Matrices Definition, Properties, Types, Formulas, Solved Examples

Introduction to Matrices

An m × n matrix is usually written as:

\(\begin{array}{l}A=\left[ \begin{matrix} {{a}_{11}} & {{a}_{12}} & ….. & {{a}_{1n}} \\ {{a}_{21}} & {{a}_{22}} & ….. & {{a}_{2n}} \\ \vdots & \vdots & \vdots & \vdots \\ {{a}_{m1}} & {{a}_{m2}} & ….. & {{a}_{mn}} \\ \end{matrix} \right]\end{array} \)

In brief, the above matrix is represented by A = [aij] mxn. The numbers a11, a12, ….. etc., are known as the elements of the matrix A, where aij belongs to the ith row and jth column and is called the (i, j)th element of the matrix A = [aij].

Important Formulas for Matrices 

If A and B are square matrices of order n, and In is a corresponding unit matrix, then

(a) A(adj.A) = | A | In = (adj A) A

(b) | adj A | = | A |n-1 (Thus A (adj A) is always a scalar matrix)

(c) adj (adj.A) = | A |n-2 A

\(\begin{array}{l}(e)\ |adj\,(adj.A)|=|A{{|}^{{{(n-1)}^{2}}}}\end{array} \)

(f) adj (AB) = (adj B) (adj A)

(g) adj (Am) = (adj A)m,

\(\begin{array}{l}(h)\ adj (kA) = {{k}^{n-1}}(adj. A) ,k\in R\end{array} \)

\(\begin{array}{l}(i)\ adj\left( {{I}_{n}} \right)={{I}_{n}}\end{array} \)

(j) adj 0 = 0

(k) A is symmetric ⇒adj A is also symmetric

(l) A is diagonal ⇒adj A is also diagonal

(m) A is triangular ⇒adj A is also triangular

(n) A is singular ⇒| adj A | = 0

Types of Matrices

(i) Symmetric matrix: A square matrix A = [aij] is called a symmetric matrix if aij = aji, for all i, j.

(ii) Skew-symmetric matrix: when aij = – aji

(iii) Hermitian and skew – Hermitian matrix:

\(\begin{array}{l}A={{A}^{\theta }}\end{array} \)

(Hermitian matrix)(Aθ represents conjugate transpose)

\(\begin{array}{l}{{A}^{\theta }}=-A\end{array} \)

(skew-Hermitian matrix)

(iv) Orthogonal matrix: if AAT = In = ATA

(v) Idempotent matrix: if A2 = A

(vi) Involuntary matrix: if A2 = I or A-1 = A

(vii) Nilpotent matrix: A square matrix A is nilpotent; if A= 0, p is an integer.

Trace of Matrix

The trace of a square matrix is the sum of the elements on the main diagonal.

(i) tr(λA_ = λ tr(A)

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

(iii) tr(AB) = tr(BA)

Matrix Transpose

\(\begin{array}{l}(i)\ {{({{A}^{T}})}^{T}}=A \;\;\;\;\;\; \\(ii)\ {{(A\pm B)}^{T}}={{A}^{T}}\pm {{B}^{T}} \;\;\;\;\; \\(iii)\ {{(AB)}^{T}}={{B}^{T}}{{A}^{T}}\\\end{array} \)

\(\begin{array}{l}(iv)\ {{(kA)}^{T}}=k{{\left( A \right)}^{T}} \;\;\;\;\;\; \\ (v)\ {{({{A}_{1}}{{A}_{2}}{{A}_{3}}…….{{A}_{n-1}}{{A}_{n}})}^{T}}=A_{n}^{T}A_{n-1}^{t}……A_{3}^{T}A_{2}^{T}A_{1}^{T}\\\end{array} \)

\(\begin{array}{l}(vi)\ {{I}^{T}}=I \;\;\;\;\;\;\; \\ (vii) tr(A)=t({{A}^{T}})\end{array} \)

Properties of Matrix Multiplication

(i) AB ≠ BA

(ii) (AB)C = A(BC)

(iii) A.(B + C) = A.B + A.C

Adjoint of a Matrix

\(\begin{array}{l}(i)\ A(adj\,A)=(adj\,A)A=|A|{{I}_{n}} \\ (ii)\ |adj\,A|=|A{{|}^{n-1}}\end{array} \)

\(\begin{array}{l}(iii)\ (adj\,\,AB)=(adj\,\,B)(adj\,\,A)\\ (iv)\ adj\,(adj\,A)=|A{{|}^{n-2}}\end{array} \)

Inverse of a Matrix

A-1 exists if A is non-singular, i.e.,

\(\begin{array}{l}|A|\ne 0\end{array} \)

\(\begin{array}{l}(i) {{A}^{-1}}=\frac{1}{|A|}(Adj.A) \\ (ii)\ {{A}^{-1}}A={{I}_{n}}=A{{A}^{-1}} \\ (iii)\ {{({{A}^{T}})}^{-1}}={{({{A}^{-1}})}^{T}} \\ (iv)\ {{({{A}^{-1}})}^{-1}}=A \\ (v)\ |{{A}^{-1}}|=|A{{|}^{-1}}=\frac{1}{|A|}\end{array} \)

Order of a Matrix

A matrix which has m rows and n columns is called a matrix of order m x n.

For example, the order of

\(\begin{array}{l}\begin{bmatrix}\;\;\; 4\;\;\;\;\;\;\;-1\;\;\;\;\;\;\;5 & & \\ \;\;\;6\;\;\;\;\;\;\;\;\;\;8\;\;\;\;\;-7\end{bmatrix}\end{array} \)

matrix is 2 x 3.

Note: (a) The matrix is just an arrangement of certain quantities.

(b) The elements of a matrix may be real or complex numbers. If all the elements of a matrix are real, then the matrix is called a real matrix.

(c) An m x n matrix has m.n elements.

Illustration 1: Construct a 3×4 matrix A = [aij], whose elements are given by aij = 2i + 3j.

Solution: In this problem, I and j are the number of rows and columns, respectively. By substituting the respective values of rows and columns in aij = 2i + 3j, we can construct the required matrix.

Given aij = 2i + 3j

so a11 = 2+3 = 5, a12 = 2+6 = 8

Similarly, a13 = 11, a14=14, a21 = 7, a22=10, a23=13, a24=16,a31=9, a32=12, a33=15, a34=18

\(\begin{array}{l} \therefore\ A=\begin{bmatrix} 5& 8& 11& 14\\ 7& 10& 13& 16\\ 9& 12& 18& 18 \end{bmatrix}\end{array} \)

Illustration 2: Construct a 3 x 4 matrix, whose elements are given by: aij =

\(\begin{array}{l}\frac{1}{2}|-3i+j|\end{array} \)

Solution:

The method for solving this problem is the same as in the above problem.

Since

\(\begin{array}{l}{{a}_{ij}}=\frac{1}{2}|-3i+j|we\,have\end{array} \)

\(\begin{array}{l}{{a}_{11}}=\frac{1}{2}|-3(1)+1|=\frac{1}{2}|-3+1|=\frac{1}{2}|-2|=\frac{2}{2}=1\end{array} \)

\(\begin{array}{l}{{a}_{12}}=\frac{1}{2}|-3(1)+2|=\frac{1}{2}|-3+2|=\frac{1}{2}|-1|=\frac{1}{2}\end{array} \)

\(\begin{array}{l}{{a}_{13}}=\frac{1}{2}|-3(1)+3|=\frac{1}{2}|-3+3|=\frac{1}{2}(0)=0\end{array} \)

\(\begin{array}{l}{{a}_{14}}=\frac{1}{2}|-3(1)+4|=\frac{1}{2}|-3+4|=\frac{1}{2};\,\,\,\,\,\,\,\,\,\,{{a}_{21}}=\frac{1}{2}|-3(2)+1|=\frac{1}{2}|-6+1|=\frac{5}{2}\end{array} \)

\(\begin{array}{l}{{a}_{22}}=\frac{1}{2}|-3(2)+2|=\frac{1}{2}|-6+2|=\frac{4}{2}=2;\,\,\,\,\,\,\,\,\,\,{{a}_{23}}=\frac{1}{2}|-3(2)+3|=\frac{1}{2}|-6+3|=\frac{3}{2}\end{array} \)

\(\begin{array}{l}{{a}_{24}}=\frac{1}{2}|-3(2)+4|=\frac{1}{2}|-6+4|=\frac{2}{2}=1\\ Similarly\,\,{{a}_{31}}=4,{{a}_{32}}=\frac{7}{2},{{a}_{33}}=3,{{a}_{34}}=\frac{5}{2}\end{array} \)

Hence, the required matrix is given by

\(\begin{array}{l} A = \begin{bmatrix} 1& \frac{1}{2}& 0& \frac{1}{2} \\ \frac{5}{2}& 2& \frac{3}{2}& 1 \\ 4& \frac{7}{2}& 3& \frac{5}{2} \end{bmatrix}\end{array} \)

Trace of a Matrix

Let A = [aij]nxn and B = [bij]nxn and λ be a scalar,

(i) tr(λA) = λ tr(A) (ii) tr(A + B) = tr(A) + tr(B) (iii) tr(AB) = tr(BA)

Square Matrix Types

Examples of Square Matrices are as follows :

Square Matrix Types

Transpose of Matrix

The matrix obtained from a given matrix A by changing its rows into columns or columns into rows is called the transpose of matrix A and is denoted by AT or A’. From the definition, it is obvious that if the order of A is m x n, then the order of AT becomes n x m; For example, transpose of a matrix.

\(\begin{array}{l}\begin{bmatrix} a_1 &a_2 &a_3 \\ b_1& b_2& b_3 \end{bmatrix}_{2\times3} is \begin{bmatrix} a_1 & b_1\\ a_2 & b_2\\ a_3 & b_3 \end{bmatrix}_{3\times2}\end{array} \).

Properties of Transpose of Matrix

(i) (AT)T= A (ii) (A + B)T = AT+ BT (iii) (AB)T = BTAT (iv) (kA)T = k(A)T

(v) (A1A2A3 ……An-1An)T =

\(\begin{array}{l}A_{n}^{T}A_{n-1}^{T}…..A_{3}^{T}A_{2}^{T}A_{1}^{T}\end{array} \)

(vi) IT = I (vii) tr(A) = tr(AT)

Problems on Matrices

Illustration 3: If

\(\begin{array}{l}A=\begin{bmatrix} 1 &-2 &3 \\ -4 & 2 & 5 \end{bmatrix}\ and\ B=\begin{bmatrix} 1 &3 \\ -1&0 \\ 2&4 \end{bmatrix}\end{array} \)

. then prove that (AB)T = BTAT.

Solution:

By obtaining the transpose of AB, i.e., (AB)T and multiplying BT and AT, we can easily get the result.

Here, AB =

\(\begin{array}{l}\begin{bmatrix} 1 & -2 &3 \\ -4& 2& 5 \end{bmatrix}  \begin{bmatrix} 1 &3 \\ -1&0 \\ 2& 4 \end{bmatrix}=\begin{bmatrix} 1(1)-2(-1)+3(2) &1(3)-2(0)+3(4) \\ -4(1)+2(-1)+5(2) &-4(3)+2(0)+5(4) \end{bmatrix} = \begin{bmatrix} 9 &15 \\ 4& 8 \end{bmatrix}\end{array} \)

.

\(\begin{array}{l}\therefore\ (AB)^{T}=\begin{bmatrix} 9 & 4\\ 15& 8 \end{bmatrix}\\ B^{T}A^{T} =\begin{bmatrix} 1 & -1 &2 \\ 3& 0& 4 \end{bmatrix} \begin{bmatrix} 1 &-4 \\ -2&2 \\ 3& 5 \end{bmatrix}\\ =\begin{bmatrix} 1(1)-1(-2)+2(3) &1(-4)-1(2)+2(5) \\ 3(1)+0(-2)+4(3) &3(-4)+0(2)+4(5) \end{bmatrix}\\  = \begin{bmatrix} 9 &4 \\ 15 & 8 \end{bmatrix}\\ =(AB)^{T}\end{array} \)

Illustration 4: If

\(\begin{array}{l}A=\begin{bmatrix} 5 &-1 &3 \\ 0& 1& 2 \end{bmatrix}\ and\ B=\begin{bmatrix} 0 &2 &3 \\ 1& -1 &4 \end{bmatrix}\end{array} \)

. Then what is (B’)’A’ equal to?

Solution:

In this problem, we use the properties of the transpose of a matrix to get the required result.

We have =

\(\begin{array}{l}{({B}’)}'{A}’ =B{A}’=\begin{bmatrix} 0 &2 &3 \\ 1& -1 &4 \end{bmatrix}\begin{bmatrix} 5 &0 \\ -1&1 \\ 3& 2 \end{bmatrix}=\begin{bmatrix} 7 & 8\\ 18& 7 \end{bmatrix}\end{array} \)

.

Illustration 5: If the matrix

\(\begin{array}{l}A=\left[ \begin{matrix} 3-x & 2 & 2 \\ 2 & 4-x & 1 \\ -2 & -4 & -1-x \\ \end{matrix} \right]\end{array} \)

is a singular matrix, then find x. Verify whether AAT = I for that value of x.

Solution:

Using the condition of a singular matrix, i.e., |A| = 0, we get the value of x and then by substituting the value of x in matrix A and multiplying it by its transpose, we will obtain the required result.

Here, A is a singular matrix if |A| = 0, i.e.,

\(\begin{array}{l}\left| \begin{matrix} 3-x & 2 & 2 \\ 2 & 4-x & 1 \\ -2 & -4 & -1-x \\ \end{matrix} \right|=0\end{array} \)

R3 –> R3 + R2

\(\begin{array}{l}\left| \begin{matrix} 3-x & 2 & 2 \\ 2 & 4-x & 1 \\ 0 & -x & -x \\ \end{matrix} \right|=0\end{array} \)

C2 → C2-C3

\(\begin{array}{l}\left| \begin{matrix} 3-x & 0 & 2 \\ 2 & 3-x & 1 \\ 0 & 0 & -x \\ \end{matrix} \right|=0\end{array} \)

Here, x = 0, 3.

When x = 0,

\(\begin{array}{l}A = \left[ \begin{matrix} 3 & 2 & 2 \\ 2 & 4 & 1 \\ -2 & -4 & -1 \\ \end{matrix} \right]\end{array} \)

\(\begin{array}{l}\therefore A{{A}^{T}}=\left[ \begin{matrix} 3 & 2 & 2 \\ 2 & 4 & 1 \\ -2 & -4 & -1 \\ \end{matrix} \right]\left[ \begin{matrix} 3 & 2 & -2 \\ 2 & 4 & -4 \\ 2 & 1 & -1 \\ \end{matrix} \right]\end{array} \)

\(\begin{array}{l}=\left[ \begin{matrix} 17 & 16 & -16 \\ 16 & 21 & -21 \\ -16 & -21 & 21 \\ \end{matrix} \right]\ne I\end{array} \)

When x = 3, 

\(\begin{array}{l}A =\left[ \begin{matrix} 0 & 2 & 2 \\ 2 & 1 & 1 \\ -2 & -4 & -4 \\ \end{matrix} \right];\,\,\,\end{array} \)

\(\begin{array}{l}\therefore\ A{{A}^{T}}=\left[ \begin{matrix} 0 & 2 & 2 \\ 2 & 1 & 1 \\ -2 & -4 & -4 \\ \end{matrix} \right]\left[ \begin{matrix} 0 & 2 & -2 \\ 2 & 1 & -4 \\ 2 & 1 & -4 \\ \end{matrix} \right]=\left[ \begin{matrix} 8 & 4 & -16 \\ 4 & 6 & -12 \\ -16 & -12 & 36 \\ \end{matrix} \right]\ne I\end{array} \)

Note: The simple way to solve is that if A is a singular matrix, then |A| = 0 and |AT| = 0. But |I| is 1.

Hence, AAT ≠ I if |A| = 0.

Illustration 6: If the matrix

\(\begin{array}{l}A = \left[ \begin{matrix} a & b & c \\ b & c & a \\ c & a & b \\ \end{matrix} \right]\end{array} \)

where a, b, c are positive real numbers such that abc = 1 and ATA = I, then find the value of a3 + b3 + c3.

Solution:

Given: abc = 1 and ATA = I

Here,

\(\begin{array}{l}A = \left[ \begin{matrix} a & b & c \\ b & c & a \\ c & a & b \\ \end{matrix} \right].So,{{A}^{T}}=\left[ \begin{matrix} a & b & c \\ b & c & a \\ c & a & b \\ \end{matrix} \right],\end{array} \)

Interchanging rows and columns,

ATA = I (given)

\(\begin{array}{l}\begin{bmatrix}a & b &c \\ b&c & a\\ c & a & b\end{bmatrix}\begin{bmatrix}a & b &c \\ b&c & a\\ c & a & b\end{bmatrix}=\begin{bmatrix}1 & 0 &0 \\ 0&1 & 0\\ 0 & 0 & 1\end{bmatrix}\end{array} \)

Solving the above equation, we have

(a2 + b2 + c2) = 1 and ab + bc + ca = 0  . …..(i)

We know, (a + b + c)2 = (a2 + b2 + c2) + 2(ab + bc + ca)

= 1

or (a + b + c) = 1  ….(ii)

Again ,we have (a3 + b3 + c3 -3abc) = (a + b + c) (a2 + b2 + c2 -ab – bc – ca)

Since abc = 1

Using  (i) and (ii), we have

(a3 + b3 + c3 -3) = 1

or a3 + b3 + c3 =4

Illustration 7: If

\(\begin{array}{l}\left[ \begin{matrix} x+3 & z+4 & 2y-7 \\ -6 & a-1 & 0 \\ b-3 & -21 & 0 \\ \end{matrix} \right]=\left[ \begin{matrix} 0 & 6 & 3y-2 \\ -6 & -3 & 2c+z \\ 2b+4 & -21 & 0 \\ \end{matrix} \right]\end{array} \)

then find the values of a, b, c, x, y and z.

Solution:

As the two matrices are equal, their corresponding elements are also equal. Therefore, by equating the corresponding elements of the given matrices, we will obtain the values of a, b, c, x, y and z.

\(\begin{array}{l}\left[ \begin{matrix} x+3 & z+4 & 2y-7 \\ -6 & a-1 & 0 \\ b-3 & -21 & 0 \\ \end{matrix} \right]=\left[ \begin{matrix} 0 & 6 & 3y-2 \\ -6 & -3 & 2c+z \\ 2b+4 & -21 & 0 \\ \end{matrix} \right]\end{array} \)

Comparing both sides, we get [x + 3 = 0]
⇒ x = – 3

z + 4 = 6  ⇒ z = 6 – 4 ⇒ z = 2

and

2y – 7 = 3y – 2 ⇒ 2y – 3y = -2 + 7 ⇒ -y = 5 ⇒ y = -5

and

a – 1 = -3 ⇒ a = -3 + 1 ⇒ a = -2

and

b – 3 = 2b + 4 ⇒ b – 2b = 4 + 3 ⇒ -b = 7 ⇒ b = -7

and

2c + z = 0 ⇒ 2c + 2 = 0 ⇒ 2c = -2 ⇒ c = -2/2 ⇒ c = -1

Thus

a = -2, b = -7, c = -1, x = -3, y = -5 and z = 2

Illustration 8:

\(\begin{array}{l}20\left[\begin{array}{cc}1 & -\tan \theta / 2 \\ \tan \theta / 2 & 1\end{array}\right]\left[\begin{array}{cc}1 & \tan \theta / 2 \\ -\tan \theta / 2 & 1\end{array}\right]^{-1} \text { is equal to} – \\ (1) \left[\begin{array}{cc}\sin \theta & -\cos \theta \\ \cos \theta & \sin \theta\end{array}\right] \\ (2) \left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta\end{array}\right] \\ (3) \left[\begin{array}{cc}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta\end{array}\right] \\ (4) \text { None of these} \\ Solution:\\ \left[\begin{array}{cc}1 & \tan \theta / 2 \\ -\tan \theta / 2 & 1\end{array}\right]^{-1}=\frac{1}{\sec ^{2} \theta / 2}\left[\begin{array}{cc}1 & -\tan \theta / 2 \\ \tan \theta / 2 & 1\end{array}\right] \\ \therefore \quad Product \begin{array}{l} =\frac{1}{\sec ^{2} \theta / 2}\left[\begin{array}{cc} 1 & -\tan \theta / 2 \\ \tan \theta / 2 & 1 \end{array}\right]\left[\begin{array}{cc} 1 & -\tan \theta / 2 \\ \tan \theta / 2 & 1 \end{array}\right] \\ =\frac{1}{\sec ^{2} \theta / 2}\left[\begin{array}{cc} 1-\tan ^{2} \theta / 2 & -2 \tan \theta / 2 \\ 2 \tan \theta / 2 & 1-\tan ^{2} \theta / 2 \end{array}\right] \\ =\left[\begin{array}{cc} \cos ^{2} \theta / 2-\sin ^{2} \theta / 2 & -2 \sin \theta / 2 \cos \theta / 2 \\ 2 \sin \theta / 2 \cos \theta / 2 & \cos ^{2} \theta / 2-\sin ^{2} \theta / 2 \end{array}\right] \\ =\left[\begin{array}{cc} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right] \end{array}\end{array} \)

Illustration 9:

\(\begin{array}{l}\text \ If \ A=\left[\begin{array}{lll}0 & -1 & 2 \\ 2 & -2 & 0\end{array}\right], B=\left[\begin{array}{ll}0 & 1 \\ 1 & 0 \\ 1 & 1\end{array}\right] \ and \ M=A B, \ then \ M^{-1} \ is \ equal \ to -\\ (1) \left[\begin{array}{cc}2 & -2 \\ 2 & 1\end{array}\right]\\ (2) \left[\begin{array}{cc}1 / 3 & 1 / 3 \\ -1 / 3 & 1 / 6\end{array}\right]\\ (3) \left[\begin{array}{cc}1 / 3 & -1 / 3 \\ 1 / 3 & 1 / 6\end{array}\right]\\ (4) \left[\begin{array}{cc}1 / 3 & -1 / 3 \\ -1 / 3 & 1 / 6\end{array}\right]\\ Solution: M=\left[\begin{array}{lll}0 & -1 & 2 \\ 2 & -2 & 0\end{array}\right]\left[\begin{array}{ll}0 & 1 \\ 1 & 0 \\ 1 & 1\end{array}\right]=\left[\begin{array}{cc}1 & 2 \\ -2 & 2\end{array}\right]\\ |M|=6, \operatorname{adj} M=\left[\begin{array}{cc}2 & -2 \\ 2 & 1\end{array}\right] \\ M^{-1}=\frac{1}{6}\left[\begin{array}{cc}2 & -2 \\ 2 & 1\end{array}\right]=\left[\begin{array}{cc}1 / 3 & -1 / 3 \\ 1 / 3 & 1 / 6\end{array}\right]\end{array} \)

Answer: [3]

Frequently Asked Questions on Matrices

Q1

What is meant by a null matrix?

A null matrix is a matrix whose all the elements are zero.

Q2

What is the transpose of a matrix?

The transpose of a matrix is the matrix obtained by interchanging its rows into columns or columns into rows.

Q3

What is meant by a diagonal matrix?

A diagonal matrix is a matrix in which the entries outside the main diagonal are all zeros.

Q4

What is meant by a square matrix?

If the number of rows and columns in a matrix is equal, then it is called a square matrix.

Q5

Give the formula to find the inverse of a matrix A.

The inverse of a matrix A is given by A-1 = Adj A/det A. Here, adj A is the adjoint of matrix A, and det A is the determinant of A.

Q6

What do you mean by Hermitian matrix?

If a square matrix is equal to its conjugate transpose, then that matrix is called a Hermitian matrix.

Q7

What do you mean by conjugate matrix?

The conjugate matrix of matrix A is found by replacing the corresponding elements of matrix A with its conjugate complex numbers.

Q8

What do you mean by a skew-symmetric matrix?

A skew-symmetric matrix is a matrix whose transpose is equal to the negative of the matrix. If A is skew-symmetric, then AT = -A.

Er. Neeraj K.Anand is a freelance mentor and writer who specializes in Engineering & Science subjects. Neeraj Anand received a B.Tech degree in Electronics and Communication Engineering from N.I.T Warangal & M.Tech Post Graduation from IETE, New Delhi. He has over 30 years of teaching experience and serves as the Head of Department of ANAND CLASSES. He concentrated all his energy and experiences in academics and subsequently grew up as one of the best mentors in the country for students aspiring for success in competitive examinations. In parallel, he started a Technical Publication "ANAND TECHNICAL PUBLISHERS" in 2002 and Educational Newspaper "NATIONAL EDUCATION NEWS" in 2014 at Jalandhar. Now he is a Director of leading publication "ANAND TECHNICAL PUBLISHERS", "ANAND CLASSES" and "NATIONAL EDUCATION NEWS". He has published more than hundred books in the field of Physics, Mathematics, Computers and Information Technology. Besides this he has written many books to help students prepare for IIT-JEE and AIPMT entrance exams. He is an executive member of the IEEE (Institute of Electrical & Electronics Engineers. USA) and honorary member of many Indian scientific societies such as Institution of Electronics & Telecommunication Engineers, Aeronautical Society of India, Bioinformatics Institute of India, Institution of Engineers. He has got award from American Biographical Institute Board of International Research in the year 2005.

CBSE Class 12 Maths Syllabus 2025-26 with Marks Distribution

The table below shows the marks weightage along with the number of periods required for teaching. The Maths theory paper is of 80 marks, and the internal assessment is of 20 marks which totally comes out to be 100 marks.

CBSE Class 12 Maths Syllabus And Marks Distribution 2023-24

Max Marks: 80

No.UnitsMarks
I.Relations and Functions08
II.Algebra10
III.Calculus35
IV.Vectors and Three – Dimensional Geometry14
V.Linear Programming05
VI.Probability08
Total Theory80
Internal Assessment20
Grand Total100

Unit-I: Relations and Functions

1. Relations and Functions

Types of relations: reflexive, symmetric, transitive and equivalence relations. One to one and onto functions.

2. Inverse Trigonometric Functions

Definition, range, domain, principal value branch. Graphs of inverse trigonometric functions.

Unit-II: Algebra

1. Matrices

Concept, notation, order, equality, types of matrices, zero and identity matrix, transpose of a matrix, symmetric and skew symmetric matrices. Operations on matrices: Addition and multiplication and multiplication with a scalar. Simple properties of addition, multiplication and scalar multiplication. Noncommutativity of multiplication of matrices and existence of non-zero matrices whose product is the zero matrix (restrict to square matrices of order 2). Invertible matrices and proof of the uniqueness of inverse, if it exists; (Here all matrices will have real entries).

2. Determinants

Determinant of a square matrix (up to 3 x 3 matrices), minors, co-factors and applications of determinants in finding the area of a triangle. Adjoint and inverse of a square matrix. Consistency, inconsistency and number of solutions of system of linear equations by examples, solving system of linear equations in two or three variables (having unique solution) using inverse of a matrix.

Unit-III: Calculus

1. Continuity and Differentiability

Continuity and differentiability, derivative of composite functions, chain rule, derivative of inverse trigonometric functions like sin-1 x, cos-1 x and tan-1 x, derivative of implicit functions. Concept of exponential and logarithmic functions.
Derivatives of logarithmic and exponential functions. Logarithmic differentiation, derivative of functions expressed in parametric forms. Second order derivatives.

2. Applications of Derivatives

Applications of derivatives: rate of change of quantities, increasing/decreasing functions, maxima and minima (first derivative test motivated geometrically and second derivative test given as a provable tool). Simple problems (that illustrate basic principles and understanding of the subject as well as real-life situations).

3. Integrals 

Integration as inverse process of differentiation. Integration of a variety of functions by substitution, by partial fractions and by parts, Evaluation of simple integrals of the following types and problems based on them.

Fundamental Theorem of Calculus (without proof). Basic properties of definite integrals and evaluation of definite integrals.

4. Applications of the Integrals

Applications in finding the area under simple curves, especially lines, circles/ parabolas/ellipses (in standard form only)

5. Differential Equations

Definition, order and degree, general and particular solutions of a differential equation. Solution of differential equations by method of separation of variables, solutions of homogeneous differential equations of first order and first degree. Solutions of linear differential equation of the type:

dy/dx + py = q, where p and q are functions of x or constants.

dx/dy + px = q, where p and q are functions of y or constants.

Unit-IV: Vectors and Three-Dimensional Geometry

1. Vectors

Vectors and scalars, magnitude and direction of a vector. Direction cosines and direction ratios of a vector. Types of vectors (equal, unit, zero, parallel and collinear vectors), position vector of a point, negative of a vector, components of a vector, addition of vectors, multiplication of a vector by a scalar, position vector of a point dividing a line segment in a given ratio. Definition, Geometrical Interpretation, properties and application of scalar (dot) product of vectors, vector (cross) product of vectors.

2. Three – dimensional Geometry

Direction cosines and direction ratios of a line joining two points. Cartesian equation and vector equation of a line, skew lines, shortest distance between two lines. Angle between two lines.

Unit-V: Linear Programming

1. Linear Programming

Introduction, related terminology such as constraints, objective function, optimization, graphical method of solution for problems in two variables, feasible and infeasible regions (bounded or unbounded), feasible and infeasible solutions, optimal feasible solutions (up to three non-trivial constraints).

Unit-VI: Probability

1. Probability

Conditional probability, multiplication theorem on probability, independent events, total probability, Bayes’ theorem, Random variable and its probability distribution, mean of random variable.

Students can go through the CBSE Class 12 Syllabus to get the detailed syllabus of all subjects. Get access to interactive lessons and videos related to Maths and Science with ANAND CLASSES’S App/ Tablet.

Frequently Asked Questions on CBSE Class 12 Maths Syllabus 2025-26

Q1

Is Calculus an important chapter in the CBSE Class 12 Maths Syllabus?

Yes, Calculus is an important chapter in the CBSE Class 12 Maths Syllabus. It is for 35 marks which means that if a student is thorough with this chapter will be able to pass the final exam.

Q2

How many units are discussed in the CBSE Class 12 Maths Syllabus?

In the CBSE Class 12 Maths Syllabus, about 6 units are discussed, which contains a total of 13 chapters.

Q3

How many marks are allotted for internals in the CBSE Class 12 Maths syllabus?

About 20 marks are allotted for internals in the CBSE Class 12 Maths Syllabus. Students can score it with ease through constant practice.