The document provides information about matrix operations and properties. It defines what a matrix is and different types of matrices. It then discusses operations like addition, subtraction, multiplication of matrices. It also covers properties such as transpose, inverse, adjoint and determinant of a matrix. It provides examples to illustrate matrix operations and properties such as finding the inverse and determinant of given matrices.

1. Prepared By
Omar Faruk
Lecturer in Mathematics
Department of Basic Science
World University of Bangladesh
Email: omar.faruk@science.wub.edu.bd

2. Introduction
Definition
Addition and Subtraction of Matrix
Multiplication of Matrix
Solved problems on Matrix Multiplication

3. Matrix: A matrix is a rectangular array of elements or numbers arranged in rows
and columns. Matrix is denoted by first or third parentheses(bracket).
A matrix consists of m horizontal rows and n vertical columns is called 𝑚 × 𝑛
matrix, denoted by
𝐴 =
𝑎11 𝑎12 . . . . . . . 𝑎1𝑛
𝑎21 𝑎22 . . . . . . . 𝑎2𝑛
⋮ ⋮
𝑎𝑚1 𝑎𝑚2 . . . . . . . 𝑎𝑚𝑛
= (𝑎𝑖𝑗)𝑚𝑛
For the entry 𝑎𝑖𝑗, the row number is denoted by 𝑖 and the column number is
denoted by 𝑗. The numbers in a matrix are called its elements.
Example: 𝐴 =
1 5 3
2 4 1
4 3 5
is a 3 × 3 matrix.

4. The size or order of a matrix is described by its number of rows and the
number of columns.
If a matrix 𝑨 has 𝒎 rows and 𝒏 columns then the order of 𝑨 is 𝒎 × 𝒏.
Example: If 𝐴 =
2 1 1
3 4 7
then the order of A is 2 × 3.

5.  Row Matrix: A matrix having only a single row is called a row matrix.
Example: 2 3 4
 Column Matrix: A matrix having only a single column is called a row matrix.
Example:
5
2
1
 Square Matrix: A matrix having equal number of rows and columns is called
a square matrix.
Example:
2 0 −1
5 3 2
1 7 3
 Rectangular Matrix: A matrix having equal number of rows and columns is
called a square matrix.
Example:
2 0 −1
1 7 3
 Null Matrix: If all elements of a matrix is zero the matrix is called null or zero
matrix and it is shown by 𝟎.

6. Example:
0 0 0
0 0 0
0 0 0
 Diagonal Matrix: A square matrix in which all the elements except the main
diagonal are zero is called diagonal matrix.
Example:
2 0 0
0 3 0
0 0 5
 Scalar Matrix: In a diagonal matrix if all elements are equal the matrix is
called a scalar matrix.
Example:
3 0 0
0 3 0
0 0 3
 Unit/Identity Matrix: A diagonal matrix whose all elements on the main
diagonal are equal to one is called identity or unit matrix. A unit matrix is
usually shown by letter I .
Example: 𝐼 = 𝐼3 =
1 0 0
0 1 0
0 0 1

7. Transpose Matrix: If the rows and columns of a matrix A are interchanged then
the resulting matrix is called transpose of A matrix. It is denoted by 𝐴′/𝐴𝑇.
Example: 𝐴 =
3 2 6
1 5 0
4 3 2
; 𝐴′
= 𝐴𝑇
=
3 2 6
1 5 0
4 3 2
Symmetric Matrix: A matrix A is called symmetric if 𝐴𝑇 = 𝐴.
Example:
1 2
2 4
is a symmetric matrix.
Skew Symmetric Matrix: A matrix A is called skew symmetric if 𝐴 = −𝐴𝑇.
Example:
0 −1
1 0
is a skew symmetric matrix

8. Conditions for Addition and Subtraction of Matrix:
 Matrices must have same dimension.
 Add/subtract matrices element-by-element
Example: If 𝐴 =
1 3 2
4 2 3
1 5 4
and 𝐵 =
5 2 1
3 1 2
4 3 2
then
𝐴 + 𝐵 =
1 + 5 3 + 2 2 + 1
4 + 3 2 + 1 3 + 2
1 + 4 5 + 3 4 + 2
=
6 5 3
7 3 5
5 8 6
And 𝐴 − 𝐵 =
1 − 5 3 − 2 2 − 1
4 − 3 2 − 1 3 − 2
1 − 4 5 − 3 4 − 2
=
−4 1 1
1 1 1
−3 2 2
Multiplication by scalar: If 𝐴 is a matrix and 𝑘 is any scalar then
𝑘. 𝐴 = 𝑘. (𝑎𝑖𝑗)
𝑚𝑛
This means that all elements of the matrix are multiplied by the scalar 𝑘.
Example: 3
2 3
−1 1
=
6 9
−3 3

9.  Multiplication of two matrices 𝑨 and 𝑩, in the form of 𝑨 × 𝑩 or 𝑨𝑩, is
possible if the number of columns in 𝑨 is equal to the number of rows in 𝑩
.
 The result of this multiplication is another matrix 𝑪 where the number of
its rows is equal to the number of rows in 𝑨 and number of its columns is
equal to the number of columns in 𝑩; that is:
𝑨𝒎×𝒏 × 𝑩𝒏×𝒑 = 𝑪𝒎×𝒑
For example, if 𝐴 =
𝑎 𝑏 𝑐
𝑑 𝑒 𝑓
𝑔 ℎ 𝑖
and 𝑋 =
𝑥
𝑦
𝑧
then the multiplication of
A and X will be
AX =
𝑎 𝑏 𝑐
𝑑 𝑒 𝑓
𝑔 ℎ 𝑖
𝑥
𝑦
𝑧
=
𝑎𝑥 + 𝑏𝑦 + 𝑐𝑧
𝑑𝑥 + 𝑒𝑦 + 𝑓𝑧
𝑔𝑥 + ℎ𝑦 + 𝑖𝑧

10. Example: If 𝐴 =
2 3
1 2
and 𝐵 =
1 2
4 1
then find 𝐴𝐵 and 𝐵𝐴.
Solution:
𝐴𝐵 =
2 3
1 2
1 2
4 1
=
2 × 1 + 3 × 4 2 × 2 + 3 × 1
1 × 1 + 2 × 4 1 × 2 + 2 × 1
=
2 + 12 4 + 3
1 + 8 2 + 2
=
14 7
9 4
Again, 𝐵𝐴 =
1 2
4 1
2 3
1 2
=
2 + 2 3 + 4
8 + 1 12 + 2
=
4 7
9 14

11. Problem: If 𝐴 =
1 −1
0 2
, 𝐵 =
1 3 0
2 0 1
and 𝐶 =
2
3
1
then show that
(𝐴𝐵)𝐶 = 𝐴(𝐵𝐶).
Solution:
Here 𝐴𝐵 =
1 −1
0 2
1 3 0
2 0 1
=
1 − 2 3 + 0 0 − 1
0 + 4 0 + 0 0 + 2
=
−1 3 −1
4 0 2
Now 𝐴𝐵 𝐶 =
−1 3 −1
4 0 2
2
3
1
=
−2 + 9 − 1
8 + 0 + 2
=
−2 + 9 − 1
8 + 0 + 2
=
6
10

12. Again, 𝐵𝐶 =
1 3 0
2 0 1
2
3
1
=
2 + 9 + 0
4 + 0 + 1
=
11
5
Now, 𝐴 𝐵𝐶 =
1 −1
0 2
11
5
=
11 − 5
0 + 10
=
6
10
Hence 𝐴𝐵 𝐶 = 𝐴(𝐵𝐶).

13. Problem: If 𝐴 =
1 2 2
2 1 2
2 2 1
then show that 𝐴2
− 4𝐴 − 5𝐼 = 0.
Solution: Here, 𝐴2
= 𝐴. 𝐴
=
1 2 2
2 1 2
2 2 1
1 2 2
2 1 2
2 2 1
=
1 + 4 + 4 2 + 2 + 4 2 + 4 + 2
2 + 2 + 4 4 + 1 + 4 4 + 2 + 2
2 + 4 + 2 4 + 2 + 2 4 + 4 + 1
=
9 8 8
8 9 8
8 8 9
Now, 𝐴2
− 4𝐴 − 5𝐼 =
9 8 8
8 9 8
8 8 9
− 4
1 2 2
2 1 2
2 2 1
− 5
1 0 0
0 1 0
0 0 1
=
9 8 8
8 9 8
8 8 9
−
4 8 8
8 4 8
8 8 4
−
5 0 0
0 5 0
0 0 5

14. =
9 − 4 − 5 8 − 8 − 0 8 − 8 − 0
8 − 8 − 0 9 − 4 − 5 8 − 8 − 0
8 − 8 − 0 8 − 8 − 0 9 − 4 − 5
=
0 0 0
0 0 0
0 0 0
= 0
Hence 𝐴2 − 4𝐴 − 5𝐼 = 0.

15.  Idempotent Matrix: A matrix 𝐴 is called idempotent if 𝐴2
= 𝐴.
Example:
2 −3 −5
−1 4 5
1 −3 −4
 Nilpotent Matrix: A matrix 𝐴 is called nilpotent if 𝐴𝑝
= 0 where 𝑝 ∈ 𝑁.
Example:
1 −1
1 −1
 Orthogonal Matrix: A matrix 𝐴 is called orthogonal if 𝐴𝐴𝑇 = 𝐼.
Example:
1
3
2 −2 1
1 2 2
2 1 −2
 Involutory Matrix: A matrix 𝐴 is called involutary matrix if 𝐴2 = 𝐼.
Example:
4 3 3
−1 0 −1
−4 −4 −3

16. 𝐓𝐡𝐚𝐧𝐤 𝐲𝐨𝐮

17. Prepared By
Omar Faruk
Lecturer in Mathematics
Department of Basic Science
World University of Bangladesh
Email: omar.faruk@science.wub.edu.bd

18. Determinant of Matrix
Adjoint of Matrix
Inverse of Matrix
Rank of Matrix

19. The determinant of a matrix 𝐴2×2 =
𝑎 𝑏
𝑐 𝑑
denoted by det 𝐴 / 𝐴 is defined as
𝐴 =
𝑎 𝑏
𝑐 𝑑
= 𝑎𝑑 − 𝑏𝑐
 For 𝐴3×3 =
𝑎 𝑏 𝑐
𝑑 𝑒 𝑓
𝑔 ℎ 𝑖
𝐴 =
𝑎 𝑏 𝑐
𝑑 𝑒 𝑓
𝑔 ℎ 𝑖
= 𝑎 𝑒𝑖 − ℎ𝑓 − 𝑏 𝑑𝑖 − 𝑔𝑓 + 𝑐(𝑔ℎ − 𝑔𝑒)
 Example: If A=
3 1 2
2 1 4
3 2 1
then
𝐴 =
3 1 2
2 1 4
3 2 1
= 3 1 − 8 − 1 2 − 12 + 2 4 − 3
= −21 + 10 + 2 = −9

20.  The transposed matrix 𝐴 formed by the cofactors of the elements of 𝐴 is
called the adjoint of 𝐴. It is denoted by 𝐴𝑑𝑗 𝐴.
 Example: If A=
2 1 0
3 2 1
1 2 0
then the co-factors of 𝐴 are
𝐶11 = +
2 1
2 0
= 0 − 2 = −2; 𝐶12 = −
3 1
1 0
= −(0 − 1) = 1
𝐶13 = +
3 2
1 2
= 6 − 2 = 4; 𝐶21 = −
1 0
2 0
= −(0 − 0) = 0
𝐶22 = +
2 0
1 0
= 0 − 0 = 0; 𝐶23 = −
2 1
1 2
= −(4 − 1) = −3
𝐶31 = +
1 0
2 1
= 1 − 0 = 1; 𝐶32 = −
2 0
3 1
= −(2 − 0) = −2
𝐶33 = +
2 1
3 2
= 4 − 3 = 1
𝐴𝑑𝑗 𝐴 =
−2 1 4
0 0 −3
1 −2 1
𝑇
=
−2 0 1
1 0 −2
4 −3 1

21.  Singular and Non-singular Matrix: A matrix 𝐴 is called singular if 𝐴 = 0
and non-singular if |𝐴| ≠ 0.
Example:
1 2
1 2
is singular and
3 2
4 1
is non-singular Matrix.
 Inverse Matrix: A matrix 𝐵 is called inverse of a matrix 𝐴 if 𝐴𝐵 = 𝐵𝐴 = 𝐼. It
is denoted by 𝐴−1
.
𝐴−1
=
1
𝐴
𝐴𝑑𝑗 𝐴
Example: Find 𝐴−1 where 𝐴 =
0 1 1
1 2 0
3 −1 4
.
Solution: Here,
𝐴 =
0 1 1
1 2 0
3 −1 4
= 0 − 1 4 − 0 + 1 −1 − 6
= −11 ≠ 0
So, 𝐴 is non-singular. Hence 𝐴 is inversible.
 A matrix is inversible only if the matrix is non-singular.

22. Now we will find the co-factors of 𝐴.
𝐶11 = +
2 0
−1 4
= 8 − 0 = 8; 𝐶12 = −
1 0
3 4
= − 4 − 0 = −4
𝐶13 = +
1 2
3 −1
= −1 − 6 = −7; 𝐶21 = −
1 1
−1 4
= − 4 + 1 = −5
𝐶22 = +
0 1
3 4
= 0 − 3 = −3; 𝐶23 = −
0 1
3 −1
= −(0 − 3) = 3
𝐶31 = +
1 1
2 0
= 0 − 2 = −2; 𝐶32 = −
0 1
1 0
= −(0 − 1) = 1
𝐶33 = +
0 1
1 2
= 0 − 1 = −1
𝐴𝑑𝑗 𝐴 =
8 −4 −7
−5 −3 3
−2 1 −1
𝑇
=
8 −5 −2
−4 −3 1
−7 3 −1

23. Hence, 𝐴−1 =
1
𝐴
𝐴𝑑𝑗 𝐴
=
1
−11
8 −5 −2
−4 −3 1
−7 3 −1

24.  The number of linearly independent rows of a matrix is called the rank of a
matrix. It is denoted by 𝝆.
Example: Find the rank of the matrix 𝐴 =
6 2 0 4
−2 −1 3 4
−1 −1 6 10
.
Solution:
6 2 0 4
−2 −1 3 4
−1 −1 6 10
~
−1 −1 6 10
−2 −1 3 4
6 2 0 4
𝑹𝟏 ⟷ 𝑹𝟑
~
1 1 −6 −10
−2 −1 3 4
6 2 0 4
𝑹𝟏
′
= (−𝟏) × 𝑹𝟏
 The number of non zero rows in echelon form will be the rank of the
matrix.
 Echelon form−
𝟏 𝒂 𝒃
𝟎 𝟏 𝒄
𝟎 𝟎 𝟏

25. ~
1 1 −6 −10
0 1 −9 −16
0 4 −36 −64
𝑹𝟐
′
= −𝟐 × 𝑹𝟏 + 𝑹𝟐
𝑹𝟑
′
= −𝟔 × 𝑹𝟏 − 𝑹𝟑
~
1 1 −6 −10
0 1 −9 −16
0 0 0 0
𝑹𝟑
′
= 𝟒 × 𝑹𝟐 − 𝑹𝟑
The matrix is in echelon form having 2 non-zero rows.
So, Rank of 𝐴, 𝜌 𝐴 = 2.

26. 𝐓𝐡𝐚𝐧𝐤 𝐲𝐨𝐮

27. Prepared By
Omar Faruk
Lecturer in Mathematics
Department of Basic Science
World University of Bangladesh
Email: omar.faruk@science.wub.edu.bd

28. Solved Problem on Inverse Matrix
Cramer’s Rule

29.  If 𝑨 =
𝟏 −𝟏 𝟏
𝟑 𝟏 𝟒
−𝟐 𝟑 𝟓
then show that 𝑨−𝟏
𝑨 = 𝑰.
Solution: Here,
𝐴 =
1 −1 1
3 1 4
−2 3 5
= 1 5 − 12 − −1 15 + 8 + 1 9 + 2
= 27 ≠ 0
So, 𝐴 is non-singular. Hence 𝐴 is inversible.
Now we will find the co-factors of 𝐴.
𝐶11 = +
1 4
3 5
= 5 − 12 = −7; 𝐶12 = −
3 4
−2 5
= − 15 − −8 = −23
𝐶13 = +
3 1
−2 3
= 9 − −2 = 11; 𝐶21 = −
−1 1
3 5
= − −5 − 3 = 8

30. 𝐶22 = +
1 1
−2 5
= 5 − −2 = 7; 𝐶23 = −
1 −1
−2 3
= − 3 − 2 = −1
𝐶31 = +
−1 1
1 4
= −4 − 1 = −5; 𝐶32 = −
1 1
3 4
= − 4 − 3 = −1
𝐶33 = +
1 −1
3 1
= 1 − (−3) = 4
𝐴𝑑𝑗 𝐴 =
−7 −23 11
8 7 −1
−5 −1 4
𝑇
=
−7 8 −5
−23 7 −1
11 −1 4

31. Hence, 𝐴−1
=
1
𝐴
𝐴𝑑𝑗 𝐴
=
1
27
−7 8 −5
−23 7 −1
11 −1 4
Now, 𝐴−1
𝐴 =
1
27
−7 8 −5
−23 7 −1
11 −1 4
1 −1 1
3 1 4
−2 3 5
=
1
27
−7 + 24 + 10 7 + 8 − 15 −7 + 32 − 25
−23 + 21 + 2 23 + 7 − 3 −23 + 28 − 5
11 − 3 − 8 −11 − 1 + 12 11 − 4 + 20

32. =
1
27
27 0 0
0 27 0
0 0 27
=
1 0 0
0 1 0
0 0 1
= 𝑰
Hence, 𝑨−𝟏
𝑨 = 𝑰.

33.  Solve the following system of equation by Cramer’s rule.
𝑥 + 𝑦 + 𝑧 = 3
𝑥 + 2𝑦 + 3𝑧 = 4
𝑥 + 4𝑦 + 9𝑧 = 6
Solution:
𝐷 =
1 1 1
1 2 3
1 4 9
= 1 18 − 12 − 1 9 − 3 + 1 4 − 2 = 2
𝐷𝑥 =
𝟑 1 1
𝟒 2 3
𝟔 4 9
= 3 18 − 12 − 1 36 − 18 + 1 16 − 12 = 4
𝐷𝑦 =
1 𝟑 1
1 𝟒 3
1 𝟔 9
= 1 36 − 18 − 3 9 − 3 + 1 6 − 4 = 2

34. 𝐷𝑧 =
1 1 𝟑
1 2 𝟒
1 4 𝟔
= 1 12 − 16 − 1 6 − 4 + 3 4 − 2 = 0
∴ 𝒙 =
𝑫𝒙
𝑫
=
4
2
= 2
𝒚 =
𝑫𝒚
𝑫
=
2
2
= 1
𝐳 =
𝑫𝒛
𝑫
=
0
2
= 0
∴ 𝒙, 𝒚, 𝒛 = (𝟐, 𝟏, 𝟎)

35. 𝐓𝐡𝐚𝐧𝐤 𝐲𝐨𝐮

36. Prepared By
Omar Faruk
Lecturer in Mathematics
Department of Basic Science
World University of Bangladesh
Email: omar.faruk@science.wub.edu.bd

37. Solution of System of Equations by Inverse Matrix
Characteristic Vector and Root

38.  Solve the following system of equation by Inverse method
𝒙 − 𝟑𝒚 + 𝟐𝒛 = 𝟑
𝟑𝒙 + 𝟐𝒚 − 𝒛 = 𝟐
𝟐𝒙 − 𝒚 + 𝒛 = 𝟒
Solution: The given system of equation can be written in Matrix-form as
𝟏 −𝟑 𝟐
𝟑 𝟐 −𝟏
𝟐 −𝟏 𝟏
𝒙
𝒚
𝒛
=
𝟑
𝟐
𝟒
Let,
𝐴 =
1 −3 2
3 2 −1
2 −1 1
, B =
3
2
4
and X =
𝑥
𝑦
𝑧
Then the equation reduces to
𝑨𝑿 = 𝑩

39. Now we have to find 𝐴−1
.
Here,
𝐴 =
1 −3 2
3 2 −1
2 −1 1
= 1 2 − 1 − −3 3 + 2 + 2 −3 − 4
= 2 ≠ 0
So, 𝐴 is non-singular. Hence 𝐴−1
exists.
Now we will find the co-factors of 𝐴.
𝐶11 = +
2 −1
−1 1
= 2 − 1 = 1; 𝐶12 = −
3 −1
2 1
= − 3 − −2 = −5
𝐶13 = +
3 2
2 −1
= −3 − 4 = −7; 𝐶21 = −
−3 2
−1 1
= − −3 − (−2 = 1
∴ 𝑿 = 𝑨−𝟏
𝑩

40. 𝐶22 = +
1 2
2 1
= 1 − 4 = −3; 𝐶23 = −
1 −3
2 −1
= − −1 − −6 = −5
𝐶31 = +
−3 2
2 −1
= 3 − 4 = −1; 𝐶32 = −
1 2
3 −1
= − −1 − 6 = 7
𝐶33 = +
1 −3
3 2
= 2 − (−9) = 11
𝑨𝒅𝒋 𝑨 =
1 −5 −7
1 −3 −5
−1 7 11
𝑇
=
1 1 −1
−5 −3 7
−7 −5 11

41. Hence, 𝑨−𝟏
=
𝟏
𝑨
𝑨𝒅𝒋 𝑨
=
1
2
1 1 −1
−5 −3 7
−7 −5 11
Now, 𝑿 = 𝑨−𝟏
𝑩
⇒
𝑥
𝑦
𝑧
=
1
2
1 1 −1
−5 −3 7
−7 −5 11
3
2
4
=
1
2
3 + 2 − 4
−15 − 6 + 28
−21 − 10 + 44

42. =
1
2
1
7
13
=
1
2
7
2
13
2
Hence, 𝒙 =
𝟏
𝟐
𝒚 =
𝟕
𝟐
𝒛 =
𝟏𝟑
𝟐

43.  A non-zero vector 𝑿 is defined as characteristic vector or Eigen vector of a
matrix 𝐴 if there exists a number 𝝀 such that 𝑨𝑿 = 𝝀𝑿 where 𝝀 is defined as
characteristic root or eigen value corresponding to the characteristic vector 𝑿.
 The matrix 𝑨 − 𝝀𝑰 is called the characteristic matrix of 𝐴.
 The determinant 𝑨 − 𝝀𝑰 is called the characteristic polynomial of 𝐴.
 The equation 𝑨 − 𝝀𝑰 = 𝟎 is called the characteristic equation of 𝐴.
Example: If 𝐴 =
1 3 2
4 2 3
1 5 4
then
𝟏 − 𝝀 3 2
4 𝟐 − 𝝀 3
1 5 𝟒 − 𝝀
= 0 is characteristic
equation of 𝐴.

44. 𝐓𝐡𝐚𝐧𝐤 𝐲𝐨𝐮

45. Prepared By
Omar Faruk
Lecturer in Mathematics
Department of Basic Science
World University of Bangladesh
Email: omar.faruk@science.wub.edu.bd

46. Characteristic Root / Eigen value

47.  Find the Characteristic Root or Eigen value of the matrix 𝐴 =
2 2 1
1 3 1
1 2 2
.
Solution: We know, the characteristic equation is,
𝑨 − 𝝀𝑰 = 𝟎
⇒
2 2 1
1 3 1
1 2 2
− 𝜆
1 0 0
0 1 0
0 0 1
= 0
⇒
2 2 1
1 3 1
1 2 2
−
𝜆 0 0
0 𝜆 0
0 0 𝜆
= 0
⇒
2 − 𝜆 2 1
1 3 − 𝜆 1
1 2 2 − 𝜆
= 0

48. ⇒ 2 − 𝜆 3 − 𝜆 2 − 𝜆 − 2 − 2 1. 2 − 𝜆 − 1 + 1 2 − 3 − 𝜆 . 1 = 0
⇒ 2 − 𝜆 6 − 3𝜆 − 2𝜆 + 𝜆2
− 2 − 2 2 − 𝜆 − 1 + 2 − 3 + 𝜆 = 0
⇒ 2 − 𝜆 𝜆2
− 5𝜆 + 4 − 2 1 − 𝜆 + 𝜆 − 1 = 0
⇒ 2 − 𝜆 𝜆2
− 4𝜆 − 𝜆 + 4 − 2 1 − 𝜆 + 𝜆 − 1 = 0
⇒ 2 − 𝜆 𝜆 𝜆 − 4 − 1 𝜆 − 4 − 2 1 − 𝜆 + 𝜆 − 1 = 0
⇒ 2 − 𝜆 𝜆 − 4 𝜆 − 1 − 2 1 − 𝜆 + 𝜆 − 1 = 0
⇒ 2 − 𝜆 𝜆 − 4 𝜆 − 1 + 2 𝜆 − 1 + 𝜆 − 1 = 0
⇒ 𝜆 − 1 2 − 𝜆 𝜆 − 4 + 2 + 1 = 0
⇒ 𝜆 − 1 2𝜆 − 8 − 𝜆2 + 4𝜆 + 3 = 0

49. ⇒ 𝜆 − 1 −𝜆2
+ 6𝜆 − 5 = 0
⇒ 𝜆 − 1 − 𝜆2
− 6𝜆 + 5 = 0
⇒ 𝜆 − 1 𝜆2 − 6𝜆 + 5 = 0
⇒ 𝜆 − 1 𝜆2 − 5𝜆 − 𝜆 + 5 = 0
⇒ 𝜆 − 1 𝜆 𝜆 − 5 − 1 𝜆 − 5 = 0
⇒ 𝜆 − 1 𝜆 − 5 𝜆 − 1 = 0
∴ 𝜆 − 1 = 0 𝜆 − 5 = 0 𝜆 − 1 = 0
⇒ 𝜆 = 1 ⇒ 𝜆 = 5 ⇒ 𝜆 = 1
∴ The characteristic root or eigen values are, 𝝀 = 𝟏, 𝟏, 𝟓.

50.  Find the Characteristic Root or Eigen value of the matrix 𝐴 =
1 1 3
1 5 1
3 1 1
.
Solution: We know, the characteristic equation is,
𝑨 − 𝝀𝑰 = 𝟎
⇒
1 1 3
1 5 1
3 1 1
− 𝜆
1 0 0
0 1 0
0 0 1
= 0
⇒
1 1 3
1 5 1
3 1 1
−
𝜆 0 0
0 𝜆 0
0 0 𝜆
= 0
⇒
1 − 𝜆 1 3
1 5 − 𝜆 1
3 1 1 − 𝜆
= 0

51. ⇒ 1 − 𝜆 5 − 𝜆 1 − 𝜆 − 1 − 1 1. 1 − 𝜆 − 3 + 3 1 − 3. 5 − 𝜆 = 0
⇒ 1 − 𝜆 5 − 5𝜆 − 𝜆 + 𝜆2 − 1 − 1 − 𝜆 − 3 + 3 1 − 15 + 3𝜆 = 0
⇒ 1 − 𝜆 𝜆2
− 6𝜆 + 4 − −𝜆 − 2 + 3 3𝜆 − 14 = 0
⇒ 𝜆2
− 6𝜆 + 4 − 𝜆3
+ 6𝜆2
− 4𝜆 + 𝜆 + 2 + 9𝜆 − 42 = 0
⇒ −𝜆3
+ 7𝜆2
− 36 = 0
⇒ 𝜆3
− 7𝜆2
+ 36 = 0
Now,
𝑓 𝜆 = 𝜆3 − 7𝜆2 + 36
𝒇 −𝟐 = −2 3
− 7. −2 2
+ 36
= 0

52. Hence,
𝜆3 − 7𝜆2 + 36 = 0
⇒ 𝜆3
+ 2𝜆2
− 9𝜆2
− 18𝜆 + 18𝜆 + 36 = 0
⇒ 𝜆2 𝜆 + 2 − 9𝜆 𝜆 + 2 + 18 𝜆 + 2 = 0
⇒ 𝜆 + 2 𝜆2
− 9𝜆 + 18 = 0
⇒ (𝜆 + 2)(𝜆2
− 6𝜆 − 3𝜆 + 18) = 0
⇒ 𝜆 + 2 𝜆 𝜆 − 6 − 3 𝜆 − 6 = 0
⇒ 𝜆 + 2 𝜆 − 6 𝜆 − 3 = 0
∴ 𝜆 + 2 = 0 𝜆 − 6 = 0 𝜆 − 3 = 0
⇒ 𝜆 = −2 ⇒ 𝜆 = 6 ⇒ 𝜆 = 3
∴ The characteristic root or eigen values are, 𝝀 = −𝟐, 𝟑, 𝟔.

53. 𝐓𝐡𝐚𝐧𝐤 𝐲𝐨𝐮

54. Prepared By
Omar Faruk
Lecturer in Mathematics
Department of Basic Science
World University of Bangladesh
Email: omar.faruk@science.wub.edu.bd

55. Cayley-Hamilton Theorem
Application of Cayley-Hamilton Theorem

56.  Cayley-Hamilton Theorem: Every square matrix satisfies it’s own
characteristic equation.
 Explanation: If 𝑨 is an 𝑚 × 𝑛 matrix (where 𝑚 = 𝑛) and 𝑰 is the identity
matrix then the characteristic polynomial of 𝑨 is defined as
If we replace 𝝀 with the matrix 𝑨 then the polynomial will be zero matrix.
 Example: Verify Cayley-Hamilton theorem for 𝑨 =
𝟐 𝟐 𝟏
𝟏 𝟑 𝟏
𝟏 𝟐 𝟐
.
Solution: The characteristic polynomial of 𝐴 is,
𝒑 𝝀 = 𝑨 − 𝝀𝑰
𝒑 𝑨 = 𝟎
𝒑 𝝀 = 𝑨 − 𝝀𝑰

57. =
2 2 1
1 3 1
1 2 2
− 𝜆
1 0 0
0 1 0
0 0 1
=
2 2 1
1 3 1
1 2 2
−
𝜆 0 0
0 𝜆 0
0 0 𝜆
=
2 − 𝜆 2 1
1 3 − 𝜆 1
1 2 2 − 𝜆
= 2 − 𝜆 3 − 𝜆 2 − 𝜆 − 2 − 2 1. 2 − 𝜆 − 1 + 1 2 − 3 − 𝜆 . 1
= 2 − 𝜆 6 − 3𝜆 − 2𝜆 + 𝜆2 − 2 − 2 2 − 𝜆 − 1 + 2 − 3 + 𝜆
= 2 − 𝜆 𝜆2
− 5𝜆 + 4 − 2 1 − 𝜆 + 𝜆 − 1
= 2𝜆2
− 10𝜆 + 8 − 𝜆3
+ 5𝜆2
− 4𝜆 − 2 + 2𝜆 + 𝜆 − 1

58. = −𝜆3
+ 7𝜆2
− 11𝜆 + 5
∴ The characteristic polynomial of 𝐴 is
𝒑 𝝀 = −𝝀𝟑 + 𝟕𝝀𝟐 − 𝟏𝟏𝝀 + 𝟓
Now, replacing 𝜆 with 𝐴, we get,
𝒑 𝑨 = −𝑨𝟑 + 𝟕𝑨𝟐 − 𝟏𝟏𝑨 + 𝟓𝑰
Now, 𝑨𝟐
=
𝟐 𝟐 𝟏
𝟏 𝟑 𝟏
𝟏 𝟐 𝟐
.
𝟐 𝟐 𝟏
𝟏 𝟑 𝟏
𝟏 𝟐 𝟐
=
4 + 2 + 1 4 + 6 + 2 2 + 2 + 2
2 + 3 + 1 2 + 9 + 2 1 + 3 + 2
2 + 2 + 2 2 + 6 + 4 1 + 2 + 4

59. =
7 12 6
6 13 6
6 12 7
And 𝑨𝟑 = 𝑨𝟐. 𝑨
=
7 12 6
6 13 6
6 12 7
.
2 2 1
1 3 1
1 2 2
=
14 + 12 + 6 14 + 36 + 12 7 + 12 + 12
12 + 13 + 6 12 + 39 + 12 6 + 13 + 12
12 + 12 + 7 12 + 36 + 14 6 + 12 + 14
=
32 62 31
31 63 31
31 62 32

60. ∴ 𝒑 𝑨 = −𝑨𝟑
+ 𝟕𝑨𝟐
− 𝟏𝟏𝑨 + 𝟓𝑰
= −
32 62 31
31 63 31
31 62 32
+ 7
7 12 6
6 13 6
6 12 7
− 11
2 2 1
1 3 1
1 2 2
+ 5
1 0 0
0 1 0
0 0 1
=
−32 −62 −31
−31 −63 −31
−31 −62 −32
+
49 84 42
42 91 42
42 84 49
−
22 22 11
11 33 11
11 22 22
+
5 0 0
0 5 0
0 0 5
=
−32 + 49 − 22 + 5 −62 + 84 − 22 + 0 −31 + 42 − 11 + 0
−31 + 42 − 11 + 0 −63 + 91 − 33 + 5 −31 + 42 − 11 + 0
−31 + 42 − 11 + 0 −62 + 84 − 22 + 0 −32 + 49 − 22 + 5
=
0 0 0
0 0 0
0 0 0
= 0
∴ 𝒑 𝑨 = 𝟎
Hence the Cayley-Hamilton theorem is verified.

61.  Find 𝑨−𝟏
by using Cayley-Hamilton theorem where 𝑨 =
𝟕 𝟐 −𝟐
−𝟔 −𝟏 𝟐
𝟔 𝟐 −𝟏
.
Solution: The characteristic polynomial of 𝐴 is,
𝒑 𝝀 = 𝑨 − 𝝀𝑰
=
7 2 −2
−6 −1 2
6 2 −1
− 𝜆
1 0 0
0 1 0
0 0 1
=
7 2 −2
−6 −1 2
6 2 −1
−
𝜆 0 0
0 𝜆 0
0 0 𝜆
=
7 − 𝜆 2 −2
−6 −1 − 𝜆 2
6 2 −1 − 𝜆

62. = 7 − 𝜆 −1 − 𝜆 −1 − 𝜆 − 4 − 2 −6. −1 − 𝜆 − 12 − 2 −12 −

63. Now, 𝑨𝟐
=
𝟕 𝟐 −𝟐
−𝟔 −𝟏 𝟐
𝟔 𝟐 −𝟏
.
𝟕 𝟐 −𝟐
−𝟔 −𝟏 𝟐
𝟔 𝟐 −𝟏
=
49 − 12 − 12 14 − 2 − 4 −14 + 4 + 2
−42 + 6 + 12 −12 + 1 + 4 12 − 2 − 2
42 − 12 − 6 12 − 2 − 2 −12 + 4 + 1
=
25 8 −8
−24 −7 8
24 8 −7
And 𝑨𝟑 = 𝑨𝟐. 𝑨
=
25 8 −8
−24 −7 8
24 8 −7
.
7 2 −2
−6 −1 2
6 2 −1

64. =
175 − 48 − 48 56 − 14 − 16 −56 + 16 + 14
−150 + 24 + 48 −48 + 7 + 16 48 − 8 − 14
150 − 48 − 24 48 − 14 − 8 −48 + 16 + 7
=
79 26 −26
−78 −25 26
78 26 −25
∴ 𝒑 𝑨 = −𝑨𝟑 + 𝟓𝑨𝟐 − 𝟕𝑨 + 𝟑𝑰
= −
79 26 −26
−78 −25 26
78 26 −25
+ 5
25 8 −8
−24 −7 8
24 8 −7
− 7
7 2 −2
−6 −1 2
6 2 −1
+ 3
1 0 0
0 1 0
0 0 1
=
−79 −26 26
78 25 −26
−78 −26 25
+
125 40 −40
−120 −35 40
120 40 −35
−
49 14 −14
−42 −7 14
42 14 −7
+
3 0 0
0 3 0
0 0 3

65. =
−79 + 125 − 49 + 3 −26 + 40 − 14 + 0 26 − 40 + 14 + 0
78 − 120 + 42 + 0 25 − 35 + 7 + 3 −26 + 40 − 14 + 0
−78 + 120 − 42 + 0 −26 + 40 − 14 + 0 25 − 35 + 7 + 3
=
0 0 0
0 0 0
0 0 0
= 0
∴ 𝒑 𝑨 = 𝟎
∴ −𝑨𝟑
+ 𝟓𝑨𝟐
− 𝟕𝑨 + 𝟑𝑰 = 𝟎 … … … … … . . 𝒊
Hence the Cayley-Hamilton theorem is verified.
Now multiplying 𝑨−𝟏
with both sides of (𝒊), we get
−𝐴−1𝐴3 + 5𝐴−1𝐴2 − 7𝐴−1𝐴 + 3𝐴−1𝐼 = 0
⇒ −𝐴2 +5𝐴 − 7𝐼 + 3𝐴−1 = 0
⇒ 3𝐴−1
= 𝐴2
− 5𝐴 + 7𝐼

66. ⇒ 𝐴−1
=
1
3
𝐴2
− 5𝐴 + 7𝐼
⇒ 𝐴−1
=
1
3
25 8 −8
−24 −7 8
24 8 −7
− 5
7 2 −2
−6 −1 2
6 2 −1
+ 7
1 0 0
0 1 0
0 0 1
⇒ 𝐴−1
=
1
3
25 8 −8
−24 −7 8
24 8 −7
−
35 10 −10
−30 −5 10
30 10 −5
+
7 0 0
0 7 0
0 0 7
⇒ 𝐴−1 =
1
3
25 − 35 + 7 8 − 10 + 0 −8 + 10 + 0
−24 + 30 + 0 −7 + 5 + 7 8 − 10 + 0
24 − 30 + 0 8 − 10 + 0 −7 + 5 + 7
∴ 𝑨−𝟏
=
𝟏
𝟑
−𝟑 −𝟐 𝟐
𝟔 𝟓 −𝟐
−𝟔 −𝟐 𝟓

67. 𝐓𝐡𝐚𝐧𝐤 𝐲𝐨𝐮

68. Prepared By
Omar Faruk
Lecturer in Mathematics
Department of Basic Science
World University of Bangladesh
Email: omar.faruk@science.wub.edu.bd

69. Special Types of Matrices
Involutory Matrix
Idempotent Matrix
Nilpotent Matrix
Orthogonal Matrix

70. Involutary Matrix: A matrix 𝑨 is called involutary matrix if 𝑨𝟐
= 𝑰.
Example:
4 3
−5 −4
is an involutary matrix.
 Show that A =
4 3 3
−1 0 −1
−4 −4 −3
is an involutary matrix.
Solution: Here,
𝐴2
= 𝐴. 𝐴
=
4 3 3
−1 0 −1
−4 −4 −3
.
4 3 3
−1 0 −1
−4 −4 −3
=
16 − 3 − 12 12 + 0 − 12 12 − 3 − 9
−4 + 0 + 4 −3 + 0 + 4 −3 + 0 + 3
−16 + 4 + 12 −12 + 0 + 12 −12 + 4 + 9

71. =
1 0 0
0 1 0
0 0 1
= 𝑰
Since 𝑨𝟐 = 𝑰, so 𝑨 is an involutary matrix.
 Idempotent Matrix: A matrix 𝑨 is called idempotent if 𝑨𝟐
= 𝑨.
Example:
2 −2 −4
−1 3 4
1 −2 −3
is an idempotent matrix.
 Show that 𝑨 =
𝟐 −𝟑 −𝟓
−𝟏 𝟒 𝟓
𝟏 −𝟑 −𝟒
is an idempotent matrix.
Solution: Here,
𝐴2
= 𝐴. 𝐴

72. =
2 −3 −5
−1 4 5
1 −3 −4
.
2 −3 −5
−1 4 5
1 −3 −4
=
4 + 3 − 5 −6 − 12 + 15 −10 − 15 + 20
−2 − 4 + 5 3 + 16 − 15 5 + 20 − 20
2 + 3 − 4 −3 − 12 + 12 −5 − 15 + 16
=
2 −3 −5
−1 4 5
1 −3 −4
= 𝑨
Since 𝑨𝟐
= 𝑨, so 𝐴 is an idempotent matrix.
 Nilpotent Matrix: A matrix 𝐴 is called nilpotent if 𝑨𝒑 = 𝟎 where 𝒑 ∈ 𝑵.
Example:
1 −1
1 −1
is a nilpotent matrix.

73.  Show that 𝑨 =
𝟏 −𝟑 −𝟒
−𝟏 𝟑 𝟒
𝟏 −𝟑 −𝟒
is a nilpotent matrix.
Solution: Here,
𝐴2
= 𝐴. 𝐴
=
𝟏 −𝟑 −𝟒
−𝟏 𝟑 𝟒
𝟏 −𝟑 −𝟒
.
𝟏 −𝟑 −𝟒
−𝟏 𝟑 𝟒
𝟏 −𝟑 −𝟒
=
𝟏 + 𝟑 − 𝟒 −𝟑 − 𝟗 + 𝟏𝟐 −𝟒 − 𝟏𝟐 + 𝟏𝟔
−𝟏 − 𝟑 + 𝟒 𝟑 + 𝟗 − 𝟏𝟐 𝟒 + 𝟏𝟐 − 𝟏𝟔
𝟏 + 𝟑 − 𝟒 −𝟑 − 𝟗 + 𝟏𝟐 −𝟒 − 𝟏𝟐 + 𝟏𝟔
=
𝟎 𝟎 𝟎
𝟎 𝟎 𝟎
𝟎 𝟎 𝟎
= 0
Since 𝑨𝟐
= 𝟎, so 𝐴 is a nilpotent matrix.

74.  Orthogonal Matrix: A matrix 𝑨 is called orthogonal if 𝑨𝑨𝑻 = 𝑰.
Example:
1
3
−1 2 2
2 −1 2
2 2 −1
is an orthogonal matrix.
 Show that 𝑨 =
𝟏
𝟑
−𝟏 𝟐 𝟐
𝟐 −𝟏 𝟐
𝟐 𝟐 −𝟏
is an orthogonal matrix.
Solution: Here,
𝐴𝑇
=
1
3
−1 2 2
2 −1 2
2 2 −1
∴ 𝐴. 𝐴𝑇 =
1
3
−1 2 2
2 −1 2
2 2 −1
.
1
3
−1 2 2
2 −1 2
2 2 −1

75. =
1
9
1 + 4 + 4 −2 − 2 + 4 −2 + 4 − 2
−2 − 2 + 4 4 + 1 + 4 4 − 2 − 2
−2 + 4 − 2 4 − 2 − 2 4 + 4 + 1
=
1
9
9 0 0
0 9 0
0 0 9
=
1 0 0
0 1 0
0 0 1
= 𝑰
Since 𝑨𝑨𝑻 = 𝑰, so 𝐴 is an orthogonal matrix.

76.  If 𝐴 =
1 2 3
−2 5 −1
2 3 4
and 𝐵 =
−1 5 3
7 −2 1
2 0 −3
then show that 𝑨𝑩 𝑻 = 𝑩𝑻𝑨𝑻.
Solution: Here,
𝐴𝐵 =
1 2 3
−2 5 −1
2 3 4
−1 5 3
7 −2 1
2 0 −3
=
−1 + 14 + 6 5 − 4 + 0 3 + 2 − 9
2 + 35 − 2 −10 − 10 + 0 −6 + 5 + 3
−2 + 21 + 8 10 − 6 + 0 6 + 3 − 12
=
19 1 −4
35 −20 2
27 4 −3

77. ∴ 𝐴𝐵 𝑇
=
19 1 −4
35 −20 2
27 4 −3
𝑇
=
19 35 27
1 −20 4
−4 2 −3
Now, 𝐴𝑇
=
1 2 3
−2 5 −1
2 3 4
𝑇
=
1 −2 2
2 5 3
3 −1 4
𝐵𝑇
=
−1 5 3
7 −2 1
2 0 −3
𝑇
=
−1 7 2
5 −2 0
3 1 −3
∴ 𝐵𝑇
𝐴𝑇
=
−1 7 2
5 −2 0
3 1 −3
1 −2 2
2 5 3
3 −1 4

78. =
−1 + 14 + 6 2 + 35 − 2 −2 + 21 + 8
5 − 4 + 0 −10 − 10 + 0 10 − 6 + 0
3 + 2 − 9 −6 + 5 + 3 6 + 3 − 12
=
19 35 27
1 −20 4
−4 2 −3
= 𝑨𝑩 𝑻
∴ 𝑨𝑩 𝑻= 𝑩𝑻𝑨𝑻.

79. 𝐓𝐡𝐚𝐧𝐤 𝐲𝐨𝐮