Matrices

INTRODUCTION
What is a matrix?
A matrix is a rectangular array of elements.
Example :
1 2 3
−2 0 1
4 7 8
,
𝑎 𝑒
𝑐 𝑏
𝑑 𝑓
(i.e) A matrix is a rectangular arrangement of numbers in rows and columns.
Elements
Each number in the array is called an entry or an element of the matrix.
Rows and Columns of a matrix: Rows run horizontally and columns run vertically.

Application of Matrices in real life
Matrices are applied in the study of electrical circuits, quantum
mechanics and optics.
Matrices are helps in the calculation of battery power outputs, resistor
conversion of electrical energy into another useful energy.
In encryption, matrices are used to scramble data for security purpose
to encode and to decode this data we need matrices. There is a key
which helps encode and decode data which is generated by matrices.
Games especially 3D, matrices are used to alter the object, in 3d space.
They use the 3d matrix to 2d matrix to convert it into the different
objects as per requirement. Matrix can help make animations in a more
precise and perfect.
In Robotics and Automation, matrices are the base elements for
the robot movements. The movements of the robots are programmed
with the calculation of matrices' rows and columns.
Matrices uses to study the trends of a business, shares and create
business models etc.,
In Construction, we can see some buildings are straight but sometimes
architects try to change the outer structure of the building like the
famous Burj Khalifa etc. This can be done with matrices.
In Geology, Matrices are used for taking seismic surveys.

Dimensions (or) order of a matrix:
The dimensions or order of a matrix gives the number of rows followed by the
number of columns in a matrix.
(i.e) The dimensions of a matrix are stated “𝑚 𝑥 𝑛” where ‘𝑚’ is the number of rows
and ‘𝑛’ is the number of columns.
Example:1
6 10
4 3
8 1
This matrix is having 3 rows and 2 columns .
Hence the dimension (Or) order of the matrix is 3 X 2
( i.e) it should be read as 3 by 2
Example : 2
𝑎 𝑏 𝑐
𝑑 𝑒 𝑓
𝑔 ℎ 𝑖
This matrix is having 3 rows and 3 columns .
Hence the dimension (Or) order of the matrix is 3 X 3

Matrix Algebra
Properties of Addition
Let A, B and C be 𝑚 𝑥 𝑛 matrices
1. 𝐴 + 𝐵 = 𝐵 + 𝐴 - Commutative
2. 𝐴 + (𝐵 + 𝐶) = (𝐴 + 𝐵) + 𝐶 - Associative
3.There is a unique 𝑚 𝑥 𝑛 matrix O with 𝐴 + 𝑂 = 𝐴 - Additive identity
4.For any 𝑚 𝑥 𝑛 matrix A there is an 𝑚 𝑥 𝑛 matrix B (called –A (Or) Additive Inverse) with
𝐴 + 𝐵 = 𝑂 - Additive inverse
Properties of Matrix Multiplication
Let A, B and C be matrices of dimensions such that the following are defined. Then
1. 𝐴(𝐵𝐶) = (𝐴𝐵)𝐶 - Associative
2. 𝐴(𝐵 + 𝐶) = 𝐴𝐵 + 𝐴𝐶 - Left Distributive
3. (𝐴 + 𝐵)𝐶 = 𝐴𝐶 + 𝐵𝐶 - Right Distributive
4.There is unique matrix 𝐼𝑛 with 𝐼𝑛 𝐴 = 𝐴 𝐼𝑛 = 𝐴 - Multiplicative identity
Properties of Scalar Multiplication
Let r and s be real numbers and A and B be matrices. Then
1.𝑟(𝑠𝐴) = (𝑟𝑠)𝐴
2.(𝑟 + 𝑠)𝐴 = 𝑟𝐴 + 𝑠𝐴
3.𝑟(𝐴 + 𝐵) = 𝑟𝐴 + 𝑟𝐵
4.(𝑟𝐵) = 𝑟(𝐴𝐵) = (𝑟𝐴)𝐵

Types of Matrices
Row Matrix : A matrix with single row is called row matrix
Example:
A = [1 2 3 5] is a row matrix of order 1 X 4
Column Matrix : A matrix with single column is called column matrix.
Example:
𝐴 =
2
0
1
is a column matrix of order 3 X 1
Zero (or) Null Matrix : A matrix with all its elements are zero
Example :
0 0
0 0
0 0
is a zero matrix of order 3 X 2
Singleton Matrix: A matrix with only one element is called singleton matrix.
Example:
1 , 𝑎 , etc

Square Matrix: If in a matrix the number of rows is equal to number of columns , then the matrix is called Square matrix
Example:
𝐴 =
1 0 2
−1 3 4
1 2 5
is a square matrix of order 3 X 3,
𝐴 =
2 1
0 1
is a square matrix of order 2 X 2
Diagonal of a matrix
The diagonal of a matrix consists of elements from the top left corner to the bottom right corner of a square matrix, which is called
the main diagonal, leading diagonal or principle diagonal.
(i.e) the main diagonal consist the elements whose row number and column number are equal.
The other diagonal is called secondary diagonal
Example:
6 3 4
1 5 1
8 2 9
,
The main diagonal consists the elements 6,5,9
The secondary diagonal consists the elements 4,5,8

Diagonal Matrix:
If in a square matrix all the elements except the elements in the main diagonal are zero, then it is
called diagonal matrix
Example:
A=
1 0 0
0 5 0
0 0 2
is a diagonal matrix of order 3 X 3
C =
−1 0
0 5
is a diagonal matrix of order 2 X 2
B = [9] is a diagonal matrix of order 1 X 1
Scalar Matrix :
If in a diagonal matrix all the elements in the main diagonal are equal, then it is called a scalar
matrix.
Example:
𝐴 =
−3 0 0
0 −3 0
0 0 −3
is a scalar matrix of order 3 X 3
𝐵 =
2 0
0 2
is a scalar matrix of order 2 X 2

Identity (or) Unit Matrix :
If all the elements in the main diagonal of a diagonal matrix is equal to 1, then it is called an identity (or) unit matrix.
Note: The identity matrix of order ‘𝑛 𝑋 𝑛’ is denoted by 𝐼𝑛
Example:
𝐼3 =
1 0 0
0 1 0
0 0 1
is the identity matrix of order 3 X 3
𝐼2 =
1 0
0 1
is the identity matrix of order 2 X 2
Upper Triangular Matrix:
In a square matrix the elements below the main diagonal are zero , then it is called upper triangular matrix
Example:
A =
1 3 4
0 5 1
0 0 3
is an upper triangular matrix of order 3 X 3
Lower Triangular Matrix:
In a square matrix the elements above the main diagonal are zero , then it is called lower triangular matrix
Example:
A =
1 0 0
2 5 0
−7 −1 3
is a lower triangular matrix of order 3 X 3

Equal Matrices:
Two matrices are considered equal if they have the same number of rows and columns (the
same dimensions) and all their corresponding elements are exactly the same.
Example:
1.𝐴 =
0 0
0 0
0 0
&𝐵 =
0 0
0 0
are not equal, since A’s order is 3 X 2 and B’s order is 2 X 2
2. If 𝐴 =
𝑎 𝑏
𝑐 𝑑
& 𝐵 =
−1 2
−3 5
are equal, then 𝑎 = −1, 𝑏 = 2, 𝑐 = −3 & 𝑑 = 5
Transpose of Matrix:
Let 𝐴 be any given matrix. The matrix obtained by interchanging rows and columns of A
is called the transpose of A and is denoted by 𝐴’ or 𝐴𝑇
.
Example:
If 𝐴 =
1 −1 2
3 1 1
−2 4 5
8 10 7
, then 𝐴𝑇
=
1 3 −2 8
−1 1 4 10
2 1 5 7
Note : If order of 𝐴 is 𝑚 𝑋 𝑛 then the order of 𝐴𝑇
is 𝑛 𝑋 𝑚

Properties of Transpose of Matrices
1.The transpose of transposed matrix is equal to the matrix itself, i.e. (A’)’ = A
(or) 𝑨𝑻 𝑻
= 𝑨
Example:
If 𝐴 =
1 0 2
−2 −1 0
, then 𝐴𝑇 =
1 −2
0 −1
2 0
Hence 𝐴𝑇 𝑇 =
1 0 2
−2 −1 0
= 𝐴
2. The transpose of the sum of the two matrices is equal to the transpose of the matrices, i.e.
(A+B)’ = A’+B’ (or) 𝑨 + 𝑩 𝑻 = 𝑨𝑻 + 𝑩𝑻.
Example:
If 𝐴 =
0 2
−1 4
& 𝐵 =
−4 2
1 0
, then 𝐴𝑇 =
0 −1
2 4
, 𝐵𝑇 =
−4 1
2 0
, 𝐴 + 𝐵 =
0 2
−1 4
+
−4 2
1 0
=
−4 4
0 4
& 𝐴𝑇
+ 𝐵𝑇
=
0 −1
2 4
+
−4 1
2 0
=
−4 0
4 4
Hence 𝐴 + 𝐵 𝑇
=
−4 0
4 4
= 𝐴𝑇
+ 𝐵𝑇

3. The transpose of the product of two matrices is equal to the product of their transposes in the reverse order,
i.e. (AB)’ = B’A’. (or) 𝑨𝑩 𝑻
= 𝑩𝑻
𝑨𝑻
.
Example:
If 𝐴 =
2
1
& 𝐵 = −3 2 , then 𝐴𝐵 =
2
1
−3 2 =
−6 4
−3 2
, 𝐴𝑇
= 2 1 , 𝐵𝑇
=
−3
2
&
𝐵𝑇
𝐴𝑇
=
−3
2
2 1 =
−6 −3
4 2
Hence 𝐴𝐵 𝑇 =
−6 −3
4 2
= 𝐵𝑇𝐴𝑇
4. The transpose of a matrix times a scalar (k) is equal to the constant times the transpose of the matrix, i.e.
(kA)’ = k A’ (or) 𝒌𝑨 𝑻 = 𝒌 𝑨𝑻.
Example:
If 𝐴 =
2 4
−1 0
0 5
and k = 2 , then 𝑘𝐴 = 2
2 4
−1 0
0 5
=
4 8
−2 0
0 10
, 𝐴𝑇 =
2 −1 0
4 0 5
& 𝑘𝐴𝑇
= 2
2 −1 0
4 0 5
=
4 −2 0
8 0 10
Hence 𝑘𝐴 𝑇
=
4 −2 0
8 0 10
= 𝑘𝐴𝑇

Determinant of a Matrix
The determinant of a matrix 𝐴 is a special number that can be calculated from a square matrix. It
is denoted by 𝐴 (or) det(𝐴)
Calculating the Determinant
The determinant of a 2 x 2 matrix A =
𝑎 𝑏
𝑐 𝑑
is given by
det 𝐴 = 𝐴 =
𝑎 𝑏
𝑐 𝑑
= 𝑎𝑑 − 𝑏𝑐
Example :
If 𝐴 =
2 −1
4 3
, then det A =
2 −1
4 3
= 2 3 − 4 −1 = 6 − 4 = 6 + 4 = 10
The determinant of 3 X 3 matrix 𝐴 =
𝑎 𝑏 𝑐
𝑑 𝑒 𝑓
𝑔 ℎ 𝑖
is given by
det 𝐴 = 𝐴 =
𝑎 𝑏 𝑐
𝑑 𝑒 𝑓
𝑔 ℎ 𝑖
= 𝑎
𝑒 𝑓
ℎ 𝑖
− 𝑏
𝑑 𝑓
𝑔 𝑖
+ 𝑐
𝑑 𝑒
𝑔 ℎ
= 𝑎 𝑒𝑖 − ℎ𝑓 − 𝑏 𝑑𝑖 − 𝑔𝑓 + 𝑐(𝑑ℎ − 𝑔𝑒) = 𝑎𝑒𝑖 − 𝑎ℎ𝑓 − 𝑏𝑑𝑖 + 𝑏𝑔𝑓 + 𝑐𝑑𝑔 − 𝑐𝑔𝑒

Shortcut to find the determinant of a 3 X 3 matrix

Example:
The determinant of 𝐴 =
2 −5 3
0 7 −2
−1 4 1
is given by
𝐷𝑒𝑡 𝐴 = 𝐴 =
2 −5 3
0 7 −2
−1 4 1
= 2
7 −2
4 1
− −5
0 −2
−1 1
+3
0 7
−1 4
= 2 7 1 − 4 −2 + 5 0 1 − −1 −2 + 3[ 0 4 − −1 7 ]
= 2 7 + 8 + 5 0 − 2 + 3 0 + 7 = 30 − 10 + 21 = 41
Using shortcut :
2 −5 3
0 7 −2
−1 4 1
Part I
𝐻𝑒𝑟𝑒 2 7 1 + −5 −2 −1 + 3 0 4
= 14 − 10 + 0 = 4

𝐻𝑒𝑟𝑒 (−1)(7)(3) + (4)(−2)(2) + (1)(0)(−5)
= −21 − 16 + 0 = −37
Part - II
2 −5 3
0 7 −2
−1 4 1
= (𝑷𝒂𝒓𝒕 – 𝑰) − ( 𝑷𝒂𝒓𝒕 – 𝑰𝑰)
= 𝟒 – − 𝟑𝟕 = 𝟒 + 𝟑𝟕 = 𝟒𝟏

Singular & Non-Singular Matrix :
If a square matrix A is said to be a singular matrix, then 𝐴 = 0
If a square matrix A is said to be a non-singular matrix, then 𝐴 ≠ 0
Example:
1. Consider 𝐴 =
1 1
1 1
det(A) =
1 1
1 1
= 1 − 1 = 0
Hence the given matrix 𝐴 is singular
2. Consider 𝐴 =
1 2
1 −1
det(A) =
1 2
1 −1
= −1 − 2 = −3 ≠ 0
Hence the given matrix 𝐴 is non-singular

Minor of a Matrix:
Minor of a matrix A is the determinant of some smaller square matrix, cut down from A by
removing one or more of its rows and columns. It is denoted by 𝑀𝑖𝑗 (minor of the element which
is in 𝑖, 𝑗 𝑡ℎ position)
Example:
Consider the matrix A =
2 0 5
1 3 4
−2 −1 6
The minor of 2 ( 1,1 𝑡ℎ position) is 𝑀11 =
3 4
1 0
The minor of -1 ( 3,2 𝑡ℎ position) is 𝑀32 =
2 5
1 4
Cofactor of a Matrix:
A cofactor is the number you get when you remove the column and row of a designated element
in a matrix, which is just a numerical grid in the form of a rectangle or a square.
The cofactor is always preceded by a positive (+) or negative (-) sign, depending whether the
element is in a + or - position.
To write the cofactor matrix of a 3 X 3 matrix, we need the following cofactor sign matrix
+ − +
− + −
+ − +
The cofactor matrix is
+𝑀11 −𝑀12 +𝑀13
−𝑀21 +𝑀22 −𝑀23
+𝑀31 −𝑀32 +𝑀33

Example :
Find the Cofactor matrix of A =
1 −1 3
2 5 4
3 7 −2
To find the minor elements
𝑀11 =
5 4
7 −2
= −10 − 28 = −38
𝑀12 =
2 4
3 −2
= −4 − 12 = −16
𝑀13 =
2 5
3 7
= 14 − 15 = −1
𝑀21 =
−1 3
7 −2
= 2 − 21 = −19
𝑀22 =
1 3
3 −2
= −2 − 9 = −11
𝑀23 =
1 −1
3 7
= 7 + 3 = 10
𝑀31 =
−1 3
5 4
= −4 − 15 = −19
𝑀32 =
1 3
2 4
= 4 − 6 = −2
𝑀33 =
1 −1
2 5
= 5 + 2 = 7
The Cofactor matrix is
+𝑀11 −𝑀12 +𝑀13
−𝑀21 +𝑀22 −𝑀23
+𝑀31 −𝑀32 +𝑀33
=
+(−38) −(−16) +(−1)
−(−19) +(−11) −(10)
+(−19) −(−2) +(7)
=
−38 16 −1
19 −11 −10
−19 2 7

Adjoint of a Matrix:
If A is any square matrix, then the adjoint of A is defined as the transpose of the corresponding co
factor matrix of A.
It is denoted by Adj. A (or) adj.A
Adj.A = Transpose of the cofactor matrix of A
Example:
Find the adjoint matrix of A =
1 −1 3
2 5 4
3 7 −2
From the previous problem, the co-factor matrix of A is given by
−38 16 −1
19 −11 −10
−19 2 7
Hence adj.A = Transpose of the cofactor matrix
=
−38 16 −1
19 −11 −10
−19 2 7
𝑇
=
−38 19 −19
16 −11 2
−1 −10 7
Inverse Matrix: (Invertible Matrix)
The inverse of a non-singular matrix A, denoted by 𝐴−1, is the matrix so that the product of A and
A-1 is the Identity matrix.
(i.e.) 𝐴. 𝐴−1 = 𝐴−1𝐴 = 𝐼
A square matrix that has an inverse is called invertible.

Note : To have an inverse , the matrix must be non-singular.
Method to find the inverse of a matrix A:
Step 1. Find 𝐴
Step 2. Write the co-factor matrix of A
Step 3. Write the adjoint matrix of A ( i.e.) adj.A = transpose of cofactor matrix of A.
Step 4. Inverse of A = 𝐴−1 =
1
𝐴
𝑎𝑑𝑗. 𝐴
Example : Find the inverse of A =
1 −1 3
2 5 4
3 7 −2
Here 𝐴 =
1 −1 3
2 5 4
3 7 −2
= 1 −10 − 28 − −1 −4 − 12 + 3(14 − 15)
= −38 − 16 − 3 = −57 ≠ 0
Hence the given matrix is non-singular.
From the previous problem, we know that adj.A =
−38 19 −19
16 −11 2
−1 −10 7
Therefore the inverse of A = 𝐴−1 =
1
𝐴
𝑎𝑑𝑗. 𝐴 =
1
−57
−38 19 −19
16 −11 2
−1 −10 7
=
−38
−57
19
−57
−19
−57
16
−57
−11
−57
2
−57
−1
−57
−10
−57
7
−57
=
2
3
−
1
3
1
3
−
16
57
11
57
−
2
57
1
57
10
57
−
7
57

Properties of Inverse Matrix:
1.If 𝐴 is nonsingular, then so is 𝐴−1
and 𝐴−1 −1
= 𝐴
2.If 𝐴 and 𝐵 are nonsingular matrices, then 𝐴𝐵 is nonsingular and 𝐴𝐵 −1 = 𝐵−1𝐴−1
3. If 𝐴,𝐵 and 𝐶 are nonsingular matrices, then 𝐴𝐵𝐶 is nonsingular and 𝐴𝐵𝐶 −1
= 𝐶−1
𝐵−1
𝐴−1
4.If 𝐴 is nonsingular then 𝐴𝑇 −1
= 𝐴−1 𝑇
5.If 𝐴 and 𝐵 are matrices with 𝐴𝐵 = 𝐼𝑛, then 𝐴 and 𝐵 are inverses of each other.
6. If 𝐴 is a square matrix where 𝑛 > 0, then 𝐴−1 𝑛 = 𝐴−𝑛 ,where 𝐴−𝑛 = 𝐴𝑛 −1
7. If 𝐴 is a square matrix and k is any scalar , then 𝑘𝐴 −1 =
1
𝑘
𝐴−1
Note : Shortcut to find the inverse of 2 X 2 matrix:
Step 1. Find 𝐴
Step 2. Inter change the elements in the main diagonal.
Step 3. Change the sign of the elements in the secondary diagonal.
Step 4. Then multiply the matrix by
1
𝐴

Symmetric Matrix
A square matrix A is said to be symmetric, if 𝐴 = 𝐴𝑇
The entries of a symmetric matrix are symmetric with respect to the main diagonal.
Example :
1.Consider 𝐴 =
1 4 −5
4 0 11
−5 11 −2
𝐴𝑇
=
1 4 −5
4 0 11
−5 11 −2
Clearly 𝐴 = 𝐴𝑇 , hence 𝐴 is symmetric
Note : the numbers in the both sides of main diagonal are equal
Properties of Symmetric Matrix
1.Addition and difference of two symmetric matrices results in symmetric matrix,
(i.e.) If A and B are two symmetric matrices, then 𝐴 ± 𝐵 is also a symmetric matrix
2.If A and B are two symmetric matrices and they follow the commutative property,
(i.e.) AB =BA, then the product of A and B is symmetric.
3.If matrix A is symmetric then 𝐴𝑛 is also symmetric, where 𝑛 is an integer.
4.If A is a symmetric matrix then 𝐴−1 is also symmetric.

Skew-Symmetric Matrix (or) Anti-Symmetric Matrix
A square matrix A is said to be symmetric, if 𝐴 = −𝐴𝑇
Example :
1. Prove that 𝑨 =
𝟎 −𝟐 𝟏𝟎
𝟐 𝟎 𝟒
−𝟏𝟎 −𝟒 𝟎
is skew- symmetric.
Solution:
𝐴𝑇 =
0 2 −10
−2 0 −4
10 4 0
⇒ −𝐴𝑇=
0 −2 10
2 0 4
−10 −4 0
Clearly 𝐴 = −𝐴𝑇 , hence 𝐴 is skew-symmetric
2. If 𝑨 =
𝟎 𝒚 −𝟒
−𝟓 𝟎 𝟖
𝒙 + 𝒚 −𝟖 𝟎
is skew-symmetric , then find the values of x and y.
Solution:
Given 𝐴 =
0 𝑦 −4
−5 0 8
𝑥 + 𝑦 −8 0
is skew-symmetric.

⇒ 𝐴 = −𝐴𝑇
------- (1)
Now 𝐴𝑇
=
0 −5 𝑥 + 𝑦
𝑦 0 −8
−4 8 0
(1) ⇒
0 𝑦 −4
−5 0 8
𝑥 + 𝑦 −8 0
= −
0 −5 𝑥 + 𝑦
𝑦 0 −8
−4 8 0
⇒
0 𝑦 −4
−5 0 8
𝑥 + 𝑦 −8 0
=
0 5 −(𝑥 + 𝑦)
−𝑦 0 8
4 −8 0
⇒ 𝑦 = 5 & 𝑥 + 𝑦 = 4
⇒ 𝑥 = 4 − 𝑦 = 4 − 5 = −1
3. If 𝑨 =
𝟎 𝒙 + 𝒚 𝟔
𝟑 𝟎 𝟗
𝒙 𝟗 𝟎
is skew-symmetric , then find the values of x and y.
Solution:
Given 𝐴 =
0 𝑥 + 𝑦 6
3 0 9
𝑥 9 0
is skew-symmetric.
⇒ 𝐴 = −𝐴𝑇
------- (1)
Now 𝐴𝑇
=
0 3 𝑥
𝑥 + 𝑦 0 9
6 9 0
(1) ⇒
0 𝑥 + 𝑦 6
3 0 9
𝑥 9 0
= −
0 3 𝑥
𝑥 + 𝑦 0 9
6 9 0
⇒
0 𝑥 + 𝑦 6
3 0 9
𝑥 9 0
= −(
0 −3 −𝑥
𝑥 + 𝑦) 0 −9
−6 −9 0
⇒ 𝑥 + 𝑦 = 3 & 𝑥 = −6
⇒ 𝑦 = 3 − 𝑥 = 3 + 6 = 9

Power of Matrix:
1. If 𝐴2
= 𝐴, then the matrix 𝐴 is called idempotent.
Example:
Consider 𝐴 =
3 −6
1 −2
𝐴2 = 𝐴. 𝐴 =
3 −6
1 −2
3 −6
1 −2
=
3 . 3 + (−6)(1) 3 −6 + (−6)(−2)
1 3 + (−2)(1) 1 −6 + (−2)(−2)
=
3 −6
1 −2
= 𝐴
Hence 𝐴 is idempotent.
Property of an idempotent Matrix:
If 𝐴 is idempotent and 𝐼 is the identity matrix of same order , then (𝐴 − 𝐼) is also idempotent.
2. If 𝐴2 = 𝐼, then the matrix 𝐴 is called involuntory.
Example:
Consider 𝐴 =
4 −1
15 −4
𝐴2 = 𝐴. 𝐴 =
4 −1
15 −4
4 −1
15 −4
=
4 4 + (−1)(15) 4 −1 + (−1)(−4)
15 4 + (−4)(15) 15 −1 + (−4)(−4)
=
1 0
0 1
= 𝐼
Hence 𝐴 is involuntory.
Property of an involuntory Matrix:
If 𝐴 is involuntary , then 𝐴 = 1 𝑜𝑟 − 1
If 𝐴 and 𝐵 are involuntory and 𝐴𝐵 = 𝐵𝐴, then 𝐴𝐵 is also involuntory.
If 𝐴 is involuntary , then
1
2
(𝐴 + 𝐼) is involuntory.

3. If 𝐴𝑘+1 = 𝐴, then the matrix 𝐴 is called periodic with period k.
Example:
Consider 𝐴 =
0 1
1 0
𝐴2
= 𝐴. 𝐴 =
0 1
1 0
0 1
1 0
=
1 0
0 1
𝐴3 = 𝐴2𝐴 =
1 0
0 1
0 1
1 0
=
0 1
1 0
= 𝐴
(i.e.) 𝐴3 = 𝐴
⟹ 𝐴2+1 = 𝐴
⟹ 𝐴 is periodic with period 2.
4. If 𝐴𝑘 = 0, then the matrix 𝐴 is called nilpotent.
Example:
Consider 𝐴 =
0 0
1 0
𝐴2 = 𝐴. 𝐴 =
0 0
1 0
0 0
1 0
=
0 0
0 0
Hence 𝐴 is nilpotent.

Trace of a matrix A
If A is a square matrix, then the Trace of A = Sum of the elements in the main diagonal of A
Example :
Find the trace of the matrix 𝐴 =
2 1 0
5 4 8
10 −3 4
The Trace of 𝐴 = 2 + 4 + 4 = 10
Eigen Value and Eigen Vector:
The German word "eigen" roughly translates as "own" , "belonging to", "proper", "characteristic“.
If A is a square matrix, then there exists a vector X and a scalar 𝜆 such that 𝐴𝑋 = 𝜆𝑋.
The scalar 𝜆 is called eigen value of A and X is called corresponding eigen vector.

Example:
An eigenvector does not change direction in a transformation:

Almost all vectors change direction, when they are multiplied by a matrix A. Certain exceptional
vectors X are in the same direction as AX. Those are the “eigenvectors”. Multiply an eigenvector
by A, and the vector AX is a number 𝜆 times the original X.
The basic equation is AX = 𝜆X. The number 𝜆 is an eigenvalue of A.
⟹ 𝐴𝑋 = 𝜆𝐼𝑋, Where 𝐼 is the identity matrix
⟹ 𝐴𝑋 − 𝜆𝐼𝑋 = 0
⟹ 𝐴 − 𝜆𝐼 𝑋 = 0
If X is non-zero then we can solve for 𝜆 by using the determinant 𝐴 − 𝜆𝐼 = 0
The equation 𝐴 − 𝜆𝐼 = 0 is called characteristic equation of the matrix A.
Method to find eigen values and eigen vectors:
Step 1. Write the characteristic equation 𝐴 − 𝜆𝐼 = 0
Step 2. Solve the characteristic equation and find the values of 𝜆 (for 2X2 matrix , there will be
two values for 𝜆, for 3X3 matrix, there will be three values for 𝜆, etc.,)
The values for 𝜆 are called eigen values of the matrix A.
Step 3. Substitute the values of 𝜆 in 𝐴 − 𝜆𝐼 𝑋 = 0 and find its corresponding matrix X
The corresponding matrix X is called eigen vector.

Example:
Find the eigen values and eigen vectors of 𝑨 =
−𝟔 𝟑
𝟒 𝟓
Solution:
Given 𝐴 =
−6 3
4 5
To find the eigen values:
The characteristic equation is 𝐴 − 𝜆𝐼 = 0, where 𝜆 is any scalar (eigen value).
⟹
−6 3
4 5
− 𝜆
1 0
0 1
= 0
⟹
−6 − 𝜆 3
4 5 − 𝜆
= 0
⟹ −6 − 𝜆 5 − 𝜆 − 4 3 = 0
⟹ − 6 + 𝜆 5 − 𝜆 − 12=0
⟹ 6 + 𝜆 5 − 𝜆 + 12 = 0
⟹ 30 − 𝜆2
+ 5𝜆 − 6𝜆 + 12 = 0
⟹ −𝜆2 − 𝜆 + 42 = 0
⟹ 𝜆2
+ 𝜆 − 42 = 0
⟹ 𝜆 − 6 𝜆 + 7 = 0
⟹ 𝜆 = 6 , −7
Hence the eigen values are 𝜆1 = 6 , 𝜆2 = −7.

To find the eigen vector corresponding to the eigen value 𝝀𝟏 = 𝟔:
Let us take the eigen vector be 𝑋1 =
𝑥1
𝑥2
We know that 𝐴 − 𝜆𝐼 𝑋 = 0
⟹ 𝐴 − 𝜆1𝐼 𝑋1 = 0
⟹
−6 3
4 5
− 6
1 0
0 1
𝑥1
𝑥2
=
0
0
⟹
−6 3
4 5
−
6 0
0 6
𝑥1
𝑥2
=
0
0
⟹
−12 3
4 −1
𝑥1
𝑥2
=
0
0
We will get
−12𝑥1 + 3𝑥2 = 0 ⟹ −4𝑥1 + 𝑥2 = 0 ⟹ 4𝑥1 − 𝑥2 = 0 -----(1)
and 4𝑥1 − 𝑥2 = 0 ------(2)
The equations (1) & (2) are same
So take 𝑥1 = 1, using this in (1), we will get 4 1 − 𝑥2 = 0
⟹ 𝑥2 = 4
Hence the eigen vector is 𝑋1 =
1
4

To find the eigen vector corresponding to the eigen value 𝝀𝟏 = −𝟕:
𝑥1
𝑥2
⟹ 𝐴 − 𝜆1𝐼 𝑋1 = 0
⟹
−6 3
4 5
+ 7
1 0
0 1
𝑥1
𝑥2
=
0
0
⟹
−6 3
4 5
+
7 0
0 7
𝑥1
𝑥2
=
0
0
⟹
1 3
4 12
𝑥1
𝑥2
=
0
0
We will get
𝑥1 + 3𝑥2 = 0 -----(1)
and 4𝑥1 + 12𝑥2 = 0 ⟹ 𝑥1 + 3𝑥2 = 0 ------(2)
The equations (1) & (2) are same
So take 𝑥1 = 1, using this in (1), we will get 1 + 3𝑥2 = 0
⟹ 𝑥2 = −
1
3
1
−
1

Example:
𝟑 𝟏
𝟎 𝟐
Solution:
Given 𝐴 =
3 1
0 2
To find the eigen values:
The characteristic equation is 𝐴 − 𝜆𝐼 = 0, where 𝜆 is any scalar (eigen value).
⟹
3 1
0 2
− 𝜆
1 0
0 1
= 0
⟹
3 − 𝜆 1
0 2 − 𝜆
= 0
⟹ 3 − 𝜆 2 − 𝜆 − 0 1 = 0
⟹ 3 − 𝜆 2 − 𝜆 =0
⟹ 𝜆 = 3 , 𝜆 = 2
Hence the eigen values are 𝜆1 = 3 , 𝜆2 = 2.

To find the eigen vector corresponding to the eigen value 𝝀𝟏 = 𝟑:
𝑥1
𝑥2
⟹ 𝐴 − 𝜆1𝐼 𝑋1 = 0
⟹
3 1
0 2
− 3
1 0
0 1
𝑥1
𝑥2
=
0
0
⟹
3 1
0 2
−
3 0
0 3
𝑥1
𝑥2
=
0
0
⟹
3 − 3 1 − 0
0 − 0 2 − 3
𝑥1
𝑥2
=
0
0
⟹
0 1
0 −1
𝑥1
𝑥2
=
0
0
⟹ 𝑥2 = 0
Take 𝑥1 = 1
1

To find the eigen vector corresponding to the eigen value 𝝀𝟐 = 𝟐:
𝑥1
𝑥2
⟹ 𝐴 − 𝜆2𝐼 𝑋2 = 0
⟹
3 1
0 2
− 2
1 0
0 1
𝑥1
𝑥2
=
0
0
⟹
3 1
0 2
−
2 0
0 2
𝑥1
𝑥2
=
0
0
⟹
3 − 2 1 − 0
0 − 0 2 − 2
𝑥1
𝑥2
=
0
0
⟹
1 1
0 0
𝑥1
𝑥2
=
0
0
⟹ 𝑥1 + 𝑥2 = 0 ⟹ 𝑥1 = −𝑥2
Take 𝑥1 = 1 ⟹ 𝑥2 = −1
1
−1

Note : To find solution for two variables linear equations.
In homogeneous system of linear equations, there are infinite
solutions, where you have
1.Only one equation with two variables (i.e) 𝑎𝑥 + 𝑏𝑦 = 0
2. Two equations with no constant terms.
(i.e) 𝑎1 𝑥 + 𝑏1 𝑦 = 0 & 𝑎2 𝑥 + 𝑏2 𝑦 = 0
In both cases, we have to choose a convenient value for 𝑥1 = 1
(say), using this you will get the value of 𝑥2

Shortcut method to write the characteristic equation of a 3X3 matrix A:
If A is a square matrix of order 3, then its characteristic equation can be written as
𝝀𝟑 − 𝑺𝟏𝝀𝟐 + 𝑺𝟐𝝀 − 𝑺𝟑 = 𝟎 ,
Where
𝑺𝟏 = 𝐒𝐮𝐦 𝐨𝐟 𝐭𝐡𝐞 𝐞𝐥𝐞𝐦𝐞𝐧𝐭𝐬 𝐢𝐧 𝐭𝐡𝐞 𝐦𝐚𝐢𝐧 𝐝𝐢𝐚𝐠𝐨𝐧𝐚𝐥 𝐨𝐟 𝐀,
𝑺𝟐 = Sum of the 2 X 2 minors of the diagonal elements of A
𝑺𝟑 = |𝑨|
Example :
−𝟐 −𝟒 𝟐
−𝟐 𝟏 𝟐
𝟒 𝟐 𝟓
Solution:
To find the characteristic equation:
We know that the characteristic equation is given by
𝜆3
− 𝑆1𝜆2
+ 𝑆2𝜆 − 𝑆3 = 0 ,
Where
𝑆1 = Sum of the elements in the main diagonal of 𝐴,
𝑆2 = Sum of the 2 X 2 minors of the diagonal elements of A
𝑆3 = |𝐴|

𝑆1 = −2 + 1 + 5 = 4
𝑆2 = Minor of (-2) + Minor of (1) + Minor of (5)
=
1 2
2 5
+
−2 2
4 5
+
−2 −4
−2 1
= 1 5 − 2 2 + −2 5 − 4 2 + −2 1 − (−2)(−4)
= 5 − 4 − 10 − 8 − 2 − 8 = −27
𝑆3 = 𝐴 =
−2 −4 2
−2 1 2
4 2 5
= −2 5 − 4 − −4 −10 − 8 + 2 −4 − 4 = −2 − 72 − 16 = −90
Hence the characteristic equation is 𝜆3 − 𝑆1𝜆2 + 𝑆2𝜆 − 𝑆3 = 0
⟹ 𝜆3 − 4𝜆2 − 27𝜆 + 90 = 0
To find the eigen values :
We have to solve the characteristic equation 𝜆3
− 4𝜆2
− 27𝜆 + 90 = 0
3 1 -4 -27 90
0 3 -3 -90
1 -1 -30 0
Hence 𝜆 = 3 is a root
The given equation reduced to (1)𝜆2+ −1 𝜆 + (−30) = 0
⟹ 𝜆2 +𝜆 − 30 = 0

⟹ 𝜆 + 5 𝜆 − 6 = 0
⟹ 𝜆 = −5, 6
Hence the values of 𝜆 = 3, −5,6
The eigen values are 𝜆1 = 3, 𝜆2 = −5, 𝜆3 = 6
To find the eigen vector corresponding to 𝝀𝟏 = 𝟑:
Let the eigen vector be 𝑋1 =
𝑥1
𝑦1
𝑧1
⟹ 𝐴 − 𝜆1𝐼 𝑋1 = 0
⟹
−2 −4 2
−2 1 2
4 2 5
− 𝜆1
1 0 0
0 1 0
0 0 1
𝑥1
𝑦1
𝑧1
=
0
0
0
⟹
−2 − 𝜆1 −4 2
−2 1 − 𝜆1 2
4 2 5 − 𝜆1
𝑥1
𝑦1
𝑧1
=
0
0
0
⟹
−2 − 3 −4 2
−2 1 − 3 2
4 2 5 − 3
𝑥1
𝑦1
𝑧1
=
0
0
0
⟹
−5 −4 2
−2 −2 2
4 2 2
𝑥1
𝑦1
𝑧1
=
0
0
0

⟹ −5𝑥1 − 4𝑦1 + 2𝑧1 = 0 --------- (1)
−2𝑥1 − 2𝑦1 + 2𝑧1 = 0 ⟹ 𝑥1 + 𝑦1 − 𝑧1 = 0 -------(2)
4𝑥1 + 2𝑦1 + 2𝑧1 = 0 ⟹ 2𝑥1 + 𝑦1 + 𝑧1 = 0 ------- (3)
Method 1: Solving (2) & (3),(Elimination method)
𝑥1 + 𝑦1 − 𝑧1 = 0
2𝑥1 + 𝑦1 + 𝑧1 = 0
-------------------------
Add 3𝑥1 + 2𝑦1 = 0
⟹ 3𝑥1 = −2𝑦1
Since this equation has infinite number of solutions, for our convenient, Take 𝑥1 = 1,
⟹ 3 1 = −2𝑦1
⟹ 𝑦1 =
3
−2
= −
3
2
Put 𝑥1 = 1, 𝑦1 = −
3
2
in (1)
⇒ −5(1) − 4 −
3
2
+ 2𝑧1 = 0
⇒ −5 + 6 + 2𝑧1 = 0
⇒ 1 + 2𝑧1 = 0
⇒ 2𝑧1 = −1 ⇒ 𝑧1 = −
1
2

Hence the eigen vector𝑋1 =
𝑥1
𝑦1
𝑧1
=
1
−
3
2
−
1
2
(or) it can be written as
2
−3
−1
Method 2: Solving (2) & (3),(cross multiplication method)
𝑥1 + 𝑦1 − 𝑧1 = 0 -------(2)
2𝑥1 + 𝑦1 + 𝑧1 = 0 ------- (3)
𝑥1 𝑦1 𝑧1
1 -1 1 1
1 1 2 1
⟹
𝑥1
( 1 1 − 1 −1 )
=
𝑦1
−1 2 − 1 1
=
𝑧1
1 1 − 2 1
⟹
𝑥1
1+1
=
𝑦1
−2−1
=
𝑧1
1−2
⟹
𝑥1
2
=
𝑦1
−3
=
𝑧1
−1
⟹ 𝑥1 = 2, 𝑦1 = −3, 𝑧1 = −1
𝑥1
𝑦1
𝑧1
=
2
−3
−1
1
−
3
2
−
1
2

𝑥1
𝑦1
𝑧1
=
1
−
3
2
−
1
2
2
−3
−1
To find the eigen vector corresponding to 𝝀𝟐 = −𝟓:
𝑥2
𝑦2
𝑧2
⟹ 𝐴 − 𝜆2𝐼 𝑋2 = 0
⟹
−2 −4 2
−2 1 2
4 2 5
− 𝜆1
1 0 0
0 1 0
0 0 1
𝑥2
𝑦2
𝑧2
=
0
0
0
⟹
−2 − 𝜆2 −4 2
−2 1 − 𝜆2 2
4 2 5 − 𝜆2
𝑥2
𝑦2
𝑧2
=
0
0
0
⟹
−2 + 5 −4 2
−2 1 + 5 2
4 2 5 + 5
𝑥2
𝑦2
𝑧2
=
0
0
0
⟹
3 −4 2
−2 6 2
4 2 10
𝑥2
𝑦2
𝑧2
=
0
0
0

⟹ 3𝑥2 − 4𝑦2 + 2𝑧2 = 0 ------- (1)
−2𝑥2 + 6𝑦2 + 2𝑧2 = 0 ⇒ −𝑥2 + 3𝑦2 + 𝑧2 = 0-------(2)
4𝑥2 + 2𝑦2 + 10𝑧2 = 0 ⇒ 2𝑥2 + 𝑦2 + 5𝑧2 = 0 -------- (3)
Solving (2) & (3),
𝑥2 𝑦2 𝑧2
3 1 -1 3
1 5 2 1
⟹
𝑥2
15−1
=
𝑦2
2+5
=
𝑧2
−1−6
⟹
𝑥2
14
=
𝑦2
7
=
𝑧2
−7
Multiply by 7, we will get
⟹
𝑥2
2
=
𝑦2
1
=
𝑧2
−1
𝑥2
𝑦2
𝑧2
=
2
1
−1
1
1
2
−
1
2

To find the eigen vector corresponding to 𝝀𝟑 = 𝟔:
𝑥3
𝑦3
𝑧3
⟹ 𝐴 − 𝜆3𝐼 𝑋3 = 0
⟹
−2 −4 2
−2 1 2
4 2 5
− 𝜆3
1 0 0
0 1 0
0 0 1
𝑥3
𝑦3
𝑧3
=
0
0
0
⟹
−2 − 𝜆3 −4 2
−2 1 − 𝜆3 2
4 2 5 − 𝜆3
𝑥3
𝑦3
𝑧3
=
0
0
0
⟹
−2 − 6 −4 2
−2 1 − 6 2
4 2 5 − 6
𝑥3
𝑦3
𝑧3
=
0
0
0
⟹
−8 −4 2
−2 −5 2
4 2 −1
𝑥3
𝑦3
𝑧3
=
0
0
0
⟹ −8𝑥3 − 4𝑦3 + 2𝑧3 = 0 ⇒ −4𝑥3 − 2𝑦3 + 𝑧3 = 0 -------(1)
−2𝑥3 − 5𝑦3 + 2𝑧3 = 0 ----- (2)
4𝑥3 + 2𝑦3 − 𝑧3 = 0 ----------(3)

Solving (1) & (2),
𝑥3 𝑦3 𝑧3
-2 1 -4 2
-5 2 -2 -5
⟹
𝑥3
−4+5
=
𝑦3
−2+8
=
𝑧3
20+4
⟹
𝑥3
1
=
𝑦3
6
=
𝑧3
24
𝑥3
𝑦3
𝑧3
=
1
6
24
Note:
1.In a system of linear equations , if all the three equations are distinct, then we can have three independent
eigen vectors.(𝑎1𝑥1 + 𝑏1𝑦1 + 𝑐1𝑧1 = 0, 𝑎2𝑥1 + 𝑏2𝑦1 + 𝑐2𝑧1 = 0, 𝑎3𝑥1 + 𝑏3𝑦1 + 𝑐3𝑧1 = 0)
To solve the linear equation ,take any two of them and use cross multiplication method to get the values of
𝑥1, 𝑦1, 𝑧1.
2.In a system of linear equations , if any two are similar among three, then we may or may not get
independent eigen vectors.(𝑎1𝑥1 + 𝑏1𝑦1 + 𝑐1𝑧1 = 0, 𝑎1𝑥1 + 𝑏1𝑦1 + 𝑐1𝑧1 = 0, 𝑎3𝑥1 + 𝑏3𝑦1 + 𝑐3𝑧1 = 0)
To solve the linear equation ,take the two which are distinct and use cross multiplication method to get the
values of 𝑥1, 𝑦1, 𝑧1.
3.In a system of linear equations , if all the three are similar then we can have infinite number of
solutions.(𝑎1𝑥1 + 𝑏1𝑦1 + 𝑐1𝑧1 = 0, 𝑎1𝑥1 + 𝑏1𝑦1 + 𝑐1𝑧1 = 0, 𝑎1𝑥1 + 𝑏1𝑦1 + 𝑐1𝑧1 = 0)
To solve the linear equation ,take 𝑥1 = 0, 𝑦1 = 1 and use the equation to find 𝑧1.

Find the Eigen values and Eigen vectors
−𝟐 𝟐 −𝟑
𝟐 𝟏 −𝟔
−𝟏 −𝟐 𝟎
Solution
W.K.T Characteristic equation is 𝜆3 − 𝑆1𝜆2 + 𝑆2𝜆 − 𝑆3 = 0
Where S1 = sum of the main diagonal = −2 + 1 + 0 = −1
𝑆1 = −1
S2 = sum of the 2x2 minors of diagonal elements
)
= minor of −2 + minor of 1 + minor of (0
=
1 −6
−2 0
+
−2 −3
−1 0
+
−2 2
2 1
= (0 − 12) + (0 − 3) + (−2 − 4)
𝑆2 = −21
𝑺3 = 𝑨 =
−2 2 −3
2 1 −6
−1 −2 0
= −2 0 − 12 − 2 0 − 6 − 3(−4 + 1)
= 24 + 12 + 9 = 45
𝑆3 = 45

The characteristic equation is 𝜆3 + 𝜆2 − 21𝜆 − 45 = 0
-3 1 1 -21 -45
0 -3 6 45
1 -2 -15 0
Hence 𝜆1 = −3 is a root
Remaining roots are given by
𝜆2 − 2𝜆 − 15 = 0
𝜆 − 5 𝜆 + 3 = 0
⟹ 𝜆2 = −3 & 𝜆3 = 5
Hence Eigen values are 𝜆 = −3, −3, 5 ( two roots are identical/equal )
To find the Eigen vectors 𝑨 − 𝝀𝑰 (𝑿) = (𝟎)
(i.e)
−2 − 𝜆 2 −3
2 1 − 𝜆 −6
−1 −2 0 − 𝜆
𝑥1
𝑥2
𝑥3
=
0
0
0
----------- (A)
When 𝜆 = −3 , (A) becomes
−2 + 3 2 −3
2 1 + 3 −6
−1 −2 0 + 3
𝑥1
𝑥2
𝑥3
=
0
0
0

1 2 −3
2 4 −6
−1 −2 3
𝑥1
𝑥2
𝑥3
=
0
0
0
The three equations are
𝑥1 + 2𝑥2 − 3𝑥3 = 0 ---------- (1)
2𝑥1 + 4𝑥2 − 6𝑥3 = 0
( dividing the equation by 2)
⟹ 𝑥1 + 2𝑥2 − 3𝑥3 = 0 -------(2)
−𝑥1 − 2𝑥2 + 3𝑥3 = 0
( multiplying the equation by −)
⟹ 𝑥1 + 2𝑥2 − 3𝑥3 = 0 -------(3)
Here all the three equation are same and two eigen values are equal
In this case
Step (1) to find the first eigen vector 𝑋1
Take 𝑥1 = 0
Put 𝑥1 = 0 in (1)
We will get 0 + 2𝑥2 − 3𝑥3 = 0
⟹ 2𝑥2 − 3𝑥3 = 0

⟹ 2𝑥2 = 3𝑥3
⟹
𝑥2
3
=
𝑥3
2
⟹ 𝑥2 = 3 & 𝑥3 = 2
Hence the first eigen vector is 𝑋1 =
𝑥1
𝑥2
𝑥3
=
0
3
2
Step (2) to find the second eigen vector 𝑿𝟐
Take 𝑥2 = 0
Put 𝑥2 = 0 in (1)
We will get 𝑥1 + 2 0 − 3𝑥3 = 0 ⟹ 𝑥1 − 3𝑥3 = 0
⟹ 𝑥1 = 3𝑥3 ⟹
𝑥1
3
= 𝑥3 ⟹
𝑥1
3
=
𝑥3
1
⟹ 𝑥1 = 3 & 𝑥3 = 1
Hence the second eigen vector is 𝑋2 =
𝑥1
𝑥2
𝑥3
=
3
0
1
To find the third Eigen vector
When 𝜆 = 5, (A) becomes
−2 − 5 2 −3
2 1 − 5 −6
−1 −2 0 − 5
𝑥1
𝑥2
𝑥3
=
0
0
0
−7 2 −3
2 −4 −6
𝑥1
𝑥2 =
0
0

−7𝑥1 +2𝑥2 −3𝑥3 = 0
2𝑥1 −4𝑥2 −6𝑥3 = 0
−𝑥1 −2𝑥2 −5𝑥3 = 0
⇒
−7𝑥1 +2𝑥2 −3𝑥3 = 0 −−− −(1)
𝑥1 −2𝑥2 −3𝑥3 = 0 −−− −(2)
𝑥1 +2𝑥2 +5𝑥3 = 0 −−− −(3)
( dividing the equation by 2)
(multiplying the equation by −)
These are three different equations, so take any two equations to solve
consider
𝑥1 − 2𝑥2 − 3𝑥3 = 0
𝑥1 + 2𝑥2 + 5𝑥3 = 0
By cross multiplication method
𝑥1 𝑥2 𝑥3
−2 −3 1
2 5 1
−2
2
𝑥1
−2 −3
2 5
=
𝑥2
−3 1
5 1
=
𝑥3
1 −2
1 2
𝑥1
−10+6
=
𝑥2
−3−5
=
𝑥3
2+2
𝑥1
−4
=
𝑥2
−8
=
𝑥3
4
Multiply by all by 4
4
𝑥1
−4
= 4
𝑥2
−8
= 4
𝑥3
4
𝑥1
−1
=
𝑥2
−2
=
𝑥3
1
Hence the third eigen vector 𝑋3 =
−1
−2
1

Find the Eigen vector of
𝟔 −𝟔 𝟓
𝟏𝟒 −𝟏𝟑 𝟏𝟎
𝟕 −𝟔 𝟒
Solution
𝑊ℎ𝑒𝑟𝑒 𝑆1 = 𝑠𝑢𝑚 𝑜𝑓 𝑡ℎ𝑒 𝑚𝑎𝑖𝑛 𝑑𝑖𝑎𝑔𝑜𝑛𝑎𝑙 = 6 − 13 + 4 = −3
𝑆1 = −3
𝑆2 = 𝑠𝑢𝑚 𝑜𝑓 𝑡ℎ𝑒 2𝑥2 𝑚𝑖𝑛𝑜𝑟𝑠 𝑜𝑓 𝑑𝑖𝑎𝑔𝑜𝑛𝑎𝑙 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑠
= 𝑚𝑖𝑛𝑜𝑟 𝑜𝑓 6 + 𝑚𝑖𝑛𝑜𝑟 𝑜𝑓 −13 + 𝑚𝑖𝑛𝑜𝑟 𝑜𝑓 (4)
=
−13 10
−6 4
+
6 5
7 4
+
6 −6
14 −13
= −52 + 60 + 24 − 35 + (−78 + 84) = 3
𝑆2 = 3
𝑆3 =
6 −6 5
14 −13 10
7 −6 4
= 6 −52 + 60 + 6 56 − 70 + 5(−84 + 91)
= 48 − 84 + 35 = 1
𝑆3 = −1
The characteristic equation is 𝜆3
+ 3𝜆2
+ 3𝜆 + 1 = 0
⟹ 𝜆 + 1 3
= 0
⟹ 𝜆 = −1, −1, −1
Here all the three eigen values are identical/equal

To find the Eigen vectors 𝑨 − 𝝀𝑰 (𝑿) = (𝟎)
(i.e)
6 − 𝜆 −6 5
14 −13 − 𝜆 10
7 −6 4 − 𝜆
𝑥1
𝑥2
𝑥3
=
0
0
0
----------- (A)
When 𝜆 = −1 , (A) becomes
6 + 1 −6 5
14 −13 + 1 10
7 −6 4 + 1
𝑥1
𝑥2
𝑥3
=
0
0
0
7 −6 5
14 −12 10
7 −6 5
𝑥1
𝑥2
𝑥3
=
0
0
0
7𝑥1 − 6𝑥2 + 5𝑥3 = 0 --------------- (1)
14𝑥1 − 12𝑥2 + 10𝑥3 = 0 ⟹ 7𝑥1 − 6𝑥2 + 5𝑥3 = 0 ( dividing the equation by 2) ------(2)
7𝑥1 − 6𝑥2 + 5𝑥3 = 0 ------- (3)
Here all three equations are same and also all the three eigen values are identical/ equal
In this case
Step (1) to find the first eigen vector
Take 𝑥1 = 0
Put 𝑥1 = 0 in (1)
7 0 − 6𝑥2 + 5𝑥3 = 0 ⟹ −6𝑥2 + 5𝑥3 = 0
⟹ −6𝑥2 = −5𝑥3
⟹ 6𝑥2 = 5𝑥3 ⟹
𝑥2
5
=
𝑥3
6
⟹ 𝑥2 = 5 & 𝑥3 = 6
Hence the first eigen vector 𝑋1 =
𝑥1
𝑥2
𝑥3
=
0
5
6

Step (2) to find the second eigen vector
Take 𝑥2 = 0
Put 𝑥2 = 0 in (1)
7𝑥1 − 6 0 + 5𝑥3 = 0 ⟹ 7𝑥1 + 5𝑥3 = 0
⟹ 7𝑥1 = −5𝑥3 ⟹
𝑥1
−5
=
𝑥3
7
⟹ 𝑥1 = −5 & 𝑥3 = 7
Hence the second eigen vector 𝑋2 =
𝑥1
𝑥2
𝑥3
=
−5
0
7
Step (3) to find the third eigen vector
Take 𝑥3 = 0
Put 𝑥3 = 0 in (1)
7𝑥1 − 6𝑥2 + 5 0 = 0 ⟹ 7𝑥1 − 6𝑥2 = 0
⟹ 7𝑥1 = 6𝑥2 ⟹
𝑥1
6
=
𝑥2
7
⟹ 𝑥1 = 6 & 𝑥2 = 7
𝑥1
𝑥2
𝑥3
=
6
7
0

Properties of eigen values and eigen vectors:
1. Sum of the eigen values of a matrix A = Sum of the elements in the main diagonal of A =
Trace of A
2. Product of eigen values of a matrix A = |𝐴|
3. Eigen values of a diagonal matrix are the elements in the main diagonal
4. Eigen values of a triangular matrix are the elements in the main diagonal.
5. If 𝜆 is the eigen value of the square matrix A, then the eigen value of 𝐴−1 is
1
𝜆
6. If 𝜆 is the eigen value of the square matrix A, then the eigen value of KA is K𝜆 , where K is any
scalar.
7. If 𝜆 is the eigen value of the square matrix A, then the eigen value of 𝐴𝑚
is 𝜆𝑚
8. If 𝜆 is the eigen value of the square matrix A, then the eigen value of A +KI is 𝜆 +K and the
eigen value of A − KI is 𝜆 − K
9. The eigen value of A and 𝐴𝑇
are equal.
10. If A and B are non-singular square matrices, then AB and BA have same eigen values.
11. The eigen values of symmetric matrix are real numbers.
12. The similar matrices have same eigen values.
13. If a real symmetric matrix of order 2 has equal eigen values then the matrix is a scalar matrix.
14. If 𝜆 is a eigen value of a orthogonal matrix 𝐴, then
1
𝜆
(𝜆 ≠ 0) is the eigen value of 𝐴.
15. If 𝐴&𝐵 are 𝑛𝑋𝑛 matrix and 𝐵 is non-singular, then 𝐴 & 𝐵−1𝐴𝐵 have same eigen values.

14. The eigen vectors corresponding to distinct eigen values of a real symmetric matrix are
orthogonal.
15. The eigen vector of a matrix is not unique.
16. If 𝜆1, 𝜆2, … 𝜆𝑛 are distinct eigen values of an 𝑛𝑋𝑛 matrix, then the corresponding eigen vectors
𝑋1, 𝑋2, … 𝑋𝑛 form a linearly independent set.
17. If two or more eigen values are equal , then it may (or) may not be possible to get linearly
independent eigen vectors corresponding to the equal eigen values.
18. Eigen vectors 𝑋1 & 𝑋2 are orthogonal , if 𝑋1𝑋2
𝑇
= 0 𝑜𝑟 𝑋1
𝑇
𝑋2 = 0.
19. If 𝑋 is a eigen vector of a matrix 𝐴, then any non-zero scalar multiple of 𝑋 is also a eigen
vector of 𝐴. 𝑖. 𝑒 if 𝑋 is a eigen vector of 𝐴 , then 𝑘𝑋 is also a eigen vector of 𝐴, where 𝑘 ≠ 0 a
scalar.
20. Any eigen vector of a matrix cannot corresponds to two different eigen values of that matrix.

Problems:
1.Find the eigen values of
𝟏 𝟎 𝟎
𝟎 −𝟐 𝟎
𝟎 𝟎 𝟓
Solution:
Since the given matrix is a diagonal matrix, the eigen values are the elements in the main diagonal
𝑖. 𝑒 𝜆1 = 1, 𝜆2 = −2 & 𝜆3 = 5
2. Find the eigen value of
𝟏 𝟑
𝟎 𝟐
Solution:
Since the given matrix is upper triangular matrix, the eigen values are the elements in the main
diagonal
𝑖. 𝑒 𝜆1 = 1, 𝜆2 = 2
𝟒 𝟎
𝟏 −𝟕
Solution:
Since the given matrix is lower triangular matrix, the eigen values are the elements in the main
diagonal
𝑖. 𝑒 𝜆1 = 4, 𝜆2 = −7

4. Find the sum & product of the eigen values of 𝐀 =
𝟏 −𝟏 𝟐
𝟎 𝟏 −𝟐
𝟑 𝟏 𝟑
Solution:
Sum of the eigen value = Sum of the diagonal elements of A = 1+1+3 =5
Product of the eigen value of A = 𝐴 = 1(3+2)+1(0+6)+2(0-3)=5+6-6 =5
𝟐 𝟑
𝟎 𝟒
corresponding to the eigen vector
𝟏
𝟎
Solution:
We know that 𝐴𝑋 = 𝜆𝑋, where A is the given matrix , 𝑋 is the eigen vector corresponding to the
eigen value 𝜆. ------- (1)
Here 𝐴 =
2 3
0 4
, X =
1
0
, to find the eigen value 𝜆
(1) ⇒
2 3
0 4
1
0
= 𝜆
1
0
⟹
2 + 0
0 + 0
=𝜆
1
0
⟹
2
0
=𝜆
1
0
⟹ 2
1
0
=𝜆
1
0
⇒ 𝜆 = 2

𝟏 𝟒
𝟑 𝟐
corresponding to the eigen vector
−𝟒
𝟑
Solution:
We know that 𝐴𝑋 = 𝜆𝑋, where A is the given matrix , 𝑋 is the eigen vector corresponding
to the eigen value 𝜆. ------- (1)
Here 𝐴 =
1 4
3 2
, X =
−4
3
, to find the eigen value 𝜆
(1) ⇒
1 4
3 2
−4
3
= 𝜆
−4
3
⟹
−4 + 12
−12 + 6
=𝜆
−4
3
⟹
8
−6
=𝜆
−4
3
⟹ 2
4
−3
=𝜆
−4
3
⇒ −2
−4
3
=𝜆
−4
3
⟹ 𝜆 = −2

7. If the eigen value of A are 1,-1,2, then find the eigen values of 𝑨−𝟏
Solution:
We know that if 𝜆 is the eigen value of A, then
1
𝜆
is the eigen value of 𝐴−1
Given 1,-1,2 are the eigen values of A
⟹
1
1
,
1
−1
,
1
2
are the eigen values of 𝐴−1
𝑖. 𝑒 1, −1,
1
2
8. If the eigen value of A are 1,2,3, then find the eigen values of 𝑨𝟑
Solution:
We know that if 𝜆 is the eigen value of A, then 𝜆𝑚
is the eigen value of 𝐴𝑚
Given 1,2,3 are the eigen values of A
Hence 13, 23, 33 are the eigen values of 𝐴3
⟹ 1,8,27 are the eigen values of 𝐴3
9. If the eigen value of A are -1, 4, 5, then find the eigen values of 𝟓𝑨
Solution:
We know that if 𝜆 is the eigen value of 𝐴, then 𝑘𝜆 is the eigen value of 𝑘𝐴
Given -1,4,5 are the eigen values of A
Hence 5 −1 , 5 4 , 5(5) are the eigen values of 5𝐴
⟹ −5,20,25 are the eigen values of 5𝐴

10. If the eigen value of A are 2, -2, 3, then find the eigen values of 𝑨 + 𝟐𝑰
Solution:
We know that if 𝜆 is the eigen value of 𝐴, then 𝜆 + 𝑘 is the eigen value of 𝐴 + 𝑘𝐼
Given 2,-2,3 are the eigen values of A, here k = 2
Hence 2 + 2, −2 + 2, 3 + 2 are the eigen values of A + 2𝐼
⟹ 4,0,5 are the eigen values of 𝐴 + 2𝐼
11. If the eigen value of A are 1, -2, 3, then find the eigen values of 𝑨 − 𝑰
Solution:
We know that if 𝜆 is the eigen value of 𝐴, then 𝜆 − 𝑘 is the eigen value of 𝐴 − 𝑘𝐼
Given 1,-2,3 are the eigen values of A, here k=1
Hence 1 − 1, −2 − 1, 3 − 1 are the eigen values of A − 𝐼
⟹ 0, −3,2 are the eigen values of 𝐴 − 𝐼
12. If the eigen value of A are 2, 3, 4, then find the eigen values of 𝒂𝒅𝒋𝑨.
Solution:
We know that 𝐴−1 =
𝑎𝑑𝑗 𝐴
𝐴
⟹ 𝑎𝑑𝑗 𝐴 = 𝐴 𝐴−1
We know that if 𝜆 is the eigen value of 𝐴, then
1
𝜆
⇒ the eigen values of 𝑎𝑑𝑗 𝐴 is 𝐴 .
1
𝜆
Given 2,3,4 are the eigen values of A,
Hence
1
2
,
1
3
,
1
4
We know that 𝐴 = product of eigen values = (2)(3)(4) = 24
Hence 24
1
2
, 24
1
3
, 24(
1
4
) are the eigen values of 𝑎𝑑𝑗 𝐴
(𝑖. 𝑒) 12,8,6 are the eigen values of 𝑎𝑑𝑗 𝐴

13. Two eigen values of 𝑨 =
𝟐 𝟐 𝟏
𝟏 𝟑 𝟏
𝟏 𝟐 𝟐
are equal to 1 each, find the eigen values of 𝑨−𝟏
Solution:
Given two eigen values are 1,1
Let the 3rd eigen value be 𝜆
We know that Sum of the eigen values = Sum of the elements in the main diagonal
(i.e) 1 + 1 + 𝜆 = 2 + 3 + 2
⇒ 2 + 𝜆 = 7
⇒ 𝜆 = 7 − 2 = 5
⇒ the three eigen values of A are 1,1,5
Hence the eigen values of 𝐴−1 are
1
1
,
1
1
,
1
5
= 1, 1,
1
5
14. The product of two eigen values of 𝑨 =
𝟔 −𝟐 𝟐
−𝟐 𝟑 −𝟏
𝟐 −𝟏 𝟑
is 16, find the 3rd eigen value.
Solution:
Given the product of two eigen values is 16
We know that product of all eigen values = 𝐴
(i.e) 16𝜆 = 6 9 − 1 + 2 −6 + 2 + 2 2 − 6 = 6 8 + 2 −4 + 2 −4 = 48 − 16
⇒ 16𝜆=32 ⇒ 𝜆 =
32
16
= 2
Hence the 3rd eigen value is 2.

15. If 𝑨 =
𝟏 𝟐 𝟑
𝟎 𝟒 𝟓
𝟎 𝟎 𝟔
, then find the eigen values of 𝑨−𝟏
Solution:
Given matrix 𝑨 =
𝟏 𝟐 𝟑
𝟎 𝟒 𝟓
𝟎 𝟎 𝟔
is an upper triangular matrix,
The eigen values are the elements in the main diagonal = 1, 4 & 6
We know that if 𝜆 is the eigen value of A then
1
𝜆
Hence the eigen values of 𝐴−1 are
1
1
,
1
4
&
1
6
= 1,
1
4
&
1
6
16. Find the constants ‘a’ and ‘b’ such that
𝒂 𝟒
𝟏 𝒃
has 3 and -2 as its eigen values.
Solution:
Given the eigen values of A =
𝑎 4
1 𝑏
are 3 & -2
We know that sum of the eigen values = sum of the elements in the main diagonal
⇒ 𝑎 + 𝑏 = 3 − 2
⇒ 𝑎 + 𝑏 = 1 ------ (1)
We know that product of eigen values = 𝐴
⇒
𝑎 4
1 𝑏
= 3 (−2)
⇒ 𝑎𝑏 − 4 = −6 ⇒ 𝑎𝑏 = −6 + 4
⇒ 𝑎𝑏 = −2 ------- (2)

Multiply (1) by ‘b’ , we will get
𝑏(𝑎 + 𝑏) = 1. (𝑏)
⇒ 𝑎𝑏 + 𝑏2 = 𝑏
Using (2), −2 + 𝑏2 − 𝑏 = 0
𝑖. 𝑒. 𝑏2 − 𝑏 − 2 = 0
⇒ 𝑏 − 2 𝑏 + 1 = 0
⇒ 𝑏 = 2 𝑜𝑟 𝑏 = −1
From (1) , If 𝑏 = 2 , 𝑎 + 2 = 1 ⇒ 𝑎 = 1 − 2 = −1
From (1) ,If 𝑏 = −1, 𝑎 − 1 = 1 ⇒ 𝑎 = 1 + 1 = 2
16. If 2, 3 are eigen values of
𝟐 𝟎 𝟏
𝟎 𝟐 𝟎
𝒃 𝟎 𝟐
, then find the value of b.
Solution:
Given 𝐴 =
2 0 1
0 2 0
𝑏 0 2
& two eigen values are 2 & 3
To find the 3rd eigen value:
⇒ 2 + 3 + 𝜆 = 2 + 2 + 2

⇒ 5 + 𝜆 = 6
⇒ 𝜆 = 6 − 5 = 1
To find the value of ‘b’:
We know that product of eigen values = 𝐴
(2)(3)(1) = 𝐴
⇒ 2 4 − 0 − 0 + 1 0 − 2𝑏 = 6
⟹ 8 − 2𝑏 = 6
⟹ −2𝑏 = 6 − 8 = −2
⟹ 𝑏 = 1
16. Find the sum, product of the eigen values of a 𝟑 𝑿 𝟑 matrix A whose
characteristic equation is 𝝀𝟑 − 𝟕𝝀𝟐 + 𝟑𝟔 = 𝟎
Solution:
Given characteristic equation is 𝜆3 − 7𝜆2 + 36 = 0 ----- (1)
We know that characteristic of a 3 𝑋 3 matrix A is given by 𝜆3 − 𝑆1𝜆2 + 𝑆2𝜆 + 𝑆3 = 0 ----- (2)
Where 𝑆1 = Sum of the eigen values = sum of the elements in the main diagonal.
𝑆2 = Sum of the values of the minors of the elements in the main diagonal.
𝑆3 = 𝐴 = product of the eigen values.
Comparing (1) & (2) , we will get
𝑆1 = 7 = sum of the eigen values
𝑆3 = 36 = product of the eigen values

17. Find the sum, product of the eigen values of
𝟐 𝟏 𝟐
𝟏 𝟑 𝟏
𝟐 𝟏 −𝟔
Solution:
Sum of the eigen values = sum of the elements in the main diagonal = 2 + 3 − 6 = −1
𝐴 = product of the eigen values = 2 −18 − 1 − 1 −6 − 2 + 2 1 − 6
= 2 −19 − 1 −8 + 2(−5)
= −38 + 8 − 10 = −40
18. Find the eigen values of 𝑨−𝟏
, if 𝐀 =
𝟐 𝟏 𝟎
𝟎 𝟑 𝟒
𝟎 𝟎 𝟒
Solution:
We know that the eigen values of an upper triangular matrix is the elements in the main diagonal.
Here eigen values of A are 2,3 & 4
We know that if 𝜆 is the eigen value of 𝐴 , then
1
𝜆
Hence
1
2
,
1
3
&
1
4
19. If 3, 6 are the eigen values of
𝟏 𝟏 𝟑
𝟏 𝟓 𝟏
𝟑 𝟏 𝟏
, write down all the eigen values of 𝑨−𝟏
Solution:
Given 3,6 are the eigen values of
1 1 3
1 5 1
3 1 1
We know that sum of the eigen values = sum of the elements in the main diagonal.
𝑖. 𝑒 𝜆 + 3 + 6 = 1 + 5 + 1
⇒ 𝜆 + 9 = 7
⇒ 𝜆 = 7 − 9 = −2

We know that if 𝜆 is the eigen value of 𝐴 , then
1
𝜆
Hence
1
3
,
1
6
&
1
−2
20. If 2, -1, -3 are the eigen values of 3X 3 matrix A, then find the eigen values of 𝐀𝟐
− 𝟐𝐈
Solution:
Given 2,-1,-3 are the eigen values of A
We know that if 𝜆 is the eigen value of a matrix A, then the eigen value of 𝐴2
is 𝜆2
Hence 22
, −1 2
, −3 2
are the eigen values of 𝐴2
𝑖. 𝑒. 4,1,9 are the eigen values of 𝐴2
We know that if 𝜆 is the eigen value of a matrix,𝐴2
, then 𝜆 − 2 is the eigen value of 𝐴2
− 2𝐼
Therefore (4-2), (1-2), (9-2) are the eigen values of 𝐴2
− 2𝐼
𝑖. 𝑒. 2, −1 , 7 are the eigen values of 𝐴2
− 2𝐼
21. If the eigen value of A are 1, 2, 3, then find the eigen values of 𝒂𝒅𝒋𝑨.
Solution:
We know that 𝐴−1
=
𝑎𝑑𝑗 𝐴
𝐴
⟹ 𝑎𝑑𝑗 𝐴 = 𝐴 𝐴−1
We know that if 𝜆 is the eigen value of 𝐴, then
1
𝜆
⇒ the eigen values of 𝑎𝑑𝑗 𝐴 is 𝐴 .
1
𝜆
Given 1,2,3 are the eigen values of A,
Hence
1
1
,
1
2
,
1
3
= 1,
1
2
,
1
3
We know that 𝐴 = product of eigen values = (1)(2)(3) = 6
Hence 6 1 , 6
1
2
, 6(
1
3
) are the eigen values of 𝑎𝑑𝑗 𝐴
(𝑖. 𝑒) 6, 3, 2 are the eigen values of 𝑎𝑑𝑗 𝐴

22. If two eigen values of 𝑨 =
𝟖 −𝟔 𝟐
−𝟔 𝟕 −𝟒
𝟐 −𝟒 𝟑
are 3 and 0, then what is the 3rd eigen value? What is
the product of the eigen values ?
Solution:
Given 3, 0 are the two eigen values of 𝐴 =
8 −6 2
−6 7 −4
2 −4 3
We know that Sum of the eigen values = Sum of the elements in the main diagonal
⇒ 3 + 0 + 𝜆 = 8 + 7 + 3
⇒ 𝜆 + 3 = 18
⇒ 𝜆 = 18 − 3 = 15
The 3rd eigen value is 15
Product of the eigen values = (3)(0)(15) =0
23. What is the eigen values of 𝑨 + 𝟑𝑰 , if the eigen values of A =
𝟏 −𝟐
−𝟓 𝟒
are 6 & -1.
Solution:
Given the eigen values of A =
1 −2
−5 4
are 6 & -1
We know that if 𝜆 is the eigen value of 𝐴, then 𝜆+3 is the eigen value of 𝐴 + 3𝐼
⇒ the eigen values of 𝐴 + 3𝐼 are (6+3) & (6-1)
⇒ the eigen values of 𝐴 + 3𝐼 are 9 & 5

24. Find the eigen values of 𝟑𝑨 + 𝟐𝑰, 𝐰𝐡𝐞𝐫𝐞 𝐀 =
𝟓 𝟒
𝟎 𝟐
Solution:
We know that the eigen values of an upper triangular matrix is the elements in the main diagonal.
Here eigen values of A are 5 & 2
We know that if 𝜆 is the eigen value of 𝐴, then k𝜆 + 2 is the eigen value of k𝐴 + 2𝐼, k any scalar.
Here k = 3,
Hence the eigen values of 3𝐴 + 2𝐼 are 3(5)+2 & 3(2)+2
𝑖. 𝑒. the eigen values of 3𝐴 + 2𝐼 are 17 & 8

Cayley- Hamilton Theorem
Statement of Cayley- Hamilton Theorem:
Every square matrix satisfies its own characteristic equation.
(i.e) If A is a square matrix of order 3, then its characteristic equation is given by
𝜆3
− 𝑆1𝜆2
+ 𝑆2𝜆 − 𝑆3 = 0 , ---------- (1)
where 𝑆1 = Sum of the elements in the main diagonal of A,
𝑆2 = Sum of the 2 X 2 minors of the diagonal elements of A
and 𝑆3 = |𝐴|
According to Cayley-Hamilton Theorem , A satisfies the equation (1)
(i.e) 𝐴3 − 𝑆1𝐴2 + 𝑆2𝐴 − 𝑆3 = 0 ----------- (2)
Note :
1.Using Cayley-Hamilton theorem we can find inverse and higher powers of
matrices.
2.Cayley- Hamilton theorem can be applied only for square matrix.

To find inverse of a matrix of order 3, by using Cayley-Hamilton Theorem:
Step1
Find the characteristic equation of the given matrix A
𝑖. 𝑒 𝜆3 − 𝑆1𝜆2 + 𝑆2𝜆 − 𝑆3 = 0
Step 2
Use Cayley – Hamilton Theorem and write the equation
𝐴3 − 𝑆1𝐴2 + 𝑆2𝐴 − 𝑆3 = 0
Step 3
Multiply the above equation by 𝐴−1
, then we will get
𝐴2 − 𝑆1𝐴 + 𝑆2 − 𝑆3𝐴−1 = 0
⇒ 𝐴−1 =
1
𝑆3
𝐴2 − 𝑆1𝐴 + 𝑆2
To find the power 𝑨𝟒 :
Step1
Find the characteristic equation of the given matrix A
𝑖. 𝑒 𝜆3 − 𝑆1𝜆2 + 𝑆2𝜆 − 𝑆3 = 0
Step 2
Use Cayley – Hamilton Theorem and write the equation
𝐴3
− 𝑆1𝐴2
+ 𝑆2𝐴 − 𝑆3 = 0
Step 3
Multiply the equation 𝐴3 − 𝑆1𝐴2 + 𝑆2𝐴 − 𝑆3 = 0 by 𝐴
We will get 𝐴4 − 𝑆1𝐴3 + 𝑆2𝐴2 − 𝑆3𝐴 = 0 ⇒ 𝐴4 = 𝑆1𝐴3 − 𝑆2𝐴2 + 𝑆3𝐴

Problems:
1.If 𝑨 =
𝟏 𝟎
𝟏 𝟕
, then find the value of k for which 𝑨𝟐
= 𝟖𝑨 + 𝒌𝑰
Solution:
Given 𝐴 =
1 0
1 7
We know that the Characteristic equation is 𝜆2 − 𝑆1𝜆 + 𝑆2 = 0---(1)
Where 𝑆1 = Sum of the elements in the main diagonal = 1+7 =8
𝑆2 = 𝐴 = 7
Hence (1) ⇒ 𝜆2
− 8𝜆 + 7 = 0 ----- (2)
By Cayley – Hamilton theorem, every square matrix satisfies its characteristic equation.
∴ 𝐴 satisfies equation (2), ⇒ 𝐴2
− 8𝐴 + 7𝐼 = 0
⇒ 𝐴2
= 8𝐴 − 7𝐼
⇒ 𝑘 = 7
2. A square matrix A has its characteristic equation as 𝝀𝟐
− 𝟐𝝀 + 𝟏 = 𝟎 , then find its inverse.
Solution:
Given 𝜆2 − 2𝜆 + 1 = 0 is the characteristic equation of A.
By Cayley- Hamilton theorem, we will get 𝐴2 − 2𝐴 + 𝐼 = 0 ----- (1)
To find inverse of A, multiply (1) by 𝐴−1
𝐴−1
𝐴2
− 2𝐴 + 𝐼 = 0 ⇒ 𝐴−1
𝐴2
− 2𝐴−1
𝐴 + 𝐴−1
𝐼 = 0
⇒ 𝐴 − 2𝐼 + 𝐴−1 = 0
⇒ 𝐴−1 = 2𝐼 − 𝐴

3.If 𝑨 =
𝟏 𝟐
𝟐 −𝟏
, then find 𝑨𝟖
Solution:
Given 𝐴 =
1 2
2 −1
We know that the Characteristic equation is 𝜆2
− 𝑆1𝜆 + 𝑆2 = 0---(1)
Where 𝑆1 = Sum of the elements in the main diagonal = 1-1 = 0
𝑆2 = 𝐴 = −1 − 4 = −5
Hence (1) ⇒ 𝜆2
− 0 𝜆 − 5 = 0 ⇒ 𝜆2
− 5 = 0 ----- (2)
− 5𝐼 = 0
⇒ 𝐴2
= 5𝐼 ------- (3)
Now 𝐴8
= 𝐴2 4
= 5𝐼 4
= 54
𝐼4
= 625𝐼
4.If 𝑨 =
−𝟏 𝟏
𝟏 𝟏
, then find 𝑨𝟏𝟎
Solution:
Given 𝐴 =
−1 1
1 1
We know that the Characteristic equation is 𝜆2
− 𝑆1𝜆 + 𝑆2 = 0---(1)
Where 𝑆1 = Sum of the elements in the main diagonal = 1-1 = 0
𝑆2 = 𝐴 = −1 − 1 = −2
Hence (1) ⇒ 𝜆2
− 0 𝜆 − 2 = 0 ⇒ 𝜆2
− 2 = 0 ----- (2)
− 2𝐼 = 0
⇒ 𝐴2
= 2𝐼 ------- (3)
Now 𝐴10
= 𝐴2 5
= 2𝐼 5
= 25
𝐼5
= 32𝐼

Orthogonal matrix
A square matrix with real numbers is said to be an orthogonal matrix, if its transpose is equal
to its inverse matrix.
(i.e.) If the product of a square matrix and its transpose gives an identity matrix, then the
square matrix is known as an orthogonal matrix.
A square matrix A is said to be orthogonal , if 𝐴𝐴𝑇
= 𝐴𝑇
𝐴 = 𝐼
(i.e.) 𝐴𝑇
= 𝐴−1
Properties of an Orthogonal matrix:
1. The orthogonal matrix has all real elements in it.
2. All identity matrices are orthogonal matrices.
3. The product of two orthogonal matrices is also an orthogonal matrix.
4. The collection of the orthogonal matrix of order n x n, in a group, is called an orthogonal
group and is denoted by ‘O’.
5. The transpose of the orthogonal matrix is also orthogonal. Thus, if matrix A is orthogonal,
then is 𝐴𝑇 is also an orthogonal matrix.
6. In the same way, the inverse of the orthogonal matrix, which is 𝐴−1 is also an orthogonal
matrix.
7. The determinant of the orthogonal matrix A has a value of ±1. (i.e.) 𝐴 = ±1
8.It is symmetric in nature
9.If the matrix is orthogonal, then its transpose and inverse are equal
10.The eigenvalues of the orthogonal matrix also have a value of ±1, and its eigenvectors
would also be orthogonal and real.

Examples:
1. Check whether 𝑨 =
𝟎 𝟏
𝟏 𝟎
is orthogonal ?
Solution:
Given 𝐴 =
0 1
1 0
⇒ 𝐴𝑇
=
0 1
1 0
Now 𝐴𝐴𝑇
=
0 1
1 0
0 1
1 0
=
0 + 1 0 + 0
0 + 0 1 + 0
=
1 0
0 1
= 𝐼
Hence 𝐴 is orthogonal.
2. Determine if the matrix 𝑨 =
−𝟏 𝟎
𝟎 𝟏
is orthogonal or not.
Solution:
Given 𝐴 =
−1 0
0 1
⇒ 𝐴𝑇 =
−1 0
0 1
Now 𝐴𝐴𝑇 =
−1 0
0 1
−1 0
0 1
=
(−1)(−1) + 0 0 + 0
0 + 0 0 + 1
=
1 0
0 1
= 𝐼

3.Determine if the matrix 𝑨 =
𝟏
𝟑
𝟏 −𝟐 𝟐
𝟐 −𝟏 −𝟐
𝟐 𝟐 𝟏
is orthogonal or not
Solution:
Given 𝐴 =
1
3
1 −2 2
2 −1 −2
2 2 1
Now 𝐴 =
1
3
1
3
1
3
1 −1 + 4 + 2 2 + 4 + 2 4 + 2 =
1
27
3 + 12 + 12 =
27
27
= 1
4.Determine if the matrix 𝑨 =
𝟑
𝟐
−
𝟏
𝟐
𝟏
𝟐
𝟑
𝟐
is orthogonal or not (Or)
Determine if the matrix 𝑨 =
𝒔𝒊𝒏𝜽 −𝒄𝒐𝒔𝜽
𝒄𝒐𝒔𝜽 𝒔𝒊𝒏𝜽
is orthogonal or not
Solution:
Given 𝐴 =
3
2
−
1
2
1
2
3
2
Now 𝐴 =
3
2
−
1
2
1
2
3
2
=
3
2
3
2
− −
1
2
1
2
=
3
4
+
1
4
=
4
4
= 1

(Or)
Given 𝐴 =
𝑠𝑖𝑛𝜃 −𝑐𝑜𝑠𝜃
𝑐𝑜𝑠𝜃 𝑠𝑖𝑛𝜃
Now 𝐴 =
𝑠𝑖𝑛𝜃 −𝑐𝑜𝑠𝜃
𝑐𝑜𝑠𝜃 𝑠𝑖𝑛𝜃
= sin2 𝜃 + cos2 𝜃 = 1
Mutually Orthogonal / Pairwise Orthogonal
Two matrices 𝑋1 & 𝑋2 are said to be Mutually (Or) Pairwise Orthogonal ,
If 𝑋1 𝑋2
𝑇
= 𝑋2 𝑋1
𝑇
= 0
Example:
𝑿𝟏 =
𝟎
𝟎
−𝟏
, 𝑿𝟐 =
𝟏
𝟐
𝟏
𝟏
𝟎
, 𝑿𝟑 =
𝟏
𝟐
𝟏
−𝟏
𝟎
are mutually orthogonal.
Solution:
Given 𝑋1 =
0
0
−1
, 𝑋2 =
1
2
1
1
0
, 𝑋3 =
1
2
1
−1
0
Now 𝑋1𝑋2
𝑇
=
0
0
−1
1
2
1 1 0 =
1
2
0 0 0
0 0 0
0 0 0
=
0 0 0
0 0 0
0 0 0
Similarly 𝑋1𝑋3
𝑇
= 𝑋2𝑋3
𝑇
= 𝑋2𝑋1
𝑇
= 𝑋3𝑋1
𝑇
= 𝑋3𝑋2
𝑇
= 0
Hence 𝑋1, 𝑋2, 𝑋3 are pairwise / mutually orthogonal.

Find the characteristic vector of
𝟐 𝟎 𝟏
𝟎 𝟐 𝟎
𝟏 𝟎 𝟐
& verify that they are mutually(pairwise)
orthogonal.
Solution
𝑊ℎ𝑒𝑟𝑒 𝑆1 = 𝑠𝑢𝑚 𝑜𝑓 𝑡ℎ𝑒 𝑚𝑎𝑖𝑛 𝑑𝑖𝑎𝑔𝑜𝑛𝑎𝑙 = 2 + 2 + 2 = 6
𝑆1 = 6
𝑆2 = 𝑠𝑢𝑚 𝑜𝑓 𝑡ℎ𝑒 2𝑥2 𝑚𝑖𝑛𝑜𝑟𝑠 𝑜𝑓 𝑑𝑖𝑎𝑔𝑜𝑛𝑎𝑙 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑠
= 𝑚𝑖𝑛𝑜𝑟 𝑜𝑓 2 + 𝑚𝑖𝑛𝑜𝑟 𝑜𝑓 2 + 𝑚𝑖𝑛𝑜𝑟 𝑜𝑓 (2)
=
2 0
0 2
+
2 1
1 2
+
2 0
0 2
= 4 + (4 − 1) + 4 = 11
𝑆2 = 11
𝑆3 =
2 0 1
0 2 0
1 0 2
= 2 4 − 0 − 0 0 − 0 + 1(0 − 2)
= 8 − 2 = 6
𝑆3 = 6
The characteristic equation is 𝜆3 − 6𝜆2 + 11𝜆 − 6 = 0

1 1 -6 11 -6
0 1 -5 6
1 -5 6 0
Hence 𝜆1 = 1 is a root
Remaining roots are given by 𝜆2 − 5𝜆 + 6 = 0
𝜆 − 2 𝜆 − 3 = 0
⟹ 𝜆2 = 2 & 𝜆3 = 3
Hence the eigen values are 𝜆 = 1,2,3
To find the Eigen vectors
𝐴 − 𝜆𝐼 (𝑋) = (0)
(i.e)
2 − 𝜆 0 1
0 2 − 𝜆 0
1 0 2 − 𝜆
𝑥1
𝑥2
𝑥3
=
0
0
0
----------- (A)
To find the first characteristic/eigen vector 𝑿𝟏
When 𝜆 = 1 , (A) becomes
2 − 1 0 1
0 2 − 1 0
1 0 2 − 1
𝑥1
𝑥2
𝑥3
=
0
0
0
1 0 1
0 1 0
1 0 1
𝑥1
𝑥2
𝑥3
=
0
0
0

𝑥1 + 0𝑥2 + 𝑥3 = 0 ------ (i)
0𝑥1 + 𝑥2 + 0𝑥3 = 0 -----(ii)
𝑥1 + 0𝑥2 + 𝑥3 = 0 ------ (iii)
Here two equations (i) & (iii) are identical
So consider the two non-identical equations (i) & (ii)
𝑥1 + 0𝑥2 + 𝑥3 = 0
0𝑥1 + 𝑥2 + 0𝑥3 = 0
Using cross multiplication method
𝑥1 𝑥2 𝑥3
0 1 1
1 0 0
0
1
𝑥1
0 1
1 0
=
𝑥2
1 1
0 0
=
𝑥3
1 0
0 1
𝑥1
−1
=
𝑥2
0
=
𝑥3
1
Hence the first eigen vector/ characteristic vector 𝑋1 =
𝑥1
𝑥2
𝑥3
=
−1
0
1
To find the second characteristic/eigen vector 𝑿𝟐

2 − 2 0 1
0 2 − 2 0
1 0 2 − 2
𝑥1
𝑥2
𝑥3
=
0
0
0
0 0 1
0 0 0
1 0 0
𝑥1
𝑥2
𝑥3
=
0
0
0
0𝑥1 + 0𝑥2 + 𝑥3 = 0 --- (i)
0𝑥1 + 0𝑥2 + 0𝑥3 = 0 ---- (ii)
𝑥1 + 0𝑥2 + 0𝑥3 = 0 ---- (iii)
Consider (i) & (iii)
𝑥1 𝑥2 𝑥3
0 1 0
0 0 1
0
0
𝑥1
0 1
0 0
=
𝑥2
1 0
0 1
=
𝑥3
0 0
1 0
𝑥1
0
=
𝑥2
1
=
𝑥3
0
Hence the second eigen vector/ characteristic vector 𝑋2 =
𝑥1
𝑥2
𝑥3
=
0
1
0
To find the third characteristic/eigen vector 𝑿𝟑

2 − 3 0 1
0 2 − 3 0
1 0 2 − 3
𝑥1
𝑥2
𝑥3
=
0
0
0
−1 0 1
0 −1 0
1 0 −1
𝑥1
𝑥2
𝑥3
=
0
0
0
−𝑥1 + 0𝑥2 + 𝑥3 = 0 ------ (i)
0𝑥1 − 𝑥2 + 0𝑥3 = 0 -------(ii)
𝑥1 + 0𝑥2 − 𝑥3 = 0 ------- (iii)
Here (i) & (iii) are identical
So consider the non-identical equations (i) & (ii)
−𝑥1 + 0𝑥2 + 𝑥3 = 0
0𝑥1 − 𝑥2 + 0𝑥3 = 0
𝑥1 𝑥2 𝑥3
0 1 −1
−1 0 0
0
−1
𝑥1
0 1
−1 0
=
𝑥2
1 −1
0 0
=
𝑥3
−1 0
0 −1
𝑥1
=
𝑥2
=
𝑥3

Hence the third eigen vector/ characteristic vector 𝑋3 =
𝑥1
𝑥2
𝑥3
=
1
0
1
To check the mutually / pairwise orthogonal, the condition is 𝑿𝟏𝑿𝟐
𝑻
𝒐𝒓 𝑿𝟐
𝑻
𝑿𝟏 =
𝑿𝟐𝑿𝟑
𝑻
𝒐𝒓 𝑿𝟑
𝑻
𝑿𝟐 = 𝑿𝟑𝑿𝟏
𝑻
𝒐𝒓 𝑿𝟏
𝑻
𝑿𝟑 = 𝟎
Clearly 𝑋2
𝑇
𝑋1 =
0
1
0
𝑇
−1
0
1
= 0 1 0
−1
0
1
= −1𝑋0 + 0𝑋1 + 1𝑋0 = [0]
Similarly 𝑋3
𝑇
𝑋2 = 𝑋1
𝑇
𝑋3 = [0]
Hence 𝑋1 , 𝑋2 & 𝑋3 are mutually / pairwise orthogonal.

Orthogonal Reduction of symmetric matrix or Diagonalisation of
symmetric matrix by orthogonal transformation.
A square matrix matrix A is called orthogonally diagonalisable if there is an
orthogonal matrix U and a diagonal matrix D for which 𝐴 = 𝑈𝐷𝑈−1 = 𝑈𝐷𝑈𝑇.
If A is a 3 x 3 real symmetric matrix. Then there is an orthogonal matrix P such that
PTAP is a diagonal matrix.
Suppose A is invertible and orthogonal diagonalisable. Show that A−1 is also
orthogonal diagonalisable.
Properties
1. An orthogonal matrix is orthogonally diagonalisable.
2. An orthogonal matrix is invertible.
3. An invertible matrix is orthogonal.
4. If a matrix is diagonalizable then it is symmetric.
5. A (real) 𝑛 𝑋 𝑛 matrix A is orthogonally diagonalizable if and only if A is
symmetric.
6. A matrix A is said to be orthogonal if its columns are mutually orthogonal.

Normalized eigen vector
Normalized eigenvector is nothing but an eigenvector having unit length. It can be found by
simply dividing each component of the vector by the length of the vector. By doing so, the vector
is converted into the vector of length one.
(i.e) If 𝑋 =
𝑎1
𝑏1
𝑐1
is an eigen vector of a matrix A,
then the normalized eigen vector is given by
𝑎1
𝐿
𝑏1
𝐿
𝑐1
𝐿
where 𝐿 = 𝑎1
2
+ 𝑏1
2
+ 𝑐1
2
is the length of the eigen vector X
Example: If 𝑿 =
𝟐
−𝟓
𝟏
, then find the normalized eigen vector
Solution:
Length of the eigen vector = 22 + −5 2 + 12 = 4 + 25 + 1 = 30
Hence the normalized eigen vector =
2
30
−
5
30
1
30

Model Matrix:
Producer to find Model Matrix.
If A is a square matrix of order 3
Step (1)
Find the characteristic equation of A
Step (2)
Find the eigen values of A say 𝜆1 , 𝜆2 𝑎𝑛𝑑 𝜆3
Step (3)
Find the eigen vectors corresponding to 𝜆1 , 𝜆2 𝑎𝑛𝑑 𝜆3,
let it be 𝑋1 =
𝑎1
𝑏1
𝑐1
, 𝑋2 =
𝑎2
𝑏2
𝑐2
𝑎𝑛𝑑 𝑋3 =
𝑎3
𝑏3
𝑐3
and each 𝑋𝑖′𝑠 are pairwise orthogonal
(i.e) pairwise orthogonal means 𝑋1𝑋2
𝑇
= 𝑋2𝑋3
𝑇
= 𝑋3𝑋1
𝑇
= 0 (or) vice-versa
Step (4)
Model matrix is given by P =
𝑎1 𝑎2 𝑎3
𝑏1 𝑏2 𝑏3
𝑐1 𝑐2 𝑐3

Normalized Model Matrix (Or) Normalized eigen vector matrix
Producer to find Normalized Model Matrix.
If A is a square matrix of order 3
Step (1)
Find the Model matrix is given by P =
𝑎1 𝑎2 𝑎3
𝑏1 𝑏2 𝑏3
𝑐1 𝑐2 𝑐3
Step (2)
Find the length of each column vector
Let it be 𝐿1 = 𝑎1
2
+ 𝑏1
2
+ 𝑐1
2
𝐿2 = 𝑎2
2
+ 𝑏2
2
+ 𝑐2
2
𝐿3 = 𝑎3
2
+ 𝑏3
2
+ 𝑐3
2
Step (3)
The normalized model matrix is given by N =
𝑎1
𝐿1
𝑎2
𝐿2
𝑎3
𝐿3
𝑏1
𝐿1
𝑏2
𝐿2
𝑏3
𝐿3
𝑐1
𝐿1
𝑐2
𝐿2
𝑐3
𝐿3
Since N is orthogonal , 𝑁𝑇 = 𝑁−1

Diagonalisation of a matrix by Orthogonal reduction (Or) Orthogonal transformation.
Producer for Diagonalisation of a matrix by Orthogonal reduction
Step (1)
Find the Normalized model matrix N
Step (2)
The diagonal matrix is given by 𝐷 = 𝑁−1 𝐴 𝑁 = 𝑁𝑇𝐴𝑁
which is called orthogonal reduction or orthogonal transformation.
Quadratic Form
A homogeneous polynomial of the second degree with one or more variables is called quadratic
form.
Example: 𝑥1
2
+ 2𝑥1𝑥2 − 3𝑥2
2
,
4𝑥1
2
+ 2𝑥2
2
− 5𝑥3
2
+ 8𝑥1𝑥2 + 3𝑥2𝑥3
The general quadratic form is given by
𝑎11𝑥1
2
+ 𝑎12𝑥1𝑥2 + ⋯ + 𝑎1𝑛𝑥1𝑥𝑛 + 𝑎21𝑥2𝑥1 + 𝑎22𝑥2
2
+ ⋯ + 𝑎2𝑛𝑥2𝑥𝑛 + ⋯ + 𝑎𝑛1𝑥𝑛𝑥1+. . +𝑎𝑛𝑛𝑥𝑛
2

Quadratic form a Matrix
Let A denote an 𝑛 𝑋 𝑛 symmetric matrix with real numbers and let 𝑋 denote an 𝑛 𝑋 1 column
vector.
Then 𝑄 = 𝑋𝑇𝐴 𝑋 is said to be a quadratic form.
Example :
1.Find the quadratic form of 𝑨 =
𝟏 𝟐
𝟐 𝟏
Solution:
Let 𝑋 =
𝑥1
𝑥2
The Quadratic form 𝑄 = 𝑋𝑇𝐴 𝑋 =
𝑥1
𝑥2
T
1 2
2 1
𝑥1
𝑥2
= 𝑥1 𝑥2
1 2
2 1
𝑥1
𝑥2
= 𝑥1 + 2𝑥2 2𝑥1 + 𝑥2
𝑥1
𝑥2
= 𝑥1 𝑥1 + 2𝑥2 + 𝑥2 2𝑥1 + 𝑥2 = 𝑥1
2
+ 2𝑥1𝑥2 + 2𝑥1𝑥2 + 𝑥2
2
= 𝑥1
2
+ 4𝑥1𝑥2 + 𝑥2
2

2.Find the quadratic form of 𝑨 =
𝟏 𝟎 𝟎
𝟎 𝟐 𝟎
𝟎 𝟎 𝟒
Solution:
Given𝐴 =
1 0 0
0 2 0
0 0 4
Let 𝑋 =
𝑥1
𝑥2
𝑥3
The Quadratic form 𝑄 = 𝑋𝑇𝐴 𝑋 =
𝑥1
𝑥2
𝑥3
T
1 0 0
0 2 0
0 0 4
𝑥1
𝑥2
𝑥3
= 𝑥1 𝑥2 𝑥3
1 0 0
0 2 0
0 0 4
𝑥1
𝑥2
𝑥3
= 𝑥1 2𝑥2 4𝑥3
𝑥1
𝑥2
𝑥3
= 𝑥1
2
+ 2𝑥2
2
+ 4𝑥3
2

Producer to find the Matrix of Quadratic form 𝒂𝟏𝟏𝒙𝟏
𝟐
+ 𝒂𝟏𝟐𝒙𝟏𝒙𝟐 + ⋯ + 𝒂𝟑𝟑𝒙𝟑
𝟐
Let the matrix be (Q.F), Q =
𝑎11 𝑎12 𝑎13
𝑎21 𝑎22 𝑎23
𝑎31 𝑎32 𝑎33
Step (1)
To find main diagonal elements (𝑎11, 𝑎22 & 𝑎33)
𝑎11 = Coefficient of 𝑥1
2
(OR) Coefficient of 𝑥2
2
(OR) Coefficient of 𝑦2
2
(OR)Coefficient of 𝑧2
Step(2)
To find the elements in 𝑎12 & 𝑎12 positions
𝑎12 = 𝑎21 =
1
2
(Coefficient of 𝑥12 + Coefficient of 𝑥21)
(OR)
𝑎12 = 𝑎21 =
1
2
(Coefficient of 𝑥𝑦)

Step (3)
𝑎13 = 𝑎31 =
1
2
(OR)
𝑎13 = 𝑎31 =
1
2
(Coefficient of 𝑥𝑧)
Step (4)
𝑎23 = 𝑎32 =
1
2
(OR)
𝑎23 = 𝑎32 =
1
2
(Coefficient of 𝑦𝑧)
Note :
If you don’t have particular terms then take the coefficients of the terms as zero
Example : consider 𝑥1
2
+ 𝑥3
2
− 2𝑥2𝑥3
Here the terms 𝑥2
2
, 𝑥1𝑥2, 𝑥2𝑥1, 𝑥3𝑥2, 𝑥1𝑥3, 𝑥3𝑥1 are not available
Consider the coefficients of these terms as zero

Example:
1.Find the matrix corresponding quadratic form 2x1
2+5x2
2+4x1x2+2x3x1
Solution:
Given quadratic form is 2𝑥1
2
+ 5𝑥2
2
+ 4𝑥1𝑥2 + 2𝑥3𝑥1
Since we have three variables 𝑥1, 𝑥2& 𝑥3, let the matrix be Q =
𝑎11 𝑎12 𝑎13
𝑎21 𝑎22 𝑎23
𝑎31 𝑎32 𝑎33
Here 𝑎11 = coefficient of 𝑥1
2
= 2
𝑎22 = coefficient of 𝑥2
2
= 5
2
= 0
𝑎12 = 𝑎21 =
1
2
(Coefficient of 𝑥12 + Coefficient of 𝑥21) =
1
2
4 + 0 = 2
𝑎13 = 𝑎31 =
1
2
1
2
0 + 2 = 1
𝑎32 = 𝑎23 =
1
2
1
2
0 + 0 = 0
Hence Q =
𝑎11 𝑎12 𝑎13
𝑎21 𝑎22 𝑎23
𝑎31 𝑎32 𝑎33
=
2 2 1
2 5 0
1 0 0

2.Find the matrix corresponding quadratic form 2𝒙𝟐 + 𝟖𝒛𝟐 + 𝟒𝒙𝒚 + 𝟏𝟎𝒙𝒛 − 𝟐𝒚𝒛
Solution:
+ 8𝑧2
+ 4𝑥𝑦 + 10𝑥𝑧 − 2𝑦𝑧
Since we have three variables 𝑥, 𝑦 & 𝑧, let the matrix be Q =
𝑎11 𝑎12 𝑎13
𝑎21 𝑎22 𝑎23
𝑎31 𝑎32 𝑎33
Here 𝑎11 = coefficient of 𝑥2= 2
𝑎22 = coefficient of 𝑦2 = 0
𝑎33 = coefficient of 𝑧2
= 8
𝑎12 = 𝑎21 =
1
2
(Coefficient of 𝑥𝑦 + Coefficient of 𝑦𝑥) =
1
2
4 + 0 = 2
𝑎13 = 𝑎31 =
1
2
(Coefficient of 𝑥𝑧 + Coefficient of 𝑧𝑥) =
1
2
10 + 0 = 5
𝑎32 = 𝑎23 =
1
2
(Coefficient of 𝑦𝑧 + Coefficient of 𝑧𝑦) =
1
2
−2 + 0 = −1
Hence Q =
𝑎11 𝑎12 𝑎13
𝑎21 𝑎22 𝑎23
𝑎31 𝑎32 𝑎33
=
2 2 5
2 0 −1
5 −1 8

Nature, Index and signature of Quadratic form:
Index of the quadratic form
The number of positive terms in the canonical is called index of the quadratic form
(or) The index of the quadratic form is equal to the number of positive Eigen values of the matrix
of quadratic form.
Signature of the quadratic form
The difference between number of positive terms and the negative terms in the canonical form is
called signature of the quadratic form.
(or) The signature of the quadratic form is equal to the difference between the number of positive
Eigen values and the number of negative Eigen values of the matrix of quadratic form.
Nature of Quadratic form
Positive definite – If all eigen values of Q are positive
Negative definite – If all eigen value of Q are negative
Positive semi-definite – If all eigen values of Q are positive and at least one eigen value is zero
Negative semi-definite – If all eigen values of Q are negative and at least one eigen value is zero
Indefinite – If the eigen values are positive as well as negative
Rank of a matrix A
The number of terms in the canonical form is the rank of A
(or) The rank of the quadratic form is equal to the number of non zero Eigen values of the matrix
of quadratic form.

Another method to find the nature of quadratic form:
If the general form of the quadratic form 𝑎𝑥1
2
+ 𝑏𝑥2
2
+ 𝑐𝑥3
2
+ 2𝑑𝑥1𝑥2 + 2𝑒𝑥2𝑥3 + 2𝑓𝑥3𝑥1
The matrix of the quadratic form is
𝑎 𝑑 𝑓
𝑑 𝑏 𝑒
𝑓 𝑒 𝑐
Let 𝐷1 = 𝑎, 𝐷2 =
𝑎 𝑑
𝑑 𝑏
, 𝐷3 =
𝑎 𝑑 𝑓
𝑑 𝑏 𝑒
𝑓 𝑒 𝑐
Nature of Quadratic form
Positive definite – If 𝐷1, 𝐷2, 𝐷3 > 0
Negative definite – If 𝐷1, 𝐷2, 𝐷3 < 0
Positive semi-definite – If 𝐷1, 𝐷2, 𝐷3 ≥ 0
Negative semi-definite – If 𝐷1, 𝐷2, 𝐷3 ≤ 0
Indefinite – If 𝐷1, 𝐷2, 𝐷3 ≤ 0 and 𝐷1, 𝐷2, 𝐷3 ≤ 0

Problems:
1.Find the rank, index, signature and nature of the quadratic form
𝟏𝟎𝒙𝟏
𝟐
+ 𝟐𝒙𝟐
𝟐
+ 𝟓𝒙𝟑
𝟐
+ 𝟔𝒙𝟐𝒙𝟑 − 𝟏𝟎𝒙𝟑𝒙𝟏 − 𝟒𝒙𝟏𝒙𝟐
Solution:
2
+ 2𝑥2
2
+ 5𝑥3
2
+ 6𝑥2𝑥3 − 10𝑥3𝑥1 − 4𝑥1𝑥2
Let the matrix of the Q.F be 𝑄 =
𝑎11 𝑎12 𝑎13
𝑎21 𝑎22 𝑎23
𝑎31 𝑎32 𝑎33
2
= 10
2
= 2
2
= 5
𝑎12 = 𝑎21=
1
2
(coefficient of 𝑥1𝑥2+coefficient of 𝑥2𝑥1) =
1
2
−4 + 0 = −2
𝑎13 = 𝑎31=
1
2
1
2
0 − 10 = −5
𝑎23 = 𝑎32=
1
2
1
2
6 + 0 = 3
Therefore the matrix of the Q.F be 𝑄 =
10 −2 −5
−2 2 3
−5 3 5

The characteristic equation of Q is 𝜆3 − 𝑆1𝜆2 + 𝑆2𝜆 − 𝑆3 = 0
Where 𝑆1 = sum of the diagonal elements = 10+2+5=17
𝑆2 = sum of the minors of the diagonal elements =
2 3
3 5
+
10 −5
−5 5
+
10 −2
−2 2
= 10 − 9 + 50 − 25 + 20 − 4 = 42
𝑆3 = 𝑄 = 10 10 − 9 + 2 −10 + 15 − 5 −6 + 10 = 10 + 10 − 20 = 0
∴ The characteristic equation of Q is 𝜆3 − 17𝜆2 + 42𝜆 − 0 = 0
⇒ 𝜆(𝜆2 − 17𝜆+42) = 0
⇒ 𝜆 = 0 (or) 𝜆2 − 17𝜆+42 = 0
⇒ 𝜆 = 0, 𝜆 − 14 𝜆 − 3 = 0
⇒ The eigen values are 𝜆1 = 0, 𝜆2 = 14, 𝜆3 = 3
Index = number of positive eigen values = 2
Signature = difference between number of positive and negative eigen values = 2-0 =2
Rank = number of eigen values = 2
Nature = here two eigen values are positive and one is zero , hence nature is semi-positive
definite.

𝟐𝒙𝟏
𝟐
+ 𝒙𝟐
𝟐
+ 𝒙𝟑
𝟐
− 𝟒𝒙𝟐𝒙𝟑 − 𝟐𝒙𝟑𝒙𝟏 + 𝟐𝒙𝟏𝒙𝟐
Solution:
2
+ 𝑥2
2
+ 𝑥3
2
− 4𝑥2𝑥3 − 2𝑥3𝑥1 + 2𝑥1𝑥2
𝑎11 𝑎12 𝑎13
𝑎21 𝑎22 𝑎23
𝑎31 𝑎32 𝑎33
2
= 2
2
= 1
2
= 1
𝑎12 = 𝑎21=
1
2
1
2
2 + 0 = 1
𝑎13 = 𝑎31=
1
2
1
2
0 − 2 = −1
𝑎23 = 𝑎32=
1
2
1
2
−4 + 0 = −2
2 1 −1
1 1 −2
−1 −2 1

The characteristic equation of Q is 𝜆3
− 𝑆1𝜆2
+ 𝑆2𝜆 − 𝑆3 = 0
Where 𝑆1 = sum of the diagonal elements = 2+1+1=4
1 −2
−2 1
+
2 −1
−1 1
+
2 1
1 1
= 1 − 4 + 2 − 1 + 2 − 1 = −1
𝑆3 = 𝑄 = 2 1 − 4 − 1 1 − 2 − 1 −2 + 1 = −6 + 1 + 1 = −4
∴ The characteristic equation of Q is 𝜆3 − 4𝜆2 − 𝜆 + 4 = 0
1 1 -4 -1 4
0 1 -3 -4
1 -3 -4 0
𝜆 = 1 is a root
The equation reduced to 𝜆2 − 3𝜆 − 4 = 0
⇒ (𝜆 + 1) 𝜆 − 4 = 0
⇒ The eigen values are 𝜆1 = 1, 𝜆2 = −1, 𝜆3 = 4
Signature = difference between number of positive and negative eigen values = 2- 1 = 1
Nature = here two eigen values are positive and one is negative , hence nature is indefinite.

𝟔𝒙𝟏
𝟐
+ 𝟑𝒙𝟐
𝟐
+ 𝟑𝒙𝟑
𝟐
− 𝟐𝒙𝟐𝒙𝟑 + 𝟒𝒙𝟑𝒙𝟏 − 𝟒𝒙𝟏𝒙𝟐
Solution:
Given quadratic form is6𝑥1
2
+ 3𝑥2
2
+ 3𝑥3
2
− 2𝑥2𝑥3 + 4𝑥3𝑥1 − 4𝑥1𝑥2
𝑎11 𝑎12 𝑎13
𝑎21 𝑎22 𝑎23
𝑎31 𝑎32 𝑎33
2
= 6
2
= 3
2
= 3
𝑎12 = 𝑎21=
1
2
1
2
−4 + 0 = −2
𝑎13 = 𝑎31=
1
2
1
2
0 + 4 = 2
𝑎23 = 𝑎32=
1
2
1
2
−2 + 0 = −1
6 −2 2
−2 3 −1
2 −1 3

The characteristic equation of Q is 𝜆3
− 𝑆1𝜆2
+ 𝑆2𝜆 − 𝑆3 = 0
Where 𝑆1 = sum of the diagonal elements = 6 + 3 + 3 = 12
3 −1
−1 3
+
6 2
2 3
+
6 −2
−2 3
= 9 − 1 + 18 − 4 + 18 − 4 = 36
𝑆3 = 𝑄 = 6 9 − 1 + 2 −6 + 2 + 2 2 − 6 = 48 − 8 − 8 = 32
∴ The characteristic equation of Q is 𝜆3 − 12𝜆2 + 36𝜆 − 32 = 0
2 1 -12 36 -32
0 2 -20 32
1 -10 16 0
𝜆 = 2 is a root
The equation reduced to 𝜆2
− 10𝜆 + 16 = 0
⇒ (𝜆 − 8) 𝜆 − 2 = 0
⇒ The eigen values are 𝜆1 = 2, 𝜆2 = 2, 𝜆3 = 8
Signature = difference between number of positive and negative eigen values = 3 - 0 = 3
Nature = here all eigen values are positive, hence nature is positive definite.

4.Find the rank, index, signature and nature of the quadratic form 𝒙𝟏
𝟐
+ 𝟐𝒙𝟐
𝟐
− 𝟑𝒙𝟑
𝟐
Solution:
Given quadratic form is 𝑥1
2
+ 2𝑥2
2
− 3𝑥3
2
𝑎11 𝑎12 𝑎13
𝑎21 𝑎22 𝑎23
𝑎31 𝑎32 𝑎33
2
= 1
2
= 2
2
= - 3
𝑎12 = 𝑎21=
1
2
1
2
0 + 0 = 0
𝑎13 = 𝑎31=
1
2
1
2
0 + 0 = 0
𝑎23 = 𝑎32=
1
2
1
2
0 + 0 = 0
1 0 0
0 2 0
0 0 −3
Clearly Q is a diagonal matrix , hence the eigen values are 𝜆1 = 1, 𝜆2 = 2, 𝜆3 = −3
Nature = here two eigen values are positive ad one is negative, hence nature is indefinite.

5.Find the rank, index, signature and nature of the quadratic form 2𝒙𝟏𝒙𝟐 + 𝟐𝒙𝟐𝒙𝟑 + 𝟐𝒙𝟑𝒙𝟏
Solution:
Given quadratic form is 2𝑥1𝑥2 + 2𝑥2𝑥3 + 2𝑥3𝑥1
𝑎11 𝑎12 𝑎13
𝑎21 𝑎22 𝑎23
𝑎31 𝑎32 𝑎33
2
= 0
2
= 0
2
= 0
𝑎12 = 𝑎21=
1
2
1
2
2 + 0 = 1
𝑎13 = 𝑎31=
1
2
1
2
2 + 0 = 1
𝑎23 = 𝑎32=
1
2
1
2
2 + 0 = 1
0 1 1
1 0 1
1 1 0

The characteristic equation of Q is 𝜆3 − 𝑆1𝜆2 + 𝑆2𝜆 − 𝑆3 = 0
Where 𝑆1 = sum of the diagonal elements = 0
0 1
1 0
+
0 1
1 0
+
0 1
1 0
= 0 − 1 + 0 − 1 + 0 − 1 = −3
𝑆3 = 𝑄 = 0 − 1 0 − 1 + 1 1 − 0 = 1 + 1 = 2
∴ The characteristic equation of Q is 𝜆3
− 3𝜆 − 2 = 0
-1 1 0 -3 -2
0 -1 1 2
1 -1 -2 0
𝜆 = −1 is a root
The equation reduced to 𝜆2 − 𝜆 − 2 = 0
⇒ (𝜆 − 2) 𝜆 + 1 = 0
⇒ The eigen values are 𝜆1 = −1, 𝜆2 = −1, 𝜆3 = 2
Nature = here two eigen values are negative and one is positive, hence nature is indefinite.

Canonical Form of a Matrix:
Every matrix is similar to a matrix with elements on the diagonal, and zeros and ones on the
diagonal just above the main diagonal, with zeros everywhere else. Such a representative for a set
of similar matrices is called a canonical form of a matrix.
Example:
1 0 0
0 2 0
0 0 5
,
−2 1 0
0 −1 0
0 0 6
,
1 1 0
0 2 1
0 0 −1

Conversion of Quadratic form into Canonical from by Orthogonal reduction:
Step 1. Find Q ( Quadratic form of symmetric matrix).
Step 2. Find the Characteristic equation of Q by using the following formula
𝜆3
− 𝑆1𝜆2
+ 𝑆2𝜆 − 𝑆3 = 0
Where S1 = sum of the main diagonal
S2 = sum of the 2x2 minors of diagonal elements & S3 = Q
Step 3. Solve the characteristic equation & find the three values of 𝜆, which are called Eigen values.
Step 4. Find the three Eigen vectors (𝑋1 =
𝑎1
𝑏1
𝑐1
, 𝑋2 =
𝑎2
𝑏2
𝑐2
, 𝑋3 =
𝑎3
𝑏3
𝑐3
) corresponding to three Eigen
values.
Step 5. Write the Model matrix P = 𝑋1 𝑋2 𝑋3 =
𝑎1 𝑎2 𝑎3
𝑏1 𝑏2 𝑏3
𝑐1 𝑐2 𝑐3
Step 6. Write the Normalized matrix N =
𝑎1
𝐿1
𝑎2
𝐿2
𝑎3
𝐿3
𝑏1
𝐿1
𝑏2
𝐿2
𝑏3
𝐿3
𝑐1
𝐿1
𝑐2
𝐿2
𝑐3
𝐿3
Where 𝐿1 = 𝑎1
2
+ 𝑏1
2
+ 𝑐1
2
, 𝐿2 = 𝑎2
2
+ 𝑏2
2
+ 𝑐2
2
& 𝐿3 = 𝑎3
2
+ 𝑏3
2
+ 𝑐3
2
Step 7. Write the diagonalized matrix D = 𝑵𝑻𝑸𝑵
Step 8. The orthogonal transformation X = NY reduces the quadratic from to canonical form
𝑪. 𝑭 = 𝒀𝑻𝑫𝒀 Where 𝒀 =
𝒚𝟏
𝒚𝟐
𝒚𝟑

Problems:
1. Reduce the quadratic form 𝒙𝟐
+ 𝒚𝟐
+ 𝒛𝟐
− 𝟐𝒙𝒚 − 𝟐𝒚𝒛 − 𝟐𝒛𝒙 to canonical form through
an orthogonal transformation
Solution:
+ 𝑦2
+ 𝑧2
− 2𝑥𝑦 − 2𝑦𝑧 − 2𝑧𝑥
𝑎11 𝑎12 𝑎13
𝑎21 𝑎22 𝑎23
𝑎31 𝑎32 𝑎33
Here 𝑎11 = coefficient of 𝑥2 = 1
𝑎22 = coefficient of 𝑦2
= 1
𝑎33 = coefficient of 𝑧2
= 1
𝑎12 = 𝑎21=
1
2
(coefficient of 𝑥𝑦+coefficient of 𝑦𝑥) =
1
2
−2 + 0 = −1
𝑎13 = 𝑎31=
1
2
(coefficient of 𝑥𝑧+coefficient of 𝑧𝑥) =
1
2
−2 + 0 = −1
𝑎23 = 𝑎32=
1
2
(coefficient of 𝑦𝑧+coefficient of 𝑧𝑦) =
1
2
−2 + 0 = −10
Therefore Quadratic form of the matrix is given by(Q.F)
Q =
𝟏 −𝟏 −𝟏
−𝟏 𝟏 −𝟏
−𝟏 −𝟏 𝟏
----------------------- (I)

𝑊ℎ𝑒𝑟𝑒 𝑆1 = sum of the main diagonal = 1 + 1 + 1 = 3
𝑆1 = 3
𝑆2 = sum of the 2x2 minors of diagonal elements
= minor of 1 + minor of 1 + minor of (1)
=
1 −1
−1 1
+
1 −1
−1 1
+
1 −1
−1 1
= (1 − 1) + (1 − 1) + (1 − 1) = 0
𝑆2 = 0
𝑆3 =
1 −1 −1
−1 1 −1
−1 −1 1
= 1 1 − 1 + 1 −1 − 1 − 1 1 + 1 = −2 − 2 = −4
𝑆3 = −4
The characteristic equation is 𝜆3 − 3𝜆2 + 0𝜆 + 4 = 0
-1 1 -3 0 4
0 -1 4 -4
1 -4 4 0
Hence 𝜆1 = −1 is a root
Remaining roots are given by 𝜆2 − 4𝜆 + 4 = 0
⇒ 𝜆 − 2 𝜆 − 2 = 0
⇒ 𝜆2 = 2 & 𝜆3 = 2

Hence the eigen values are 𝝀 = −𝟏, 𝟐, 𝟐 ( two roots are identical)
To find the Eigen vectors 𝐴 − 𝜆𝐼 (𝑋) = (0)
(i.e)
1 − 𝜆 −1 −1
−1 1 − 𝜆 −1
−1 −1 1 − 𝜆
𝑥1
𝑥2
𝑥3
=
0
0
0
----------- (A)
To find the first eigen vector 𝑿𝟏
When 𝜆 = −1, (A) becomes
1 + 1 −1 −1
−1 1 + 1 −1
−1 −1 1 + 1
𝑥1
𝑥2
𝑥3
=
0
0
0
2 −1 −1
−1 2 −1
−1 −1 2
𝑥1
𝑥2
𝑥3
=
0
0
0
2𝑥1 − 𝑥2 − 𝑥3 = 0 ------ (i)
−𝑥1 + 2𝑥2 − 𝑥3 = 0 ------- (ii)
−𝑥1 − 𝑥2 + 2𝑥3 = 0 -------- (iii)
All the three equations (i), (ii) & (iii) are non-identical
Hence choose any two equation to solve for 𝒙𝟏, 𝒙𝟐& 𝒙𝟑

Consider (i) & (ii)
2𝑥1 − 𝑥2 − 𝑥3 = 0 ------ (i)
−𝑥1 + 2𝑥2 − 𝑥3 = 0 ------ (ii)
𝑥1 𝑥2 𝑥3
−1 −1 2
2 −1 −1
−1
2
𝑥1
−1 −1
2 −1
=
𝑥2
−1 2
−1 −1
=
𝑥3
2 −1
−1 2
𝑥1
1+2
=
𝑥2
1+2
=
𝑥3
4−1
𝑥1
3
=
𝑥2
3
=
𝑥3
3
( multiply all by 3 )
𝑥1
1
=
𝑥2
1
=
𝑥3
1
𝑥1
𝑥2
𝑥3
=
1
1
1
To find the second eigen vector 𝑿𝟐
1 − 2 −1 −1
−1 1 − 2 −1
−1 −1 1 − 2
𝑥1
𝑥2
𝑥3
=
0
0
0
−1 −1 −1
−1 −1 −1
−1 −1 −1
𝑥1
𝑥2
𝑥3
=
0
0
0

−𝑥1 − 𝑥2 − 𝑥3 = 0 ⟹ 𝑥1 + 𝑥2 + 𝑥3 = 0 ------ (i)
−𝑥1 − 𝑥2 − 𝑥3 = 0 ⟹ 𝑥1 + 𝑥2 + 𝑥3 = 0 ------- (ii)
−𝑥1 − 𝑥2 − 𝑥3 = 0 ⟹ 𝑥1 + 𝑥2 + 𝑥3 = 0 -------- (iii)
Here all the three equations (i) , (ii) & (iii) are identical
In this case Take 𝒙𝟏 = 𝟎
Put 𝑥1 = 0 in (1) , we will get 0 + 𝑥2 + 𝑥3 = 0 ⟹ 𝑥2 + 𝑥3 = 0
⟹ 𝑥2 = −𝑥3 ⟹
𝑥2
1
=
𝑥3
−1
∴ 𝑥1 = 0 , 𝑥2 = 1, 𝑥3 = −1
𝑥1
𝑥2
𝑥3
=
0
1
−1
Since two eigen values are identical and equal to 2 ( 𝒊𝒆 𝝀 = 𝟐, 𝟐)
Here to find the third eigen vector 𝑿𝟑 , we have to use the pairwise orthogonal condition.
Take the third eigen vector 𝑋3 =
𝑥1
𝑥2
𝑥3
Pairwise orthogonal condition
𝑿𝟑𝑿𝟏
𝑻
= (𝟎) & 𝑿𝟑𝑿𝟐
𝑻
= (𝟎)
𝑿𝟑𝑿𝟏
𝑻
= (𝟎) ⟹
𝑥1
𝑥2
𝑥3
1
1
1
𝑇
= 0 ⟹
𝑥1
𝑥2
𝑥3
1 1 1 = (0)

⟹ 1. 𝑥1 + 1. 𝑥2 + 1. 𝑥3 = 0
⟹ 𝑥1 + 𝑥2 + 𝑥3 = 0 --------- (iv)
𝑿𝟑𝑿𝟐
𝑻
= 𝟎 ⟹
𝑥1
𝑥2
𝑥3
0
1
−1
𝑇
= 0 ⟹
𝑥1
𝑥2
𝑥3
0 1 −1 = (0)
⟹ 0. 𝑥1 + 1. 𝑥2 − 1. 𝑥3 = 0
⟹ 𝑥2 − 𝑥3 = 0 --------- (v)
Consider the equations (iv) & (v)
𝑥1 + 𝑥2 + 𝑥3 = 0 -------- (iv)
0𝑥1 + 𝑥2 − 𝑥3 = 0 --------- (v)
𝑥1 𝑥2 𝑥3
1 1 1
1 −1 0
1
1
𝑥1
1 1
1 −1
=
𝑥2
1 1
−1 0
=
𝑥3
1 1
0 1
𝑥1
−1−1
=
𝑥2
0+1
=
𝑥3
1−0
𝑥1
−2
=
𝑥2
1
=
𝑥3
1
𝑥1
𝑥2
𝑥3
=
−2
1
1
The Model matrix P =
1 0 −2
1 1 1
1 −1 1

Note :
Model matrix is generated by the eigen vectors 𝑿𝟏, 𝑿𝟐 & 𝑿𝟑 , (ie) P = 𝑿𝟏 𝑿𝟐 𝑿𝟑
Normalized Model Matrix N =
𝟏
𝟏𝟐 +𝟏𝟐+𝟏𝟐
𝟎
𝟎𝟐+𝟏𝟐+ −𝟏 𝟐
−
𝟐
−𝟐 𝟐+𝟏𝟐+𝟏𝟐
𝟏
𝟏𝟐 +𝟏𝟐+𝟏𝟐
𝟏
𝟎𝟐+𝟏𝟐+ −𝟏 𝟐
𝟏
−𝟐 𝟐+𝟏𝟐+𝟏𝟐
𝟏
𝟏𝟐 +𝟏𝟐+𝟏𝟐
−
𝟏
𝟎𝟐+𝟏𝟐+ −𝟏 𝟐
𝟏
−𝟐 𝟐+𝟏𝟐+𝟏𝟐
𝑁 =
1
3
0 −
2
6
1
3
1
2
1
6
1
3
−1
2
1
6
----------- (II)
𝑁𝑇 =
1
3
1
3
1
3
0
1
2
−1
2
−
2
6
1
6
1
6
------------- (III)
The diagonalized matrix D = 𝑵𝑻 𝑸 𝑵
=
1
3
1
3
1
3
0
1
2
−1
2
−
2
6
1
6
1
6
1 −1 −1
−1 1 −1
−1 −1 1
1
3
0 −
2
6
1
3
1
2
1
6
1
3
−1
2
1
6
, { by using (I) , (II) & (III) }

∴ 𝑫 =
−1 0 0
0 2 0
0 0 2
Hence the orthogonal transformation X = NY reduces the quadratic form into canonical
form
(C.F) = 𝒀𝑻 𝑫 𝒀 where 𝒀 =
𝒚𝟏
𝒚𝟐
𝒚𝟑
𝑪. 𝑭 =
𝒚𝟏
𝒚𝟐
𝒚𝟑
𝑻
−1 0 0
0 2 0
0 0 2
𝒚𝟏
𝒚𝟐
𝒚𝟑
= 𝒚𝟏 𝒚𝟐 𝒚𝟑
−1 0 0
0 2 0
0 0 2
𝒚𝟏
𝒚𝟐
𝒚𝟑
= −𝒚𝟏 𝟐𝒚𝟐 𝟐𝒚𝟑
𝒚𝟏
𝒚𝟐
𝒚𝟑
C.F = −𝒚𝟏
𝟐
+ 𝟐𝒚𝟐
𝟐
+ 𝟐𝒚𝟑
𝟐
Here C.F has both positive and negative terms , hence its nature is indefinite.

2.Reduce the quadratic form 𝒙𝟏
𝟐
+ 𝟐𝒙𝟐
𝟐
+ 𝒙𝟑
𝟐
− 𝟐𝒙𝟏𝒙𝟐 + 𝟐𝒙𝟐𝒙𝟑 to canonical form by using
orthogonal transformation
Solution
2
+ 2𝑥2
2
+ 𝑥3
2
− 2𝑥1𝑥2 + 2𝑥2𝑥3
𝑎11 𝑎12 𝑎13
𝑎21 𝑎22 𝑎23
𝑎31 𝑎32 𝑎33
2
= 1
2
= 2
2
= 1
𝑎12 = 𝑎21=
1
2
1
2
−2 + 0 = −1
𝑎13 = 𝑎31=
1
2
1
2
0 + 0 = 0
𝑎23 = 𝑎32=
1
2
1
2
2 + 0 = 1
Therefore Quadratic form of the matrix is given by(Q.F)
Q =
𝟏 −𝟏 𝟎
−𝟏 𝟐 𝟏
𝟎 𝟏 𝟏
----------------------- (I)

W.K.T Characteristic equation is 𝜆3
− 𝑆1𝜆2
+ 𝑆2𝜆 − 𝑆3 = 0
Where 𝑆1 = sum of the main diagonal = 1 + 2 + 1 = 4
𝑆1 = 4
𝑆2 = sum of the 2x2 minors of diagonal elements
= minor of 1 + minor of 2 + minor of (1)
=
2 1
1 1
+
1 0
0 1
+
1 −1
−1 2
= (2 − 1) + (1 − 0) + (2 − 1) = 3
𝑆2 = 3
𝑆3 =
1 −1 0
−1 2 1
0 1 1
= 1 2 − 1 + 1 −1 − 0 + 0(−1 − 0)
= 1 − 1 = 0
𝑆3 = 0
The characteristic equation is
𝜆3
− 4𝜆2
+ 3𝜆 + 0 = 0
𝜆3 − 4𝜆2 + 3𝜆 = 0 ⟹ 𝜆 𝜆2 − 4𝜆 + 3 = 0
⟹ 𝜆1 = 0 is a root
Remaining roots are given by
𝜆2 − 4𝜆 + 3 = 0
𝜆 − 3 𝜆 − 1 = 0
⟹ 𝜆2 = 3 & 𝜆3 = 1

Hence the eigen values are 𝝀 = 𝟎, 𝟏, 𝟑
To find the Eigen vectors
𝐴 − 𝜆𝐼 (𝑋) = (0)
(i.e)
1 − 𝜆 −1 0
−1 2 − 𝜆 1
0 1 1 − 𝜆
𝑥1
𝑥2
𝑥3
=
0
0
0
----------- (A)
To find the first eigen vector 𝑿𝟏
1 − 0 −1 0
−1 2 − 0 1
0 1 1 − 0
𝑥1
𝑥2
𝑥3
=
0
0
0
1 −1 0
−1 2 1
0 1 1
𝑥1
𝑥2
𝑥3
=
0
0
0
𝑥1 − 𝑥2 + 0𝑥3 = 0 ------ (i)
−𝑥1 + 2𝑥2 + 𝑥3 = 0 ------- (ii)
0𝑥1 + 𝑥2 + 𝑥3 = 0 -------- (iii)

Consider (i) & (ii)
𝑥1 − 𝑥2 + 0𝑥3 = 0 ------ (i)
−𝑥1 + 2𝑥2 + 𝑥3 = 0 ------ (ii)
𝑥1 𝑥2 𝑥3
−1 0 1
2 1 −1
−1
2
𝑥1
−1 0
2 1
=
𝑥2
0 1
1 −1
=
𝑥3
1 −1
−1 2
𝑥1
−1−0
=
𝑥2
0−1
=
𝑥3
2−1
𝑥1
−1
=
𝑥2
−1
=
𝑥3
1
𝑥1
1
=
𝑥2
1
=
𝑥3
−1
(multiply each by − )
𝑥1
𝑥2
𝑥3
=
1
1
−1
To find the second eigen vector 𝑿𝟐
1 − 1 −1 0
−1 2 − 1 1
0 1 1 − 1
𝑥1
𝑥2
𝑥3
=
0
0
0
0 −1 0
−1 1 1
0 1 0
𝑥1
𝑥2
𝑥3
=
0
0
0

0𝑥1 − 𝑥2 + 0𝑥3 = 0 ------ (i)
−𝑥1 + 𝑥2 + 𝑥3 = 0 ------- (ii)
0𝑥1 + 𝑥2 + 0𝑥3 = 0 ------- (iii)
Here two equations (i) & (iii) are identical
So consider the non-identical equations (i) & (ii)
0𝑥1 − 𝑥2 + 0𝑥3 = 0 ------ (i)
−𝑥1 + 𝑥2 + 𝑥3 = 0 ------- (ii)
𝑥1 𝑥2 𝑥3
−1 0 0
1 1 −1
−1
1
𝑥1
−1 0
1 1
=
𝑥2
0 0
1 −1
=
𝑥3
0 −1
−1 1
𝑥1
−1−0
=
𝑥2
0−0
=
𝑥3
0−1
𝑥1
−1
=
𝑥2
0
=
𝑥3
−1
𝑥1
1
=
𝑥2
0
=
𝑥3
−1
(multiply each by − )
𝑥1
𝑥2
𝑥3
=
1
0
1

To find the third eigen vector 𝑿𝟑
1 − 3 −1 0
−1 2 − 3 1
0 1 1 − 3
𝑥1
𝑥2
𝑥3
=
0
0
0
−2 −1 0
−1 −1 1
0 1 −2
𝑥1
𝑥2
𝑥3
=
0
0
0
−2𝑥1 − 𝑥2 + 0𝑥3 = 0 ------ (i)
−𝑥1 − 𝑥2 + 𝑥3 = 0 ------- (ii)
0𝑥1 + 𝑥2 − 2𝑥3 = 0 ------- (iii)
Consider (i) & (ii)
−2𝑥1 − 𝑥2 + 0𝑥3 = 0 ------ (i)
−𝑥1 − 𝑥2 + 𝑥3 = 0 ------- (ii)
𝑥1 𝑥2 𝑥3
−1 0 −2
−1 1 −1
−1
−1

𝑥1
−1 0
−1 1
=
𝑥2
0 −2
1 −1
=
𝑥3
−2 −1
−1 −1
𝑥1
−1−0
=
𝑥2
0+2
=
𝑥3
2−1
𝑥1
−1
=
𝑥2
2
=
𝑥3
1
𝑥1
𝑥2
𝑥3
=
−1
2
1
The Model matrix 𝑃 =
1 1 −1
1 0 2
−1 1 1
The normalized matrix 𝑁 =
1
3
1
2
−1
6
1
3
0
2
6
−1
3
1
2
−1
6
----------- (II)
⟹ 𝑵𝑻 =
1
3
1
3
−1
3
1
2
0
1
2
−1
6
2
6
1
6
-------------- (III)
The diagonalized matrix 𝑫 = 𝑵𝑻
𝑸𝑵

=
1
3
1
3
−1
3
1
2
0
1
2
−1
6
2
6
1
6
1 −1 0
−1 2 1
0 1 1
1
3
1
2
−1
6
1
3
0
2
6
−1
3
1
2
−1
6
(by using (I),(II) & (III) )
𝐷 =
0 0 0
0 1 0
0 0 3
Hence the orthogonal transformation X = NY reduces the quadratic form into canonical
form
(C.F) = 𝒀𝑻 𝑫 𝒀 where 𝒀 =
𝒚𝟏
𝒚𝟐
𝒚𝟑
𝑪. 𝑭 =
𝒚𝟏
𝒚𝟐
𝒚𝟑
𝑻
0 0 0
0 1 0
0 0 3
𝒚𝟏
𝒚𝟐
𝒚𝟑
= 𝒚𝟏 𝒚𝟐 𝒚𝟑
0 0 0
0 1 0
0 0 3
𝒚𝟏
𝒚𝟐
𝒚𝟑
= 𝟎 𝒚𝟐 𝟑𝒚𝟑
𝒚𝟏
𝒚𝟐
𝒚𝟑
C.F = 𝒚𝟐
𝟐
+ 𝟑𝒚𝟑
𝟐
Hear both terms are positive and one is zero, Hence the nature of C.F is positive semi-definite

3. Can 𝑨 =
𝟏 𝟎
𝟎 𝟏
be diagonalized? Why?
Solution:
Given 𝐴 =
1 0
0 1
which is a unit matrix & also a diagonalized matrix.
We know that
1.Diagonalization of a unit matrix is again unit matrix.
2.Diagonalization of a diagonalized matrix is possible, since 𝐴 = 𝐼−1𝐴𝐼
4. When is a quadratic form said to be singular? What is the rank then?
Solution:
The quadratic form is said to be singular, if the matrix of the quadratic form is singular
The rank of this matrix is less than or equal to its order.
(i.e) If the order of the matrix is 3, then its rank is ≤ 3
5. Write down the quadratic form corresponding to the symmetric matrix 𝑨 =
𝟐 𝟎 −𝟐
𝟎 𝟐 𝟏
−𝟐 𝟏 −𝟐
Solution:
Given 𝐴 =
2 0 −2
0 2 1
−2 1 −2

If the matrix of the Q.F be 𝑄 =
𝑎11 𝑎12 𝑎13
𝑎21 𝑎22 𝑎23
𝑎31 𝑎32 𝑎33
then the quadratic form will be
𝑎11𝑥1
2
+ 𝑎22𝑥2
2
+ 𝑎33𝑥3
2
+ 2𝑎12𝑥1𝑥2 + 2𝑎13𝑥1𝑥3 + 2𝑎23𝑥2𝑥3 = 0
where 𝑎11 = coefficient of 𝑥1
2
2
2
𝑎12 = coefficient of 𝑥1𝑥2
𝑎21= coefficient of 𝑥2𝑥1
Here 𝑎11 = 2, 𝑎22 = 2, 𝑎33 = −2 , 𝑎12 = 𝑎21 = 0, 𝑎13 = 𝑎31 = −2, 𝑎23 = 𝑎32 = 1
The quadratic form is 2𝑥1
2
+ 2𝑥2
2
− 2𝑥3
2
+ 2 0 𝑥1𝑥2 + 2 −2 𝑥1𝑥3 + 2 1 𝑥2𝑥3
=2𝑥1
2
+ 2𝑥2
2
− 2𝑥3
2
− 4𝑥1𝑥3 + 2𝑥2𝑥3

6. Write down the quadratic form of the matrix is 𝑨 =
𝟎 𝟓 −𝟏
𝟓 𝟏 𝟔
−𝟏 𝟔 𝟐
Solution:
If the matrix of the Q.F be 𝑄 =
𝑎11 𝑎12 𝑎13
𝑎21 𝑎22 𝑎23
𝑎31 𝑎32 𝑎33
then the quadratic form will be
𝑎11𝑥1
2
+ 𝑎22𝑥2
2
+ 𝑎33𝑥3
2
+ 2𝑎12𝑥1𝑥2 + 2𝑎13𝑥1𝑥3 + 2𝑎23𝑥2𝑥3 = 0
where 𝑎11 = coefficient of 𝑥1
2
2
2
Here 𝑎11 = 0, 𝑎22 = 1, 𝑎33 = 2 , 𝑎12 = 𝑎21 = 5, 𝑎13 = 𝑎31 = −1, 𝑎23 = 𝑎32 = 6
The quadratic form is 0𝑥1
2
+ 1𝑥2
2
+ 2𝑥3
2
+ 2 5 𝑥1𝑥2 + 2 −1 𝑥1𝑥3 + 2 6 𝑥2𝑥3
=𝑥2
2
+ 2𝑥3
2
+ 10𝑥1 𝑥2 − 2𝑥1𝑥3 + 12𝑥2𝑥3

7. Write down the nature of a quadratic form whose matrix is
−𝟏 𝟎 𝟎
𝟎 −𝟏 𝟎
𝟎 𝟎 −𝟐
Solution:
Given matrix
−1 0 0
0 −1 0
0 0 −2
is a diagonal matrix
The eigen values are the elements in the main diagonal
(i.e) −1, −1, −2 which are all negative.
Hence the nature of the quadratic form is negative definite.
8. If the Eigen values of matrix A of order 3X3 are 2, 3 &1, then find the determinant of A
Solution:
Given eigen values of A are 2,3 & 1
We know that det(𝐴) = 𝐴 = product of eigen values = (1)(2)(3) =6
9. For what values of ‘𝒄’ the eigen values of the Matrix
𝟏 𝟐
𝒄 𝟒
are real and unequal,
real and equal , complex conjugates?
Solution:
Given
1 2
𝑐 4
The characteristic equation is given by 𝜆2 − 𝑆1𝜆 + 𝑆2 = 0

Where 𝑆1 = Sum of the elements in the main diagonal = 1 + 4 =5
𝑆2 =
1 2
𝑐 4
= 4 − 2𝑐
Hence the characteristic equation becomes 𝜆2
− 5𝜆 + (4 − 2𝑐) = 0 ----- (1)
We know that the eigen values are the roots of the characteristic equation
Case (1): The roots of (1) are real and unequal, if 𝒃𝟐
− 𝟒𝒂𝒄 > 𝟎
𝑖. 𝑒. −5 2 − 4 1 4 − 2𝑐 > 0
⇒ 25 − 16 + 8𝑐 > 0
⇒ 9 + 8𝑐 > 0
⇒ 8𝑐 ≻ −9
⇒ 𝑐 ≻ −
9
8
Case (2): The roots of (1) are real and equal, if 𝒃𝟐 − 𝟒𝒂𝒄 = 𝟎
𝑖. 𝑒. −5 2 − 4 1 4 − 2𝑐 = 0
⇒ 25 − 16 + 8𝑐 = 0
⇒ 𝑐 = −
9
8
Case (3): The roots of (1) are complex conjugate , if 𝒃𝟐 − 𝟒𝒂𝒄 < 𝟎
𝑖. 𝑒. −5 2 − 4 1 4 − 2𝑐 < 0
⇒ 𝑐 < −
9
8

10. If −𝟏 is an Eigen value of the Matrix A =
𝟏 −𝟐
−𝟑 𝟐
, then find the Eigen values of 𝑨𝟒
.
Solution:
Given 𝐴 =
1 −2
−3 2
& one of its eigen value is −1
Let the other eigen value be 𝜆
∴ −1 + 𝜆 = 1 + 2
⇒ 𝜆 = 3+1 = 4
The eigen values of A are −1 & 4
We know that if 𝜆 is the eigen value of 𝐴, then 𝜆4
is the eigen value of 𝐴4
Hence the eigen values of 𝐴4
are −1 4
& 4 4
𝑖. 𝑒. the eigen values of 𝐴4 are 1 & 256

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