Mathematics Matrices and Determinants

Matrices and Determinants Formula Sheet

All key Matrices and Determinants formulas for JEE Main & Advanced: matrix operations, transpose, determinants, minors, cofactors, adjoint, inverse, Cramer's rule.

8 min read Updated Jun 2026 #formula sheet#quick revision#jee-main

Every must-know result from the Matrices and Determinants chapter, grouped for fast last-minute revision. Everything here is pulled from the chapter topics — use it as your one-page reference before the exam.

Matrix Basics

A matrix is a rectangular array of numbers. Element $a_{ij}$ sits in row $i$, column $j$ (remember: Row then Column).

$$\boxed{\text{Order} = m \times n \quad (m \text{ rows} \times n \text{ columns})}$$

Compact notation: $A = [a_{ij}]_{m \times n}$

Trace (sum of diagonal elements of a square matrix):

$$\boxed{\text{tr}(A) = \sum_{i=1}^{n} a_{ii}}$$

Equality of matrices:

$$\boxed{A = B \iff [a_{ij}]_{m \times n} = [b_{ij}]_{m \times n}}$$

(Same order and all corresponding elements equal.)

Memory hook: $a_{ij}$ = “RC Cola” — Row first, Column second.

Types of Matrices

TypeDefining condition
Row matrixOrder $1 \times n$
Column matrixOrder $m \times 1$
Square matrixRows = columns ($n \times n$)
Diagonal$a_{ij} = 0$ when $i \neq j$
ScalarDiagonal matrix with all diagonal elements equal
Identity $I_n$Diagonal $= 1$, rest $= 0$; $AI = IA = A$
Null / zero $O$All elements $= 0$; $A + O = A$
Upper triangular$a_{ij} = 0$ when $i > j$
Lower triangular$a_{ij} = 0$ when $i < j$
Symmetric$A = A^T$, i.e. $a_{ij} = a_{ji}$
Skew-symmetric$A = -A^T$, i.e. $a_{ij} = -a_{ji}$ (diagonal must be $0$)
High-yield distinction
Symmetric ($A = A^T$): diagonal can be anything. Skew-symmetric ($A = -A^T$): all diagonal elements are forced to $0$, since $a_{ii} = -a_{ii}$.

Matrix Algebra

Addition, Subtraction, Scalar Multiplication

Defined only for matrices of the same order:

$$\boxed{(A \pm B)_{ij} = a_{ij} \pm b_{ij}} \qquad \boxed{(kA)_{ij} = k\, a_{ij}}$$

Addition is commutative, associative, has identity $O$ and inverse $-A$.

Scalar multiplication properties: $k(lA) = (kl)A$, $\;k(A+B) = kA + kB$, $\;(k+l)A = kA + lA$, $\;1\cdot A = A$, $\;(-1)A = -A$.

Matrix Multiplication

Possible only if columns of $A$ = rows of $B$. If $A$ is $m \times n$ and $B$ is $n \times p$, then $AB$ is $m \times p$.

$$\boxed{(AB)_{ik} = \sum_{j=1}^{n} a_{ij}\, b_{jk}}$$
PropertyResult
Commutative?$AB \neq BA$ in general (NOT commutative)
Associative$(AB)C = A(BC)$
Distributive (left)$A(B+C) = AB + AC$
Distributive (right)$(A+B)C = AC + BC$
Identity$AI = IA = A$
Zero$AO = OA = O$
Scalar$k(AB) = (kA)B = A(kB)$
Two classic JEE traps
  1. $AB = O$ does not imply $A = O$ or $B = O$ (no zero-product property).
  2. $AB = AC$ does not imply $B = C$ unless $A$ is invertible ($|A| \neq 0$).

Powers of a Matrix

For a square matrix: $A^n = \underbrace{A \cdot A \cdots A}_{n \text{ times}}$

ResultCondition
$A^m A^n = A^{m+n}$always
$(A^m)^n = A^{mn}$always
$A^m B^m = (AB)^m$only if $AB = BA$
$I^n = I$always
$D^n = \text{diag}(a^n, b^n, \dots)$diagonal $D$

Nilpotent: $A^k = O$ for some $k$. Idempotent: $A^2 = A$. Involutory: $A^2 = I$.

Transpose

$$\boxed{A^T = [a_{ji}]_{n \times m}}$$
PropertyResult
Double transpose$(A^T)^T = A$
Addition$(A+B)^T = A^T + B^T$
Scalar$(kA)^T = kA^T$
Product (order reverses)$(AB)^T = B^T A^T$
Extended$(ABC)^T = C^T B^T A^T$

Orthogonal matrix: $A^T A = AA^T = I$ (so $A^{-1} = A^T$).

Symmetric / Skew-Symmetric Decomposition

Every square matrix splits uniquely into a symmetric and a skew-symmetric part:

$$\boxed{A = \underbrace{\frac{A + A^T}{2}}_{\text{symmetric}} + \underbrace{\frac{A - A^T}{2}}_{\text{skew-symmetric}}}$$

Determinants: Evaluation

Determinant is a scalar defined only for square matrices; denoted $|A|$ or $\det(A)$.

2×2:

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

3×3 (expansion along Row 1):

$$\boxed{\begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix} = a_1(b_2c_3 - b_3c_2) - b_1(a_2c_3 - a_3c_2) + c_1(a_2b_3 - a_3b_2)}$$

3×3 (Sarrus rule): sum of three “down-right” diagonal products minus three “down-left” anti-diagonal products.

$$|A| = (a_1 b_2 c_3 + b_1 c_2 a_3 + c_1 a_2 b_3) - (c_1 b_2 a_3 + a_1 c_2 b_3 + b_1 a_2 c_3)$$
Sarrus is 3×3 only
Sarrus rule works only for $3 \times 3$. For $n \geq 4$ use cofactor expansion or row reduction.

Special / Instant Determinants

MatrixDeterminant
DiagonalProduct of diagonal elements
Triangular (upper/lower)Product of diagonal elements
Identity$\|I_n\| = 1$
Zero matrix$\|O\| = 0$
Any zero row or column$\|A\| = 0$
Two identical rows/columns$\|A\| = 0$
Proportional rows/columns$\|A\| = 0$

Minors and Cofactors

Minor $M_{ij}$ = determinant of submatrix after deleting row $i$ and column $j$.

Cofactor:

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

Sign pattern (checkerboard, $+$ at $(1,1)$):

$$\begin{bmatrix} + & - & + \\ - & + & - \\ + & - & + \end{bmatrix}$$

Cofactor expansion — row (fix row $i$, sum over $j$):

$$\boxed{|A| = \sum_{j=1}^{n} a_{ij}\, C_{ij}}$$

Column (fix column $j$, sum over $i$):

$$\boxed{|A| = \sum_{i=1}^{n} a_{ij}\, C_{ij}}$$

Cross-row vanishing property (key for adjoint):

$$\boxed{\sum_{j=1}^{n} a_{ij}\, C_{kj} = 0 \quad (i \neq k)}$$
Smart expansion
Always expand along the row or column with the most zeros — every zero kills one minor calculation.

Properties of Determinants

#PropertyFormula
1Transpose$\|A^T\| = \|A\|$
2Row/column interchange$R_i \leftrightarrow R_j \Rightarrow \|A\| \to -\|A\|$
3Identical rows/columns$\|A\| = 0$
4Proportional rows/columns$R_i = kR_j \Rightarrow \|A\| = 0$
5Zero row/column$\|A\| = 0$
6Scale one row by $k$$\|A\| \to k\|A\|$
7Add multiple of a row$R_i \to R_i + kR_j \Rightarrow \|A\|$ unchanged
8Product$\|AB\| = \|A\|\,\|B\|$
9Identity / zero$\|I\| = 1,\;\; \|O\| = 0$
10Inverse$\|A^{-1}\| = \dfrac{1}{\|A\|}$
11Adjoint$\|\text{adj}(A)\| = \|A\|^{\,n-1}$

Consequences of the product rule:

$$\boxed{|A^n| = |A|^n} \qquad \boxed{|kA| = k^n |A| \;\;(n \times n)}$$
The most-used identity
$|kA| = k^n|A|$ for an $n \times n$ matrix — not just $k|A|$. E.g. for $3 \times 3$, $|3A| = 27|A|$.

Standard Determinant Patterns

Vandermonde:

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

General $n \times n$ Vandermonde: $\displaystyle \prod_{1 \le i < j \le n} (a_j - a_i)$

Cyclic determinant:

$$\begin{vmatrix} a & b & c \\ c & a & b \\ b & c & a \end{vmatrix} = -(a^3 + b^3 + c^3 - 3abc)$$$$= -(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$$

$1+x$ pattern:

$$\begin{vmatrix} 1+a & 1 & 1 \\ 1 & 1+b & 1 \\ 1 & 1 & 1+c \end{vmatrix} = abc + ab + bc + ca$$$$\begin{vmatrix} 1+x & 1 & 1 \\ 1 & 1+x & 1 \\ 1 & 1 & 1+x \end{vmatrix} = x^2(x+3)$$
Note on the cyclic determinant
The chapter writes the cofactor-expanded value of $\begin{vmatrix} a & b & c \\ c & a & b \\ b & c & a \end{vmatrix}$ as $a^3+b^3+c^3-3abc$. Expanding the standard cyclic determinant directly gives $-(a^3+b^3+c^3-3abc) = 3abc - a^3 - b^3 - c^3$; sign depends on the row arrangement, so verify the orientation in the problem before applying.

Adjoint and Inverse

Singular vs non-singular:

$$\boxed{A^{-1} \text{ exists} \iff |A| \neq 0 \;(\text{non-singular})}$$

Adjoint = transpose of the cofactor matrix:

$$\boxed{\text{adj}(A) = [C_{ij}]^T}$$

2×2 shortcut (swap diagonal, negate off-diagonal):

$$\boxed{\text{adj}\begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}}$$

Inverse formula:

$$\boxed{A^{-1} = \frac{1}{|A|}\,\text{adj}(A)}$$

2×2 inverse:

$$\boxed{\begin{bmatrix} a & b \\ c & d \end{bmatrix}^{-1} = \frac{1}{ad-bc}\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}}$$

Adjoint Properties ($n \times n$)

PropertyFormula
Magic product$A\cdot\text{adj}(A) = \text{adj}(A)\cdot A = \|A\|\, I$
Determinant of adjoint$\|\text{adj}(A)\| = \|A\|^{\,n-1}$
Adjoint of transpose$\text{adj}(A^T) = [\text{adj}(A)]^T$
Adjoint of product$\text{adj}(AB) = \text{adj}(B)\cdot\text{adj}(A)$
Adjoint of adjoint$\text{adj}(\text{adj}(A)) = \|A\|^{\,n-2} A$
Adjoint of scalar multiple$\text{adj}(kA) = k^{\,n-1}\text{adj}(A)$

Inverse Properties

PropertyFormula
Inverse of inverse$(A^{-1})^{-1} = A$
Inverse of product (reverses)$(AB)^{-1} = B^{-1}A^{-1}$
Inverse of transpose$(A^T)^{-1} = (A^{-1})^T$
Determinant of inverse$\|A^{-1}\| = \dfrac{1}{\|A\|}$
Inverse of scalar multiple$(kA)^{-1} = \dfrac{1}{k}A^{-1}$
Identity$I^{-1} = I$
Diagonal matrixreciprocal of each diagonal element
Orthogonal matrix$A^{-1} = A^T$
Involutory matrix$A^{-1} = A$ (since $A^2 = I$)
TIA reverses order
Transpose, Inverse, Adjoint all reverse the order of a product: $(AB)^T = B^T A^T$, $\;(AB)^{-1} = B^{-1}A^{-1}$, $\;\text{adj}(AB) = \text{adj}(B)\,\text{adj}(A)$. There is no simple formula for $(A+B)^{-1}$.

Linear Equations

Matrix form of a system: $\boxed{AX = B}$ ($A$ = coefficient matrix, $X$ = variables, $B$ = constants).

Matrix method (when $|A| \neq 0$):

$$\boxed{X = A^{-1}B}$$

Cramer’s rule (when $D = |A| \neq 0$):

$$\boxed{x_i = \frac{D_i}{D}}$$

where $D_i$ replaces the $i$-th column of $A$ with $B$. For 3 variables:

$$x = \frac{D_x}{D}, \quad y = \frac{D_y}{D}, \quad z = \frac{D_z}{D}$$

Consistency of a Square System $AX = B$

ConditionNatureSolutions
$\|A\| \neq 0$ConsistentUnique solution
$\|A\| = 0,\;(\text{adj }A)B \neq O$InconsistentNo solution
$\|A\| = 0,\;(\text{adj }A)B = O$ConsistentInfinitely many solutions

Homogeneous System $AX = O$

  • Always has the trivial solution $X = O$.
  • Non-trivial solutions exist $\iff |A| = 0$ (then infinitely many).
First move, every system
Compute $|A|$ first. If $|A| \neq 0$ you are done — unique solution. Only if $|A| = 0$ do you need the $(\text{adj }A)B$ check.

Geometry Application

Area of a triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$:

$$\boxed{\text{Area} = \frac{1}{2}\left|\begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix}\right|}$$

If this determinant is $0$, the three points are collinear.