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WHAT IS IT?
 Matrix algebra is a means of making calculations
upon arrays of numbers (or data).
 Most data sets are matrix-type

WHY USE IT?
 Matrix algebra makes mathematical expression
and computation easier.
 It allows you to get rid of cumbersome notation,
concentrate on the concepts involved and
understand where your results come from.


 x  y 
7,
3x  y 
5.
x  y  2z  7,
3x  y  6z  5.





 2x  y  4z  2,
HOW ABOUT SOLVING
5x  4 y  10z 
1,
Consider the following set of equations:
It is easy to show that x = 3 and y
= 4.
Matrices can help…
1.1 Matrices

DEFINITIONS - SCALAR
 a scalar is a number
⚫ (denoted with regular type: 1 or 22)

DEFINITIONS - VECTOR
 Vector: a single row or column of numbers
⚫ denoted with bold small letters
⚫ Row vector
a =
1
2 3 4
5

5

4

⚫ Column vector
1
b =
2
3

DEFINITIONS - MATRIX
 A system of m n numbers arranged in the form of an
ordered set of m rows, each consisting of an ordered
set of n numbers, is called an m x n matrix
 If there are m rows and n columns in the array, the
matrix is said to be of order m x n or (m,n) or m by n
 A matrix is an array of numbers
A =
 Denoted with a bold Capital letter
 All matrices have an order (or dimension):
that is, the number of rows  the number of columns.
So, A is 2 by 3 or (2  3).
a12
a13
a22
a2
3
a
11

a2
1

21 22



2n


m1 m
2
 mn

a1n

 a11 a12
 a
a a
A 
 
a
a a
In the matrix
• numbers aij are called elements. First subscript
indicates the row; second subscript indicates the
column. The matrix consists of mn elements
• It is called “the m  n matrix A = [aij]” or simply “the
matrix A ” if number of rows and columns are
understood.
1.1 Matrices

TYPES OF MATRICES
m = n. Ex: A =
5
 Square Matrix: A square matrix is a matrix that
has the same number of rows and columns i.e. if
4

 Row Matrix: An m x n matrix is called row
matrix if m = 1. Ex: A = 1 2 3 4 5
 Column Matrix: An m x n matrix is called row
matrix if n = 1. Ex: A = 1
2
3
1
2
3
4

TYPES OF MATRICES
 Zero Matrix: A matrix each of whose elements is
zero & is called a zero matrix. It is usually denoted
by “O”. It is also called “Null Matrix”
0 0
0
0 0
0

0




0
0

 
A 


TYPES OF MATRICES
 Diagonal Matrix: A square matrix with its all non
diagonal elements as zero. i.e if A = [aij] is a diagonal
aij
matrix, then = 0 whenever i ≠ j. Diagonal
elements are the aij elements of the square matrix A
for which i = j.
1 0 0
0 2 0
0 0 3

TYPES OF MATRICES
 Diagonal elements are said to constitute the main
diagonal or principal diagonal or simply a
diagonal.
 The diagonals which lie on a line perpendicular to the
diagonal are said to constitute secondary diagonal.
1 2
3 4
Here main diagonal consists of 1 & 4 and secondary
diagonal consists of 2 & 3

TYPES OF MATRICES
 Scalar Matrix: It’s a diagonal matrix whose all
elements are equal.
2 0 0
0 2 0
0 0 2
 Unit Matrix: It’s a scalar matrix whose all
diagonal elements are equal to unity. It is also
called a Unit Matrix or Identity Matrix. It is
denoted by In.
1 0 0
0 1 0
0 0 1

TYPES OF MATRICES
 Triangular Matrix: If every element above or
below the diagonal is zero, the matrix is said to
be a triangular matrix.
1 4
3
0 2
1
0 0
3
Upper Triangular Matrix
1 0
0
3 2
0
5 −6
3
Lower Triangular Matrix

EQUALITY OF MATRICES
 Two matrices A & B are said to be equal iff:
i. A and B are of the same order
ii. All the elements of A are equal as that of
corresponding elements of B
 Two matrices A = [aij] & B = [bij] of the
same
order are said to be equal if aij = bij
If A =
1
2
3
4
B =
𝑥
𝑦
𝑧
𝑤
If A & B are equal, then
x=1, y=2, z=3, w=4

EQUALITY OF MATRICES
(PROBLEMS FOR
PRACTICE)
Q1: If
𝑥
− 𝑦
2𝑥
− 𝑦
2𝑥
+ 𝑧
3𝑧
+ 𝑤
=
−1 5
0
13
; find x,y,z,w.
Q2: If
𝑥
2𝑥
+ 𝑧
3𝑥
− 𝑦
3𝑦
− 𝑤
=
3
4
7
2
; find x,y,z,w.
Q3: If
𝑎 + 𝑏 2
5
𝑎𝑏
=
6
5
8
2
; find a,b.

TRACE OF A MATRIX
 In a square matrix A, the sum of all the diagonal
elements is called the trace of A. It is denoted by tr
A.
⚫ Ex: If A = tr A = 1+4+1 = 6
⚫ Ex: If B =
1 2
3
0 4
5
7 6
1
1 2
3 4
tr B = 1+4 = 5

OPERATIONS ON
MATRICES
Addition/Subtraction
Scalar Multiplication
Matrix Multiplication

ADDITION AND SUBTRACTION
 Two matrices may be added (or subtracted) iff
they are the same order.
 Simply add (or subtract) the corresponding
elements. So, A + B = C

ADDITION AND SUBTRACTION (CONT.)
Where
a32
 b32  c32
a31  b31  c31
a22
 b22  c22
a21  b21  c21
a12
 b12  c12
a31 a32 
b31 b32 
a11  b11  c11
c32




c3
1



21 22

21 22

21 22

c
 
c
c12

c1
1
b
 
b
a

a
a12  b11 b12

a1
1

ADDITION /
SUBTRACTION
(PROBLEMS FOR
PRACTICE)
Q1: If A =
1 3
4
−2 4 8
3 −2
−1
and B
4 1
0
1 3
5
0 1
6
find A+B, A-B.
Q2: If A =
3 8
11
6 −3 8
and B =
1 −6
15
3 8
17
find A+B, A-B.

SCALAR MULTIPLICATION
 To multiply a scalar times a matrix, simply
multiply each element of the matrix by the scalar
quantity
 Ex: If A =
3 8 11
6 −3 8
a21 a22
 ka21
, then
10A =
30 80 110
60 −30 80
ka12

ka22


k a11 a12  
ka11

PROBLEMS FOR PRACTICE
Q1: If A =
1 3
4
−2 4 8
3 −2
−1
and B
4 1
0
1 3
5
0 1
6
find 5A+2B.
Q2: If A =
3 8
11
6 −3 8
and B =
1 −6
15
3 8
17
find 7A - 5B.
Q3: If A =
2 −2
7
4 6
3
; find matrix X such that
X+A=O where O is a null matrix.

Q4: If A =
2
3
and B =
4 7
3 4 5
3
Show that 5(A+B) = 5A + 5B.
Q5: If A =
1 2 0
4
and B =
2 1 0 3
2 4 −1 3 1 −2 2
3
find a 2 x 4 matrix “X” such that A - 2X = 3B.
Q6: If X+Y =
7
0
2
5
and X – Y =
3
0
0
3
Find X&Y.
Q7: Find additive inverse of
5 10 9
1 −3
−2

MATRIX MULTIPLICATION
If A = [aij] is a m  p matrix and B = [bij] is a p
 n matrix, then AB is defined as a m  n
matrix C = AB, where C= [cij] with
p
k
1
cij   aik bkj  ai1b1 j  ai 2b2 j  ...  aipbpj

0
Example: A 
1
1 4


2 3
,
1
2


 5
0
B  
2
3
and C =
AB.
Evaluate c 21.
 1
2
1 2 3
 2
3


0
4
 

0

21
c  0(1) 1 2  4 5  22
for 1  i  m, 1  j  n.

RULE OF MATRIX MULTIPLICATION
 Multiplication or Product of two matrices A & B is
possible iff the number of columns of A is equal to
the number of rows of B.
 The rule of the multiplication of the matrices is
row-column wise (→↓).
 The first row of AB is obtained by multiplying the 1st
row of A with 1st, 2nd & 3rd column of B.
 The second row of AB is obtained by multiplying
the 2nd row of A with 1st, 2nd & 3rd column of B.
 The third row of AB is obtained by multiplying
the 3rd row of A with 1st, 2nd & 3rd column of B.

MATRIX MULTIPLICATION
1 2
3
0 1
4
1
2


 5
0
Example: A    , B  
2
3
, Evaluate C = AB.
21
 c11  1 (1)  2  2  3  5  18
c22
1
2
c  1 2  2  3  3  0  8
12
 0  (1)  1 2  4  5  22
 0  2  1 3  4  0  3

1 2
3
1

2
3 


c

0
4



  5
0



1
2
C  AB 
1
2 3
 2
3 
18

0
1 2
2
3

4
  


0


Q1: If A = and B =
1
9
0
1
6
9
. Find AB.
Q2: If A =
2 3 2
4 3 1
1 1
−1
2 −3 4
and B =
−1 −2
−1
6 12
6
3 −2 3 5
10 5
Show that AB is a null matrix & BA is not a null
matrix.
Q3: If A =
3
2
4
1
and B =
𝑎
𝑏
3
5
Find a & b such
that AB = BA.
Q4: If A =
1
2
, B
=
3
3 4 4
5
1
, C =
1
1
2
2
Show that A(BC) = (AB)C

Q5: Find “x” such that :
1 𝑥
1
1 3 2
1
2 5 1
2
15 3 2
𝑥
= O
Q6: If A =
3
0
5
BA if exists.
and B = 4 2 −1
0
. Find AB &

Q7: A factory produces three items A, B and C.
Annual sales are given below:
If the unit price of the items are Rs. 2.50/-, Rs.
1.25/- and Rs. 1.50/- respectively, find the total
revenue in each city with the help of matrices.
City
Products
A B C
Delhi 5000 1000 20000
Mumbai 6000 10000 8000

Q8: If A =
1 3
2
0 5
7
6 4
8
Find A2 + 7A + 3I
Q9: If A =
1 2
2
2 1
2
2 2
1
Prove that A2 = 4A + 5I

Matrices A, B and C are conformable,
A + B = B + A
A + (B +C) = (A + B) +C
(A + B) = A + B, where  is a scalar
(distributive law)
(commutative law)
(associative law)
PROPERTIES OF MATRICES

Matrices A, B and C are conformable,
A(B + C) = AB + AC
(A + B)C = AC + BC
A(BC) = (AB) C
AB  BA in general
AB = 0 NOT necessarily imply A = 0 or B = 0
AB = AC NOT necessarily imply B = C
PROPERTIES OF MATRICES

TRANSPOSE OF A MATRIX
34
The matrix obtained by interchanging the
rows and columns of a matrix A is called the
transpose of A (written as AT or A` ).
The transpose of A is
2
3
5

4
6



Example: A 
1
1
4 5


2


3
6
AT
For a matrix A = [aij], its transpose AT = [bij],
where bij = aji.

PRACTICE PROBLEMS
Q1: If A =
2
3
4
5
B =
3
1
2
5
Find A′ + B′, (A+B)′, A′B′
Q2: Verify that (AB)′ = B′A′ if
If A =
1 0
2
−1 2
−3
2 0
, B = 1 −1
0 −2
Q3: If A =
1 2
, B = 5 6 , C =
3 4 7 8
1
0
0
1
Show that (ABC)′ = C′B′A′

SYMMETRIC & SKEW SYMMETRIC
MATRICES
36
A matrix A such that AT = A is called symmetric,
i.e., aji = aij for all i and j.
A + AT must be symmetric. Why?
5
is symmetric.
1 2 3

4
5



3

6


Example: A  2
A matrix A such that AT = -A is called skew-
symmetric, i.e., aji = -aij for all i and j.
A - AT must be skew-symmetric. Why?

PRACTICE PROBLEMS
1 2 3
Q1: Express 4
7
5
8
6
9
as a sum of symmetric &
skew symmetric matrix.
Q2: If A =
2
4
5
3
, then prove that
i) A+A′ is a symmetric matrix
ii) A - A′ is a skew symmetric matrix
iii) AA′ & A′A are symmetric matrices
Q3: Express
6
1
3
4
as a sum of symmetric & skew
symmetric matrix.

1.5 Determinants
a
Consider a 2  2 matrix: A 
a11

a

a12 
 21
22 
Determinant of order 2
Determinant of A, denoted | A,| is a number
and can be evaluated by
 a11a22  a12 a21
21
a12
22
a a
| A |
a11

 a11a22  a12 a21
21
a12
22
a a
| A |
a11
Determinant of order 2
easy to remember (for order 2 only)..
1 2
4
Example: Evaluate the determinant: 3
1
3
4
2
 1 4  2  3  2
1.5 Determinants
+
-

PRACTICE PROBLEMS
Q1: Find the determinant of :
i)
8
9
1
7
ii)
4
0
1
0

1.5 Determinants of order 3
1 2
3
5
8
6

Consider an example: A  4



7

9


Its determinant can be obtained by:
1 2 3
A  4 5 6  3
4
7 8 9
5
 6
1 2
 9
1
2
7 8 7 8 4 5
 33  66  93  0
You are encouraged to find the determinant
by using other rows or columns

PRACTICE PROBLEMS
Find the value of
i)
3 −
5
4
7 6 1
1 2 3
ii)
1 4 7
−2 3 4
1 4 −4

1.5 Determinants
2. |AT| = |A|
3. |AB| = |A||B|
determinant of a matrix
= that of its transpose
The following properties are true for
determinants of any order.
1. If every element of a row (column) is zero,
e.g., 1 2
 1 0  2  0  0 , then |A| = 0.
0 0

Orthogonal matrix
 A matrix A is called orthogonal if AAT = ATA =
I, i.e., AT = A-1
orthogonal.
2

 
1/ 3 1/ 6 1/
3 2 / 6
0
3 1/ 6
1/
i
s
 

1
/
2


Example: prove that A  1/
We’ll see that orthogonal matrix represents a
rotation in fact!
1.3 Types of matrices
Since, AT
3


 1/ 3 1/ 3 1/

  1/ 6 2 / 6
1/
2 0 1/
6. Hence, AAT = ATA =
I.
2


1/




Can you show
the details?

(AB)-1 = B-1A-1
(AT)T = A and (A)T = 
AT
(A + B)T = AT + BT
(AB)T = BT AT
1.4 Properties of matrix

APPLICATION OF
MATRICES
(METHOD OF SOLVING A SYSTEM OF LINEAR
EQUATIONS)
Cramer’s
Rule Q1:

PRACTICE PROBLEMS –
CRAMER’S RULE / DETERMINANT
METHOD
Q1: Solve: 2x + 3y = 5, 3x – 2y = 1
Q2: Solve: x + 3y = 2, 2x + 6y = 7
Q3: Solve: 2x + 7y = 9, 4x + 14y = 18
Q4: Solve: x + y + z = 20, 2x + y – z = 23, 3x + y + z = 46
Q5: Solve: 2x – 3y – z = 0, x + 3y – 2z = 0, x – 3y = 0
Q6: Solve: x + 4y – 2z = 3, 3x + y + 5z = 7, 2x + 3y + z = 5 Q7:
Solve: x – y + 3z = 6, x + 3y – 3z = - 4, 5x + 3y + 3z = 10

PRACTICE PROBLEMS –
CRAMER’S RULE / DETERMINANT
METHOD
Q8: Find the cost of sugar and wheat per kg if the cost of 7
kg of sugar and 3 kg of wheat is Rs. 34 and cost of 3 kg of
sugar and 7 kg of wheat is Rs. 26.
Q9: The perimeter of a triangle is 45 cm. The longest side
exceeds the shortest side by 8 cm and the sum of lengths of
the longest & shortest side is twice the length of the other
side. Find the length of the sides of triangle.
Q10: The sum of three numbers is 6. If we multiply the
third number by 2 and add the first number to the result,
we get 7. By adding second & third number to three times
the first number, we get 12. Find the numbers.

INTRODUCTION
 Cramer’s Rule is a method for solving linear
simultaneous equations. It makes use of
determinants and so a knowledge of these is
necessary before proceeding.
 Cramer’s Rule relies on determinants

COEFFICIENT MATRICES
 You can use determinants to solve a system of
linear equations.
 You use the coefficient matrix of the linear
system.
 Linear System
ax+by=e
cx+dy=f
Coeff Matrix


c d

a b


CRAMER’S RULE FOR 2X2
SYSTEM
 Let A be the coefficient matrix
 Linear System
ax+by=e
cx+dy=f
Coeff Matrix
solution:

If detA  0, then the system has exactly one
and
a e
c f
det A
y

c d
a b
= ad – bc

KEY POINTS
 The denominator consists of the coefficients of
variables (x in the first column, and y in the
second column).
 The numerator is the same as the denominator,
with the constants replacing the coefficients of the
variable for which you are solving.

EXAMPLE - APPLYING CRAMER’S RULE
ON A SYSTEM OF TWO EQUATIONS
Solve the system:
 8x+5y= 2
 2x-4y= -10
So:
and

 4
5



2
The coefficient matrix is:8 and 8  (32)  (10)  42
5
2  4
 42
2
5
10  4
x

 42
8 2
2 10
y 

42
 1
 42
2 5
 4

 8  (50)

x 
10
 42 
42
8 2
10

 80  4

 84
 2
 42  42  42
y 
2
Solution: (-1,2)

APPLYING CRAMER’S RULE
ON A SYSTEM OF TWO EQUATIONS
y
x
b
c d
b
f d
e
c f
x 
Dx
y 
Dy
D
D 
a
D 
e
D 
a

cx  dy  f
 ax  by 
e 

3x  5 y 
14
2x  3y  16
D 
2  3
 (2)(5)  (3)(3)  10  9  19
3 5
14 5
 3
 (16)(5)  (3)(14)  80  42  38
D 
16
x
 (2)(14)  (3)(16)  28  48  76
14
2 16
Dy 
3
76
 
4
D 19
D  38
x  x
  2 y

D 19
Dy

EVALUATING A 3X3
DETERMINANT
(EXPANDING ALONG THE TOP ROW)
 Expanding by Minors (little 2x2 determinants)
3
3
2
2
1
a
3
3
2
2
1
a
3
3
2
2
3
2
1
b b
a b
 c
c
a c
 b
c
b c
a b3 c3
a1 b1 c1
a b2 c2 
a
 (3)(3)  (  2)(4)
 8   23
 (1)(6)
 6  9
3
 (2)
2 0
3 1
2
3
 (3)
2
2 3 1
1 3  2
2 0 3  (1)
0
1 2 3

USING CRAMER’S
RULE
TO SOLVE A SYSTEM OF
THREE EQUATIONS
Consider the following set of linear equations
a11x1  a12 x2  a13 x3  b1
a21x1  a22 x2  a23 x3  b2
a31x1  a32 x2  a33 x3  b3

USING CRAMER’S
RULE
THREE EQUATIONS
The system of equations above can be written in a
matrix form as:
21
a33 
 x3

b3
a a
a11 a12
22
a32
a13   x1 
b1 

a
  x
  b

 23  
2


2


a3

USING CRAMER’S
RULE
THREE EQUATIONS
Define
a1
1
 A 
a21
a12 a13

a22
a32
a23



a
31

a33



x1 
b1

x   x2
 and B
 b2




x3




b3

1 2 3
If D  0, then the system has a unique
solution as shown below (Cramer's Rule).
D D
x 
D1
, x 
D2
, D
x 
D3

USING CRAMER’S RULE
TO SOLVE A SYSTEM OF THREE
EQUATIONS
where
a11 a12 a13
a22 a23
a13 a32 a33
b1
a12
a13
a22 a23
a32 a33
D  a12
b3
D1  b2
a11 a13
a23
a33
a11
a12
a22
b
1
b2
b
3
D3  a12
a13
b1
D2  a12 b2
b
3

EXAMPLE 1
Consider the following equations:
2x1  4x2  5x3  36
3x1  5x2  7x3  7 5x1 
3x2  8x3  31
 Ax  B
where
7




5
 2 4
5

 A  3 5


EXAMPLE 1

x1 
 36

x  x2
 and
B   7 
  

 x3 
3
1

D  3
2 4 5
5 7 
336
5 3 8
36 4 5
D1  7 5 7  672
31 3 8

EXAMPLE 1
2 36 5
D2  3
5
7
31
7
8
 1008
2 4 36
D3  3 5 7  1344
5 3 31
1
x 
D1

672
 2
2
x 
D2
3
x 
D3
D 336

1008
 3
D 336

1344
 4
D 336

CRAMER’S RULE - 3 X 3
 Consider the 3 equation system below with
variables x, y and z:
a1x  b1y  c1z  C1 a2
x  b2 y  c2 z  C2 a3x 
b3y  c3z  C3

CRAMER’S RULE - 3 X 3
 The formulae for the values of x, y and z are shown
below. Notice that all three have the same
denominator.
C1 1
b c1
C2 2
b c2
C b c
x  3 3 3
2 2
a1 b1
c1
a b
c
2
a3 3
b c3
y 
a3
a1 C1 c1
a2 C2 c2
C3 c3
a1 b1
c1
a2 b2
c2
z 
a3
a1 b1 C1
a2 b2 C2
b3 C3
1 1
a b c1
a2 b2 c2
a3 b3 c3

EXAMPLE 1
 Solve the system : 3x - 2y + z = 9
x + 2y - 2z = -5
x + y - 4z = -2

x 
9 2 1
5 2 2
2 1
4

23
 1 y 
1
3 9 1
1 5 2
2
3 2 1 23 3 2 1
1 2 2 1 2 2
1 1 4 1 1 4
23
4

69
 3

EXAMPLE
1
z 
1
3 2 9
1 2 5
1
2

3 2 1
1 2 2
1 1 4
0
23
 0
The solution is
(1, -3, 0)

CRAMER’S RULE
Not all systems have a definite
solution. If the determinant of the
coefficient matrix is zero, a solution
cannot be found using Cramer’s Rule
because of division by zero.
When the solution cannot be
determined, one of two conditions
exists:
⚫ The planes graphed by each equation
are
parallel and there are no solutions.
⚫ The three planes share one line (like
three pages of a book share the same
spine) or represent the same plane, in
which case there are infinite solutions.

If matrices A and B such that AB = BA = I,
then B is called the inverse of A (symbol: A-1);
and A is called the inverse of B (symbol: B-1).
The inverse of a matrix
0

 6 2
3
B  1 1

1 0

1


Show B is the the inverse of matrix A.
1 2
3
3
2
3




1

4


Example: A  1
0

1
0



0

1


Ans: Note that
Can you show the
details?
1.3 Types of matrices

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