JEE Main 2024 Question Paper Solution Discussion Live JEE Main 2024 Question Paper Solution Discussion Live

# Matrices

A rectangular array of m × n numbers (real or complex) in the form of m horizontal lines (called rows) and n vertical lines (called columns) is called a matrix of order m by n, written as m × n matrix. Such an array is enclosed by [ ] or ( ). In this article, we will learn the meaning of matrices, types of matrices, important formulas, etc.

## 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

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

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

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

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

(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

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

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

## 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)

## 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:

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

Determinants

Matrices and Determinants Previous Year Questions

## 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.