0

, some

D_{i} \neq = 0

JEE Main 2025 Jan 23 Shift 2, Mathematics Q3

The system of equations

$x + y + z = 6,$

$x + 2 y + 5 z = 9,$

$x + 5 y + λ z = μ,$

has no solution if

(A) $λ = 17, μ \neq = 18$
(B) $λ \neq = 17, μ \neq = 18$
(C) $λ = 15, μ \neq = 17$
(D) $λ = 17, μ = 18$

Solution:

$D = 111125 15 λ = (2 λ - 25) - (λ - 5) + 3 = λ - 17$
So $D = 0$ exactly when $λ = 17$ .
With $λ = 17$ , replace the third column by the right-hand side:
$D_{3} = 111125 69 μ = (2 μ - 45) - (μ - 9) + 18 = μ - 18$
With $λ = 17$ and $μ \neq = 18$ : $D = 0$ and $D_{3} \neq = 0$ , so there is no solution. (For $μ = 18$ , the third equation is $4 \times$ (second) $- 3 \times$ (first), and there are infinitely many solutions.)

Answer key (A) $λ = 17, μ \neq = 18$

PYQ 2: Classifying a system with Cramer's determinants

JEE Advanced 2023 Paper 1, Mathematics Section 4 Q1

Let $α$ , $β$ and $γ$ be real numbers. Consider the following system of linear equations

$x + 2 y + z = 7$

$x + α z = 11$

$2 x - 3 y + β z = γ$

Match each entry in List-I to the correct entries in List-II.

List-I	List-II
(P) If $β = \frac{1}{2} (7 α - 3)$ and $γ = 28$ , then the system has	(1) a unique solution
(Q) If $β = \frac{1}{2} (7 α - 3)$ and $γ \neq = 28$ , then the system has	(2) no solution
(R) If $β \neq = \frac{1}{2} (7 α - 3)$ where $α = 1$ and $γ \neq = 28$ , then the system has	(3) infinitely many solutions
(S) If $β \neq = \frac{1}{2} (7 α - 3)$ where $α = 1$ and $γ = 28$ , then the system has	(4) $x = 11$ , $y = - 2$ and $z = 0$ as a solution
	(5) $x = - 15$ , $y = 4$ and $z = 0$ as a solution

(A) (P) $\to$ (3); (Q) $\to$ (2); (R) $\to$ (1); (S) $\to$ (4)
(B) (P) $\to$ (3); (Q) $\to$ (2); (R) $\to$ (5); (S) $\to$ (4)
(C) (P) $\to$ (2); (Q) $\to$ (1); (R) $\to$ (4); (S) $\to$ (5)
(D) (P) $\to$ (2); (Q) $\to$ (1); (R) $\to$ (1); (S) $\to$ (3)

Solution:

$D = 112 20 - 3 1 α β = 7 α - 2 β - 3$
So $D = 0$ exactly when $β = \frac{1}{2} (7 α - 3)$ .
When $D = 0$ , the third left-hand side is a combination of the first two: $- \frac{3}{2} (x + 2 y + z) + \frac{7}{2} (x + α z) = 2 x - 3 y + β z$ . The same combination of the right-hand sides is $- \frac{3}{2} \cdot 7 + \frac{7}{2} \cdot 11 = 28$ .
(P) $D = 0$ and $γ = 28$ : the third equation adds nothing, so infinitely many solutions. (Q) $D = 0$ and $γ \neq = 28$ : the equations contradict, so no solution.
(R) and (S): $β \neq = \frac{1}{2} (7 α - 3)$ means $D \neq = 0$ , a unique solution. In (S), $α = 1$ and $γ = 28$ , and $x = 11$ , $y = - 2$ , $z = 0$ satisfies all three equations ( $7$ , $11$ , $28$ ), so it is that solution. $(x, y, z) = (- 15, 4, 0)$ fails the first equation ( $- 15 + 8 = - 7 \neq = 7$ ), so (5) never applies.
So P $\to$ 3, Q $\to$ 2, R $\to$ 1, S $\to$ 4.

Answer key (A) (P) $\to$ (3); (Q) $\to$ (2); (R) $\to$ (1); (S) $\to$ (4)

PYQ 3:

adj (adj M) = ∣ M ∣^{n - 2} M

JEE Main 2025 Jan 22 Shift 2, Mathematics Q4

For a $3 \times 3$ matrix $M$ , let trace $(M)$ denote the sum of all the diagonal elements of $M$ . Let $A$ be a $3 \times 3$ matrix such that $∣ A ∣ = \frac{1}{2}$ and trace $(A) = 3$ . If $B = adj (adj (2 A))$ , then the value of $∣ B ∣ + trace (B)$ equals :

(A) 56
(B) 132
(C) 174
(D) 280

Solution:

For a $3 \times 3$ matrix $M$ , $adj (adj M) = ∣ M ∣ M$ . Take $M = 2 A$ : $∣2 A ∣ = 2^{3} ∣ A ∣ = 4$ .
So $B = 4 \cdot 2 A = 8 A$ .
$∣ B ∣ = 8^{3} ∣ A ∣ = 512 \cdot \frac{1}{2} = 256$ and $tr B = 8 tr A = 24$ , so $∣ B ∣ + tr B = 280$ .

Answer key (D) 280

PYQ 4:

∣ adj A ∣ = ∣ A ∣^{n - 1}

and

A^{- 1} = \frac{adj A}{∣ A ∣}

JEE Main 2026 Jan 22 Shift 2, Mathematics Q9

If $X = x y z$ is a solution of the system of equations $A X = B$ , where $adj A = 4 - 5 1 20 - 2 253$ and $B = 402$ , then $∣ x + y + z ∣$ is equal to :

(A) 3
(B) $\frac{3}{2}$
(C) 1
(D) 2

Solution:

$∣ adj A ∣ = 4 (0 + 10) - 2 (- 15 - 5) + 2 (10 - 0) = 100$ . For $n = 3$ , $∣ adj A ∣ = ∣ A ∣^{2}$ , so $∣ A ∣ = \pm 10$ .
$X = A^{- 1} B = \frac{1}{∣ A ∣} (adj A) B = \pm \frac{1}{10} 20 - 10 10 = \pm 2 - 1 1$
Either way $∣ x + y + z ∣ = 2$ .

Answer key (D) 2

PYQ 5:

A = I + N

with

N^{2} = O

JEE Main 2026 Jan 28 Shift 2, Mathematics Q25

Let $A = [31 - 4 - 1]$ and $B$ be two matrices such that $A^{100} = 100 B + I$ . Then the sum of all the elements of $B^{100}$ is ______ .

Solution:

$N N^{2} = A - I = [21 - 4 - 2] = [0000]$
$I$ and $N$ commute, so by the binomial theorem every term from $N^{2}$ on vanishes: $A^{100} = (I + N)^{100} = I + 100 N$ .
Comparing with $A^{100} = 100 B + I$ gives $B = N$ , so $B^{100} = N^{100} = O$ and the sum of its entries is $0$ .

Answer key 0

PYQ 6: Idempotent matrix,

A^{2} = A

JEE Main 2025 Apr 2 Shift 1, Mathematics Q5

Let $A = [α 6 - 1 β]$ , $α > 0$ , such that $det (A) = 0$ and $α + β = 1$ . If $I$ denotes $2 \times 2$ identity matrix, then the matrix $(I + A)^{8}$ is:

(A) $[46 - 1 - 1]$
(B) $[257514 - 64 - 127]$
(C) $[10252024 - 511 - 1024]$
(D) $[7661530 - 255 - 509]$

Solution:

$det A = α β + 6 = 0$ and $α + β = 1$ , so $α, β$ are the roots of $t^{2} - t - 6 = 0$ ; with $α > 0$ , $α = 3$ , $β = - 2$ .
$A A^{2} = [36 - 1 - 2] = [36 - 1 - 2] = A$
so $A^{k} = A$ for every $k \geq 1$ .
$I$ and $A$ commute: $(I + A)^{8} = I + (\sum_{k = 1}^{8}^{8} C_{k}) A = I + (2^{8} - 1) A = I + 255 A$ .
$I + 255 A = [7661530 - 255 - 509]$

Answer key (D) $[7661530 - 255 - 509]$

PYQ 7: Transpose of a product, skew-symmetric matrix

JEE Main 2026 Jan 23 Shift 2, Mathematics Q24

Let $A = 0 - 2 3 20 - 1 - 3 10$ and $B$ be a matrix such that $B (I - A) = I + A$ . Then the sum of the diagonal elements of $B^{T} B$ is equal to _______.

Solution:

$A^{T} = - A$ , so $A$ is skew-symmetric. $I - A$ is invertible ( $∣ I - A ∣ = 1 + 4 + 9 + 1 = 15$ ), and $B = (I + A) (I - A)^{- 1}$ .
Transpose, reversing the order: $B^{T} = [(I - A)^{- 1}]^{T} (I + A)^{T} = (I - A^{T})^{- 1} (I + A^{T}) = (I + A)^{- 1} (I - A)$ .
Write $P = I + A$ and $Q = I - A$ . They commute ( $P Q = QP = I - A^{2}$ ), so
$B^{T} B = P^{- 1} Q P Q^{- 1} = P^{- 1} P Q Q^{- 1} = I$
The diagonal of $B^{T} B = I_{3}$ adds up to $3$ .

Answer key 3

PYQ 8: Row operations

JEE Main 2025 Jan 29 Shift 1, Mathematics Q2

Let $M$ and $m$ respectively be the maximum and the minimum values of

$f (x) = 1 + sin^{2} x sin^{2} x sin^{2} x cos^{2} x 1 + cos^{2} x cos^{2} x 4 sin 4 x 4 sin 4 x 1 + 4 sin 4 x, x \in R .$

Then $M^{4} - m^{4}$ is equal to :

(A) $1280$
(B) $1295$
(C) $1040$
(D) $1215$

Solution:

$R_{2} \to R_{2} - R_{1}$ and $R_{3} \to R_{3} - R_{1}$ leave the determinant unchanged, so $f (x)$ equals
$1 + sin^{2} x - 1 - 1 cos^{2} x 10 4 sin 4 x 01$
Expand along the first row: $f (x) = (1 + sin^{2} x) + cos^{2} x + 4 sin 4 x = 2 + 4 sin 4 x$ .
As $sin 4 x$ runs over $[- 1, 1]$ : $M = 6$ , $m = - 2$ , and $M^{4} - m^{4} = 1296 - 16 = 1280$ .

Answer key (A) $1280$

PYQ 9:

(adj A)^{- 1} = \frac{A}{∣ A ∣}

JEE Main 2026 Apr 4 Shift 1, Mathematics Q6

Let $A = 1 - 2 1 103215$ . Then the sum of all elements of the matrix $adj (adj (2 (adj A)^{- 1}))$ is equal to :

(A) $3$
(B) $4$
(C) $- 4$
(D) $- 3$

Solution:

$∣ A ∣ = 1 (0 - 3) - 1 (- 10 - 1) + 2 (- 6 - 0) = - 4$ .
$(adj A)^{- 1} = \frac{A}{∣ A ∣}$ , so $B = 2 (adj A)^{- 1} = \frac{2 A}{- 4} = - \frac{A}{2}$ .
For $n = 3$ , $adj (adj B) = ∣ B ∣ B$ , and $∣ B ∣ = (- \frac{1}{2})^{3} ∣ A ∣ = \frac{1}{2}$ . So the matrix is $\frac{1}{2} (- \frac{A}{2}) = - \frac{A}{4}$ .
The entries of $A$ add up to $12$ , so the required sum is $- 3$ .

Answer key (D) $- 3$

PYQ 10: Invertible exactly when

∣ A ∣ \neq = 0

JEE Advanced 2026 Paper 1, Mathematics Section 1 Q3

Which one of the following matrices can be obtained by performing elementary row transformations on the $3 \times 3$ identity matrix ?

(A) $111111111$
(B) $121132141$
(C) $122135148$
(D) $1 - 1 0 112123$

Solution:

Every elementary row operation can be undone, so a matrix obtained from $I$ by row operations is invertible; and every invertible matrix can be row-reduced to $I$ . So we need the option with a non-zero determinant.
(A) has equal rows: $0$ . (B): $1 (3 - 8) - 1 (2 - 4) + 1 (4 - 3) = - 2$ . (C): $1 (24 - 20) - 1 (16 - 8) + 1 (10 - 6) = 0$ . (D): $1 (3 - 4) - 1 (- 3 - 0) + 1 (- 2 - 0) = 0$ .
Only (B) is invertible.

Answer key (B) $121132141$

Every formula on this page is checked with a computer algebra system, and every worked answer is recomputed and matched with the answer key. Past papers: JEE Main 2025-2026, JEE Advanced 2022-2026; the questions were read by hand (lists checked 2026-10-15).

Learn each topic in depth

Notes, derivations, solved examples and the past-paper questions on each topic.

More maths formula pages

Practise questions like these

Concept-wise questions with instant solutions, and full mock tests, on Fundamenthol.

Practice questions →Take a mock test