ALGEBRA OF MATRICES

Table of Contents

§1. Basic Concepts

Consider the system of equations:

3x + 4y − 3z = 5
2x − 9y + 7z = 4
4x − 2y + z = 2
6x + 8y − 3z = 1

Here x, y and z are unknowns and their coefficients are all numbers. Arranging the coefficients in the order in which they occur in the equations and enclosing them in square brackets, we obtain the rectangular array of the form:

[
[ 3 4 -3 ]
[ 2 -9 7 ]
[ 4 -2 1 ]
[ 6 8 -3 ]
]

This rectangular array is an example of a matrix. The horizontal lines are called rows or row vectors, and the vertical lines are called columns or column vectors of the matrix. There are 4 rows and 3 columns in this matrix. Therefore, it is a matrix of the type 4 × 3. The difference between a matrix and a number should be clearly understood. A matrix is not a number; it has got no numerical value. It is something formed by the help of numbers. It is just an ordered collection of numbers arranged in the form of a rectangular array. Simply speaking, a matrix is a number with a proper arrangement.

We shall use capital letters to denote matrices.

Thus,

A = [ [ 3 0 ]
[ -2 1 ] ]

B = [ [ 0 0 0 ]
[ 0 0 0 ] ]

are both matrices. They are of the type 2 × 2 and 2 × 3 respectively.

§2. Matrix – Definition

A set of numbers (real or complex) arranged in the form of a rectangular array having m rows and n columns is called an m × n matrix (to be read as ‘m by n matrix’).

An m × n matrix is usually written as:

A = [
[ 5 4 6 10 ]
[ 3 8 2 11 ]
[ 4 3 7 12 ]
]

In a compact form, the above matrix is represented by A = [aᵢⱼ], where i = 1, 2, …, m and j = 1, 2, …, n. The element aᵢⱼ belongs to the i-th row and j-th column. The first suffix i always denotes the row and the second suffix j denotes the column in which the element occurs.

Field of a Matrix

A matrix over a field F is a set of m × n elements arranged in the form of a rectangular array having m rows and n columns, all the elements of the matrix belonging to the field F.

In the present treatment, the elements of a matrix shall be assumed to be complex numbers, unless stated otherwise.

Order of a Matrix

If a matrix has m rows and n columns, it is said to be of order m × n.

Examples:

• A matrix having 4 rows and 2 columns is of order 4 × 2.
• A matrix having 3 rows and 1 column is of order 3 × 1.

Example

Write the elements a₁₁, a₃₂, a₂₄ for the matrix:

A = [
[ a11 a12 … a1n ]
[ a21 a22 … a2n ]
[ ⋮ ⋮ ⋮ ]
[ am1 am2 … amn ]
]

Solution:
The matrix has 3 rows and 4 columns, hence it is of order 3 × 4.

• a₁₁ = 5 (element of first row and first column)
• a₃₂ = 3 (element of third row and second column)
• a₂₄ = 11 (element of second row and fourth column)

§3. Special Types of Matrices

(i) Square Matrix

Definition: An m × n matrix for which m = n (i.e., number of rows equals number of columns) is called a square matrix. The number m ( = n ) is called the order of the square matrix. It is also called an n‑rowed square matrix.

Thus, in a square matrix, we have the same number of rows and columns.

The elements a₁₁, a₂₂, a₃₃, …, aₙₙ are called the diagonal elements and the diagonal along which they lie is called the principal diagonal of the matrix.

Example:

A =
[
[ 0 1 2 3 ]
[ 2 3 1 0 ]
[ 5 0 1 2 ]
[ 0 0 1 2 ]
]

This is a square matrix of order 4 × 4. The elements 0, 3, 1, 2 constitute the principal diagonal of the matrix.

(ii) Unit Matrix or Identity Matrix

Definition: A square matrix in which each of whose diagonal elements is 1 and each of whose non‑diagonal elements is equal to 0 is called a unit matrix or identity matrix and is denoted by I.

A unit matrix of order n is usually written as Iₙ.

Examples:

Unit matrix of order 3:

I₃ =
[
[ 1 0 0 ]
[ 0 1 0 ]
[ 0 0 1 ]
]

Unit matrix of order 2:

I₂ =
[
[ 1 0 ]
[ 0 1 ]
]

(iii) Null Matrix or Zero Matrix

Definition: The matrix whose elements are all equal to zero is called the null matrix or zero matrix. The null matrix of type m × n is usually denoted by O.

Often, a null matrix is simply denoted by the symbol 0 and read as “zero”.

Examples:

Null matrix of order 3 × 4:

[
[ 0 0 0 0 ]
[ 0 0 0 0 ]
[ 0 0 0 0 ]
]

Null matrix of order 3 × 3:

[
[ 0 0 0 ]
[ 0 0 0 ]
[ 0 0 0 ]
]

(iv) Row Matrix and Column Matrix

Row Matrix: Any 1 × n matrix which has only one row and n columns is called a row matrix or row vector.

Example:

X = [ 2 7 −8 5 11 ] → Row matrix of type 1 × 5

Column Matrix: Any m × 1 matrix which has only one column and m rows is called a column matrix or column vector.

Example:

Y =
[
[ −9 ]
[ 1 ]
[ 3 ]
]

This is a column matrix of type 3 × 1.

(v) Submatrix of a Matrix

Definition: Any matrix obtained by omitting one or more rows and/or columns from a given m × n matrix A is called a submatrix of A.

The matrix A itself is a submatrix of A and can be obtained from A by omitting zero rows or columns.

A square submatrix of a square matrix A is called a principal submatrix if its diagonal elements are the diagonal elements of the matrix A.

Principal submatrices are obtained only by omitting corresponding rows and columns.

Example:

A =
[
[ 1 2 3 ]
[ 7 11 6 ]
[ 0 8 1 ]
]

then the matrix

[
[ 1 2 ]
[ 7 11 ]
]

is a submatrix of A. It can be obtained from A by omitting the third row and the third column.

(vi) Equality of Two Matrices

Definition: Two matrices A = [aᵢⱼ] and B = [bᵢⱼ] are said to be equal if:

They are of the same order, and
aᵢⱼ = bᵢⱼ for each pair of subscripts i and j.

If two matrices A and B are equal, we write A = B. If they are not equal, we write A ≠ B.

§4. Operations on Matrices

(i) Addition of Matrices

Definition: Two matrices can be added only when they are of the same order. If

A = [ aᵢⱼ ] and B = [ bᵢⱼ ]

are two matrices of the same order m × n, then their sum A + B is defined as the matrix whose elements are obtained by adding the corresponding elements of A and B.

Thus,

A + B = [ aᵢⱼ + bᵢⱼ ]

for all values of i = 1, 2, …, m and j = 1, 2, …, n.

Example 1

A =
[
[ 2 3 1 ]
[ 4 0 −1 ]
]

and

B =
[
[ 5 1 2 ]
[ −3 6 4 ]
]

find A + B.

Solution:

A + B =
[
[ 2+5 3+1 1+2 ]
[ 4−3 0+6 −1+4 ]
]

=
[
[ 7 4 3 ]
[ 1 6 3 ]
]

(ii) Subtraction of Matrices

Definition: If A and B are two matrices of the same order, then A − B is defined as the matrix obtained by subtracting the corresponding elements of B from A.

Thus,

A − B = [ aᵢⱼ − bᵢⱼ ]

Example 2

A =
[
[ 6 4 ]
[ 2 1 ]
]

and

B =
[
[ 1 3 ]
[ 5 2 ]
]

find A − B.

Solution:

A − B =
[
[ 6−1 4−3 ]
[ 2−5 1−2 ]
]

=
[
[ 5 1 ]
[ −3 −1 ]
]

(iii) Scalar Multiplication of a Matrix

Definition: If k is a scalar (real or complex number) and A = [ aᵢⱼ ] is a matrix, then the matrix obtained by multiplying each element of A by k is called the scalar multiple of A and is denoted by kA.

Thus,

kA = [ k aᵢⱼ ]

Example 3

A =
[
[ 1 −2 3 ]
[ 0 4 −1 ]
]

find 3A.

Solution:

3A =
[
[ 3×1 3×(−2) 3×3 ]
[ 3×0 3×4 3×(−1) ]
]

=
[
[ 3 −6 9 ]
[ 0 12 −3 ]
]

Properties of Matrix Addition

If A, B and C are matrices of the same order, then:

Commutative law: A + B = B + A
Associative law: (A + B) + C = A + (B + C)
Existence of zero matrix: A + O = A
Existence of additive inverse: A + (−A) = O

where O denotes the null matrix of the same order as A.

§5. Multiplication of Matrices

Condition for Multiplication

Let A be a matrix of order m × n and B be a matrix of order n × p. Then the product AB is defined and is a matrix of order m × p.

If the number of columns of A is not equal to the number of rows of B, then the product AB is not defined.

Definition of Matrix Multiplication

Let

A = [ aᵢⱼ ] of order m × n
B = [ bᵢⱼ ] of order n × p

Then the element in the i-th row and j-th column of the product matrix AB is obtained by multiplying the elements of the i-th row of A with the corresponding elements of the j-th column of B and then adding the products.

Thus, the element cᵢⱼ of the matrix AB is given by:

cᵢⱼ = aᵢ₁b₁ⱼ + aᵢ₂b₂ⱼ + … + aᵢₙbₙⱼ

Example 1

A =
[
[ 1 2 3 ]
[ 4 5 6 ]
]

and

B =
[
[ 1 2 ]
[ 3 4 ]
[ 5 6 ]
]

find AB.

Solution:

Here, A is of order 2 × 3 and B is of order 3 × 2, so the product AB is defined and will be of order 2 × 2.

AB =
[
[ (1×1 + 2×3 + 3×5) (1×2 + 2×4 + 3×6) ]
[ (4×1 + 5×3 + 6×5) (4×2 + 5×4 + 6×6) ]
]

=
[
[ 22 28 ]
[ 49 64 ]
]

Example 2

A =
[
[ 2 1 ]
[ 3 4 ]
]

and

B =
[
[ 1 0 ]
[ 2 1 ]
]

find AB and BA.

Solution:

AB =
[
[ (2×1 + 1×2) (2×0 + 1×1) ]
[ (3×1 + 4×2) (3×0 + 4×1) ]
]

=
[
[ 4 1 ]
[ 11 4 ]
]

BA =
[
[ (1×2 + 0×3) (1×1 + 0×4) ]
[ (2×2 + 1×3) (2×1 + 1×4) ]
]

=
[
[ 2 1 ]
[ 7 6 ]
]

Since AB ≠ BA, matrix multiplication is not commutative.

Properties of Matrix Multiplication

Let A, B and C be matrices of suitable orders. Then:

Associative law: (AB)C = A(BC)
Distributive law: A(B + C) = AB + AC
Distributive law: (A + B)C = AC + BC
Not commutative: AB ≠ BA (in general)

Multiplication by Identity Matrix

If A is a square matrix of order n, then:

IₙA = AIₙ = A

where Iₙ is the unit matrix of order n.

Multiplication by Zero Matrix

If A is any matrix and O is the zero matrix of suitable order, then:

AO = OA = O

§6. Symmetric and Skew-Symmetric Matrices

(i) Symmetric Matrix

Definition: A square matrix A is said to be a symmetric matrix if

Aᵀ = A

That is, the transpose of the matrix is equal to the matrix itself.

In a symmetric matrix, the elements satisfy the condition:

aᵢⱼ = aⱼᵢ for all i and j.

Example:

A =
[
[ 2 3 4 ]
[ 3 5 6 ]
[ 4 6 1 ]
]

Since the elements on both sides of the principal diagonal are equal, the matrix A is a symmetric matrix.

(ii) Skew-Symmetric Matrix

Definition: A square matrix A is said to be skew-symmetric if

Aᵀ = −A

This implies that:

aᵢⱼ = −aⱼᵢ for all i and j.

Also, in a skew-symmetric matrix, all the diagonal elements are zero.

Example:

A =
[
[ 0 2 −1 ]
[ −2 0 4 ]
[ 1 −4 0 ]
]

This matrix satisfies Aᵀ = −A. Hence, A is a skew-symmetric matrix.

Important Properties

Every symmetric matrix is necessarily a square matrix.
Every skew-symmetric matrix is necessarily a square matrix.
The diagonal elements of a skew-symmetric matrix are always zero.
If A is symmetric, then Aᵀ is also symmetric.
If A is skew-symmetric, then Aᵀ is also skew-symmetric.

Decomposition of a Matrix into Symmetric and Skew-Symmetric Parts

Every square matrix A can be expressed uniquely as the sum of a symmetric matrix and a skew-symmetric matrix.

That is,

A = (1/2)(A + Aᵀ) + (1/2)(A − Aᵀ)

where:

• (1/2)(A + Aᵀ) is a symmetric matrix
• (1/2)(A − Aᵀ) is a skew-symmetric matrix

Example

A =
[
[ 1 2 3 ]
[ 4 5 6 ]
[ 7 8 9 ]
]

find the symmetric and skew-symmetric matrices.

Solution:

Aᵀ =
[
[ 1 4 7 ]
[ 2 5 8 ]
[ 3 6 9 ]
]

Symmetric part:

(1/2)(A + Aᵀ) =
[
[ 1 3 5 ]
[ 3 5 7 ]
[ 5 7 9 ]
]

Skew-symmetric part:

(1/2)(A − Aᵀ) =
[
[ 0 −1 −2 ]
[ 1 0 −1 ]
[ 2 1 0 ]
]

§7. Inverse of a Matrix

Definition of Inverse Matrix

A square matrix A is said to be invertible or non-singular if there exists a square matrix B of the same order such that:

AB = BA = I

where I is the unit (identity) matrix of the same order. The matrix B is called the inverse of A and is denoted by A⁻¹.

If no such matrix exists, then A is said to be singular and has no inverse.

Necessary Condition for Existence of Inverse

A square matrix A has an inverse if and only if

|A| ≠ 0

That is, the determinant of A must be non-zero.

Inverse of a Matrix by Adjoint Method

If A is a non-singular square matrix, then its inverse is given by:

A⁻¹ = (1 / |A|) adj(A)

where adj(A) denotes the adjoint of A.

Adjoint of a Matrix

Definition: The adjoint of a square matrix A is defined as the transpose of the cofactor matrix of A and is denoted by adj(A).

Steps to Find Inverse by Adjoint Method

Find the determinant |A| of the given matrix A.
Find the cofactor of each element of A.
Arrange the cofactors in matrix form and take its transpose to obtain adj(A).
Multiply adj(A) by 1 / |A| to obtain A⁻¹.

Example 1

Find the inverse of the matrix:

A =
[
[ 1 2 ]
[ 3 4 ]
]

Solution:

Step 1: Determinant of A

|A| = (1×4 − 2×3) = 4 − 6 = −2 ≠ 0

Hence, the inverse of A exists.

Step 2: Cofactors of A

C₁₁ = 4 , C₁₂ = −3
C₂₁ = −2 , C₂₂ = 1

Step 3: Adjoint of A

adj(A) =
[
[ 4 −2 ]
[ −3 1 ]
]

Step 4: Inverse of A

A⁻¹ = (1 / −2)
[
[ 4 −2 ]
[ −3 1 ]
]

=
[
[ −2 1 ]
[ 3/2 −1/2 ]
]

Example 2

Find the inverse of the matrix:

A =
[
[ 2 1 1 ]
[ 1 2 1 ]
[ 1 1 2 ]
]

Solution:

Step 1: Determinant of A

|A| = 2(2×2 − 1×1) − 1(1×2 − 1×1) + 1(1×1 − 2×1)
= 2(3) − 1(1) + 1(−1)
= 6 − 1 − 1
= 4 ≠ 0

Hence, the inverse of A exists.

Step 2: Cofactor matrix of A

[
[ 3 −1 −1 ]
[ −1 3 −1 ]
[ −1 −1 3 ]
]

Step 3: Adjoint of A

adj(A) =
[
[ 3 −1 −1 ]
[ −1 3 −1 ]
[ −1 −1 3 ]
]

Step 4: Inverse of A

A⁻¹ = (1 / 4)
[
[ 3 −1 −1 ]
[ −1 3 −1 ]
[ −1 −1 3 ]
]

Properties of Inverse Matrices

(A⁻¹)⁻¹ = A
(AB)⁻¹ = B⁻¹ A⁻¹
(Aᵀ)⁻¹ = (A⁻¹)ᵀ
(kA)⁻¹ = (1 / k) A⁻¹ , k ≠ 0

§8. Solution of System of Linear Equations by Matrix Method

System of Linear Equations

A set of equations of the form:

a₁₁x + a₁₂y + a₁₃z = b₁
a₂₁x + a₂₂y + a₂₃z = b₂
a₃₁x + a₃₂y + a₃₃z = b₃

is called a system of linear equations in three variables.

Matrix Form of a System

The above system can be written in the matrix form as:

AX = B

where

A =
[
[ a₁₁ a₁₂ a₁₃ ]
[ a₂₁ a₂₂ a₂₃ ]
[ a₃₁ a₃₂ a₃₃ ]
]

X =
[
[ x ]
[ y ]
[ z ]
]

B =
[
[ b₁ ]
[ b₂ ]
[ b₃ ]
]

Condition for Unique Solution

The system AX = B has a unique solution if and only if:

|A| ≠ 0

In this case, the solution is given by:

X = A⁻¹B

Example 1

Solve the system of equations:

x + 2y + z = 6
2x + 3y + 2z = 11
3x + 4y + z = 8

Solution:

Step 1: Write the system in matrix form AX = B

A =
[
[ 1 2 1 ]
[ 2 3 2 ]
[ 3 4 1 ]
]

X =
[
[ x ]
[ y ]
[ z ]
]

B =
[
[ 6 ]
[ 11 ]
[ 8 ]
]

Step 2: Find |A|

|A| = 1(3×1 − 2×4) − 2(2×1 − 2×3) + 1(2×4 − 3×3)
= 1(−5) − 2(−4) + 1(−1)
= −5 + 8 − 1
= 2 ≠ 0

Hence, the system has a unique solution.

Step 3: Find A⁻¹

A⁻¹ = (1 / 2)
[
[ −5 2 1 ]
[ 4 −2 −2 ]
[ −1 2 −1 ]
]

Step 4: Find X = A⁻¹B

X = (1 / 2)
[
[ −5 2 1 ]
[ 4 −2 −2 ]
[ −1 2 −1 ]
]

×
[
[ 6 ]
[ 11 ]
[ 8 ]
]

=
[
[ 1 ]
[ 2 ]
[ 1 ]
]

Therefore,

x = 1 , y = 2 , z = 1

Important Remarks

If |A| = 0, the system may have no solution or infinitely many solutions.
Matrix method is applicable only when the inverse of coefficient matrix exists.
This method provides a systematic and compact way of solving linear equations.

§9. Homogeneous and Non-Homogeneous Systems of Linear Equations

Homogeneous System of Linear Equations

A system of linear equations is said to be homogeneous if all the constant terms are zero.

General form:

a₁₁x + a₁₂y + a₁₃z = 0
a₂₁x + a₂₂y + a₂₃z = 0
a₃₁x + a₃₂y + a₃₃z = 0

In matrix form:

AX = O

where A is the coefficient matrix, X is the column matrix of variables, and O is the null (zero) matrix.

Properties of Homogeneous System

A homogeneous system always has at least one solution, namely the trivial solution:

x = 0 , y = 0 , z = 0

A homogeneous system has a non-trivial solution if and only if:

|A| = 0

If |A| ≠ 0, then the only solution is the trivial solution.

Example 1

Discuss the solutions of the homogeneous system:

x + y + z = 0
2x + 3y + 4z = 0
3x + 4y + 5z = 0

Solution:

Coefficient matrix A =
[
[ 1 1 1 ]
[ 2 3 4 ]
[ 3 4 5 ]
]

|A| = 1(3×5 − 4×4) − 1(2×5 − 4×3) + 1(2×4 − 3×3)
= 1(15 − 16) − 1(10 − 12) + 1(8 − 9)
= −1 + 2 − 1
= 0

Since |A| = 0, the system has infinitely many non-trivial solutions.

Non-Homogeneous System of Linear Equations

A system of linear equations is said to be non-homogeneous if at least one of the constant terms is non-zero.

General form:

a₁₁x + a₁₂y + a₁₃z = b₁
a₂₁x + a₂₂y + a₂₃z = b₂
a₃₁x + a₃₂y + a₃₃z = b₃

where at least one of b₁, b₂, b₃ ≠ 0.

Consistency of a System of Equations

A system of linear equations is said to be consistent if it has at least one solution. Otherwise, it is called inconsistent.

Conditions for Consistency

For the system AX = B:

If |A| ≠ 0, the system is consistent and has a unique solution.
If |A| = 0 and the system has at least one solution, then it has infinitely many solutions.
If |A| = 0 and the system has no solution, then the system is inconsistent.

Important Remarks

Every homogeneous system is always consistent.
A non-homogeneous system may be consistent or inconsistent.
Determinant of the coefficient matrix plays a key role in deciding the nature of solutions.

§10. Rank of a Matrix

Definition of Rank

The rank of a matrix is defined as the maximum number of linearly independent rows or columns in the matrix.

The rank of a matrix A is denoted by rank(A) or ρ(A).

Rank of a Matrix by Minors

The rank of a matrix A is:

• r, if there exists at least one non-zero minor of order r, and
• every minor of order (r + 1) is zero.

Important Results

The rank of a non-zero matrix is at least 1.
The rank of a zero matrix is 0.
For a matrix of order m × n,

rank(A) ≤ min(m, n)

A square matrix A is non-singular if and only if:

rank(A) = order of A

Elementary Transformations and Rank

The rank of a matrix remains unchanged under the following elementary transformations:

Interchanging any two rows or any two columns.
Multiplying any row or column by a non-zero constant.
Adding a multiple of one row (or column) to another row (or column).

Rank of a Matrix by Elementary Transformations

To find the rank of a matrix, we apply elementary row or column transformations to reduce the matrix to a simpler form such as:

• Row echelon form, or
• Reduced row echelon form.

The number of non-zero rows in the reduced form gives the rank of the matrix.

Example 1

Find the rank of the matrix:

A =
[
[ 1 2 3 ]
[ 2 4 6 ]
[ 1 1 1 ]
]

Solution:

Applying elementary row operations:

R₂ → R₂ − 2R₁

A =
[
[ 1 2 3 ]
[ 0 0 0 ]
[ 1 1 1 ]
]

R₃ → R₃ − R₁

A =
[
[ 1 2 3 ]
[ 0 0 0 ]
[ 0 −1 −2 ]
]

Rearranging rows:

[
[ 1 2 3 ]
[ 0 −1 −2 ]
[ 0 0 0 ]
]

There are two non-zero rows.

Hence,

rank(A) = 2

Applications of Rank

To determine the consistency of a system of linear equations.
To find the number of solutions of a system.
To check whether a square matrix is singular or non-singular.
To study linear dependence and independence of rows and columns.

Important Remark

The rank of the coefficient matrix and the augmented matrix plays a crucial role in deciding the nature of solutions of a system of linear equations.

§11. Consistency of System by Rank Method

Statement of the Theorem

Consider a system of linear equations written in matrix form as:

AX = B

where A is the coefficient matrix, X is the column matrix of variables, and B is the column matrix of constants.

Let [A : B] denote the augmented matrix of the system.

Then:

The system is consistent if and only if:

rank(A) = rank([A : B])

If the system is consistent and:

• rank(A) = rank([A : B]) = number of unknowns, then the system has a unique solution.

• rank(A) = rank([A : B]) < number of unknowns, then the system has infinitely many solutions.

rank(A) ≠ rank([A : B])

then the system is inconsistent and has no solution.

Augmented Matrix

The matrix obtained by adjoining the column matrix B to the coefficient matrix A is called the augmented matrix of the system.

A =
[
[ a₁₁ a₁₂ a₁₃ ]
[ a₂₁ a₂₂ a₂₃ ]
[ a₃₁ a₃₂ a₃₃ ]
]

and

B =
[
[ b₁ ]
[ b₂ ]
[ b₃ ]
]

then the augmented matrix is:

[A : B] =
[
[ a₁₁ a₁₂ a₁₃ | b₁ ]
[ a₂₁ a₂₂ a₂₃ | b₂ ]
[ a₃₁ a₃₂ a₃₃ | b₃ ]
]

Example 1

Discuss the consistency of the system:

x + y + z = 6
2x + 2y + 2z = 12
3x + 3y + 3z = 18

Solution:

Coefficient matrix A =
[
[ 1 1 1 ]
[ 2 2 2 ]
[ 3 3 3 ]
]

Constant matrix B =
[
[ 6 ]
[ 12 ]
[ 18 ]
]

Augmented matrix [A : B] =
[
[ 1 1 1 | 6 ]
[ 2 2 2 | 12 ]
[ 3 3 3 | 18 ]
]

Applying elementary row operations:

R₂ → R₂ − 2R₁
R₃ → R₃ − 3R₁

[
[ 1 1 1 | 6 ]
[ 0 0 0 | 0 ]
[ 0 0 0 | 0 ]
]

Thus,

rank(A) = 1
rank([A : B]) = 1

Number of unknowns = 3

Since rank(A) = rank([A : B]) < number of unknowns, the system is consistent and has infinitely many solutions.

Example 2

Discuss the consistency of the system:

x + y + z = 1
2x + 2y + 2z = 2
3x + 3y + 3z = 5

Solution:

Augmented matrix [A : B] =
[
[ 1 1 1 | 1 ]
[ 2 2 2 | 2 ]
[ 3 3 3 | 5 ]
]

Applying elementary row operations:

R₂ → R₂ − 2R₁
R₃ → R₃ − 3R₁

[
[ 1 1 1 | 1 ]
[ 0 0 0 | 0 ]
[ 0 0 0 | 2 ]
]

Here,

rank(A) = 1
rank([A : B]) = 2

Since rank(A) ≠ rank([A : B]), the system is inconsistent and has no solution.

Important Remarks

Rouché–Capelli theorem provides a complete criterion for the consistency of a system of linear equations.
It is applicable to both homogeneous and non-homogeneous systems.
Rank method is especially useful when the determinant of the coefficient matrix is zero.

§12.Consistency of System by Rank Method

Statement of the Theorem

Consider a system of linear equations written in matrix form as:

AX = B

where A is the coefficient matrix, X is the column matrix of variables, and B is the column matrix of constants.

Let [A : B] denote the augmented matrix of the system.

Then:

The system is consistent if and only if:

rank(A) = rank([A : B])

If the system is consistent and:

• rank(A) = rank([A : B]) = number of unknowns, then the system has a unique solution.

• rank(A) = rank([A : B]) < number of unknowns, then the system has infinitely many solutions.

rank(A) ≠ rank([A : B])

then the system is inconsistent and has no solution.

Augmented Matrix

The matrix obtained by adjoining the column matrix B to the coefficient matrix A is called the augmented matrix of the system.

A =
[
[ a₁₁ a₁₂ a₁₃ ]
[ a₂₁ a₂₂ a₂₃ ]
[ a₃₁ a₃₂ a₃₃ ]
]

and

B =
[
[ b₁ ]
[ b₂ ]
[ b₃ ]
]

then the augmented matrix is:

[A : B] =
[
[ a₁₁ a₁₂ a₁₃ | b₁ ]
[ a₂₁ a₂₂ a₂₃ | b₂ ]
[ a₃₁ a₃₂ a₃₃ | b₃ ]
]

Example 1

Discuss the consistency of the system:

x + y + z = 6
2x + 2y + 2z = 12
3x + 3y + 3z = 18

Solution:

Coefficient matrix A =
[
[ 1 1 1 ]
[ 2 2 2 ]
[ 3 3 3 ]
]

Constant matrix B =
[
[ 6 ]
[ 12 ]
[ 18 ]
]

Augmented matrix [A : B] =
[
[ 1 1 1 | 6 ]
[ 2 2 2 | 12 ]
[ 3 3 3 | 18 ]
]

Applying elementary row operations:

R₂ → R₂ − 2R₁
R₃ → R₃ − 3R₁

[
[ 1 1 1 | 6 ]
[ 0 0 0 | 0 ]
[ 0 0 0 | 0 ]
]

Thus,

rank(A) = 1
rank([A : B]) = 1

Number of unknowns = 3

Since rank(A) = rank([A : B]) < number of unknowns, the system is consistent and has infinitely many solutions.

Example 2

Discuss the consistency of the system:

x + y + z = 1
2x + 2y + 2z = 2
3x + 3y + 3z = 5

Solution:

Augmented matrix [A : B] =
[
[ 1 1 1 | 1 ]
[ 2 2 2 | 2 ]
[ 3 3 3 | 5 ]
]

Applying elementary row operations:

R₂ → R₂ − 2R₁
R₃ → R₃ − 3R₁

[
[ 1 1 1 | 1 ]
[ 0 0 0 | 0 ]
[ 0 0 0 | 2 ]
]

Here,

rank(A) = 1
rank([A : B]) = 2

Since rank(A) ≠ rank([A : B]), the system is inconsistent and has no solution.

Important Remarks

Rouché–Capelli theorem provides a complete criterion for the consistency of a system of linear equations.
It is applicable to both homogeneous and non-homogeneous systems.
Rank method is especially useful when the determinant of the coefficient matrix is zero.

§13. Practice Problems.

Section A: Very Short Answer Questions

Define a matrix.
What is the order of a matrix having 3 rows and 5 columns?
Write the general form of an m × n matrix.
What is a square matrix?
What is a zero (null) matrix?
When is the sum of two matrices defined?
State the condition for the existence of inverse of a matrix.
What is meant by the rank of a matrix?
Define a symmetric matrix.
Define a skew-symmetric matrix.

Section B: Short Answer Questions

Write a unit matrix of order 3.
Give an example of a row matrix and a column matrix.
If A is a square matrix, write the condition for A to be non-singular.
State two properties of matrix addition.
State two properties of matrix multiplication.
What is meant by the transpose of a matrix?
When do two matrices become equal?
What is an augmented matrix?
State the condition for a homogeneous system to have non-trivial solutions.
What is the significance of rank in solving linear equations?

Section C: Long Answer Questions

Find the sum and difference of the matrices:

A =
[
[ 1 2 ]
[ 3 4 ]
]

B =
[
[ 5 6 ]
[ 7 8 ]
]

Find 3A − 2B, where:

A =
[
[ 2 0 ]
[ 1 −3 ]
]

B =
[
[ 1 4 ]
[ 2 1 ]
]

Multiply the matrices:

A =
[
[ 1 2 3 ]
[ 4 5 6 ]
]

B =
[
[ 1 0 ]
[ 0 1 ]
[ 1 1 ]
]

Find the transpose of the matrix:

A =
[
[ 2 −1 3 ]
[ 4 0 5 ]
]

Find the inverse of the matrix using adjoint method:

A =
[
[ 1 2 ]
[ 3 5 ]
]

Section D: Application-Based Questions

Solve the system of equations using matrix method:

x + y = 3
2x + 3y = 7

Discuss the consistency of the system:

x + y + z = 1
2x + 2y + 2z = 2
3x + 3y + 3z = 4

Express the given system in matrix form and find its solution:

2x − y + z = 1
3x + y − z = 5
x + 2y + z = 4

Section E: Conceptual & Proof-Based Questions

Prove that matrix multiplication is not commutative.
Show that the inverse of a symmetric matrix is also symmetric, if it exists.
Prove that (AB)⁻¹ = B⁻¹A⁻¹.
Show that every square matrix can be expressed as the sum of a symmetric and a skew-symmetric matrix.

Section F: Objective Type Questions

The rank of a zero matrix is:
(a) 0 (b) 1 (c) 2 (d) Undefined
If |A| ≠ 0, then the system AX = B has:
(a) No solution
(b) Infinite solutions
(c) Unique solution
(d) Two solutions
If A is a square matrix, then A A⁻¹ equals:
(a) 0 (b) A (c) I (d) A²
A symmetric matrix must be:
(a) Rectangular
(b) Square
(c) Zero
(d) Singular

§15. Additional Theory and Remarks

Types of Square Matrices.

Diagonal Matrix
A square matrix in which all non-diagonal elements are zero is called a diagonal matrix.

Example:
[
[ 2 0 0 ]
[ 0 5 0 ]
[ 0 0 7 ]
]

Scalar Matrix
A diagonal matrix in which all diagonal elements are equal is called a scalar matrix.

Example:
[
[ 3 0 0 ]
[ 0 3 0 ]
[ 0 0 3 ]
]

Triangular Matrices

Upper Triangular Matrix
A square matrix in which all elements below the principal diagonal are zero.

Example:
[
[ 1 2 3 ]
[ 0 4 5 ]
[ 0 0 6 ]
]

Lower Triangular Matrix
A square matrix in which all elements above the principal diagonal are zero.

Example:
[
[ 7 0 0 ]
[ 2 5 0 ]
[ 1 4 3 ]
]

Trace of a Matrix

The trace of a square matrix A is defined as the sum of its diagonal elements.

A =
[
[ a₁₁ a₁₂ ]
[ a₂₁ a₂₂ ]
]

then:

trace(A) = a₁₁ + a₂₂

Important Properties of Transpose (Additional)

(A + B)ᵀ = Aᵀ + Bᵀ
(kA)ᵀ = kAᵀ
(AB)ᵀ = BᵀAᵀ

Special Results

If A is a skew-symmetric matrix of odd order, then:

|A| = 0

If A is a square matrix, then:

A + Aᵀ is symmetric
A − Aᵀ is skew-symmetric

Additional Example

Show that A + Aᵀ is symmetric, where:

A =
[
[ 1 2 3 ]
[ 4 5 6 ]
[ 7 8 9 ]
]

Solution:

Aᵀ =
[
[ 1 4 7 ]
[ 2 5 8 ]
[ 3 6 9 ]
]

A + Aᵀ =
[
[ 2 6 10 ]
[ 6 10 14 ]
[10 14 18 ]
]

Since (A + Aᵀ)ᵀ = A + Aᵀ, the matrix is symmetric.

Important Exam Remarks

Every diagonal matrix is a square matrix.
Every scalar matrix is a diagonal matrix, but not conversely.
Trace is defined only for square matrices.
Determinant of a triangular matrix is the product of its diagonal elements.

§16. End-of-Chapter Problems and Additional Examples

Solved Numerical Examples (Additional)

Example 1

Find the inverse of the matrix, if it exists:

A =
[
[ 2 1 ]
[ 5 3 ]
]

Solution:

|A| = (2×3 − 1×5) = 6 − 5 = 1 ≠ 0

Hence, the inverse exists.

adj(A) =
[
[ 3 −1 ]
[ −5 2 ]
]

A⁻¹ =
[
[ 3 −1 ]
[ −5 2 ]
]

Example 2

Find the rank of the matrix:

A =
[
[ 1 2 3 ]
[ 2 4 6 ]
[ 3 6 9 ]
]

Solution:

Applying row operations:

R₂ → R₂ − 2R₁
R₃ → R₃ − 3R₁

[
[ 1 2 3 ]
[ 0 0 0 ]
[ 0 0 0 ]
]

Only one non-zero row exists.

rank(A) = 1

Example 3

Show that the matrix is singular:

A =
[
[ 1 2 3 ]
[ 4 5 6 ]
[ 7 8 9 ]
]

Solution:

|A| = 1(5×9 − 6×8) − 2(4×9 − 6×7) + 3(4×8 − 5×7)
= 1(45 − 48) − 2(36 − 42) + 3(32 − 35)
= −3 + 12 − 9
= 0

Hence, the matrix is singular.

Unsolved Exercises.

Exercise 1

Write the order of the following matrices:

[
[ 1 2 3 ]
[ 4 5 6 ]
]

Find A + B, where:

A =
[
[ 1 0 ]
[ 2 3 ]
]

B =
[
[ 4 1 ]
[ 0 2 ]
]

Find 2A − 3B, where:

A =
[
[ 3 1 ]
[ 2 4 ]
]

B =
[
[ 1 5 ]
[ 2 1 ]
]

Exercise 2

Multiply the matrices:

A =
[
[ 1 2 ]
[ 3 4 ]
]

B =
[
[ 2 0 ]
[ 1 2 ]
]

Find the transpose of the matrix:

A =
[
[ 1 3 5 ]
[ 2 4 6 ]
]

Find the inverse of the matrix:

A =
[
[ 2 3 ]
[ 1 4 ]
]

Exercise 3

Solve the system of equations using matrix method:

x + y + z = 6
2x + 3y + z = 11
3x + 4y + 2z = 17

Discuss the consistency of the system:

x + y + z = 2
2x + 2y + 2z = 4
3x + 3y + 3z = 7

Important Final Remarks

Matrix methods simplify lengthy algebraic computations.
Rank method is preferred when determinant is zero.
Clear stepwise presentation carries full marks in exams.
Practice numerical problems thoroughly to gain confidence.