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Matrix Handouts
Matrices
A matrix is a set of real or complex numbers (or elements) arranged in
rows and columns to form a rectangular array.
A matrix having m rows and n columns is an m x n (read m by n or
m cross n ) matrix and is referred to as having order m x n.
A matrix can be represented explicitly by enclosing the array within
large square brackets.
A matrix is any doubly subscripted array of elements arranged in rowsand columns.
Capital letters A , B ,C , X , Y, Z etc are used for matrix notation .
2 3 7
A3x3 = 8 5 0 ; A is 3 x 3 matrix
3 4 1
3 x3
2 3 0
B4x3 = 5 7 1 ; B is 4 x 3 matrix
3 2 9
4 0 3 4 x 3
In matrix A, 1st row 1st column element is denoted as a11 = 2
1st row 2nd column element is denoted as a12 = 3
1st row 3rd column element is denoted as a13 = 7
2nd row 1st column element is denoted as a21 = 8
2nd
row 2nd
column element is denoted as a22 = 5Developed by Ms. SAROJ MISHRA
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and so on ...........
Types of Matrices
(1) Vectors : A Vector is a special type of matrix in which there is only
one row or one column .(a) Row Vector : If there is only one row and more than one column in any matrix ,called
'Row Vector' .
1 x n matrix [ a1 a2 a3 ...................... an ] is a row vector .
(b) Column Vector : If there is only one column and more than one row in any matrix ,
called ' Column Vector ' .
n x 1 matrix a1
a2
a3
.
.
.
.
.
.
an
(2) Zero or Null Matrix A matrix , with every element zero , is called a null amtrix . It is
denoted by O . It need not be sqaure . In matrix theory it plays the role of zero .
0 0 0
O = 0 0 00 0 0
0 0 0 4 x 3
(3) Square Matrix : A matrix in which number of rows is equal to number of columns.
4 78 2 is a 2 x 2 square matrix .
2x2
3 7 8
9 0 4 is a 3 x 3 square matrix
1 2 5
3x3
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(4) Diagonal Matrix :A square matrix all of whose elements are zero except those
in the leading diagonal is called a diagonal matrix .
i.e. All elements except diagonal lements are zero called diagonal matrix .
1 0 0
D = 0 5 0 is a 3 x 3 diagonal matrix .
0 0 8 3x3
4 0 0 0
0 9 0 0
D = 0 0 5 0 is a 4 x 4 diagonal matrix .
0 0 0 13
4x4
(5) Upper / Lower Triangular Matrix : A square matrix all of whose elements below the
main diagonal are zero is called upper triangular .
7 3 2
0 4 9 is a 3 x 3 upper diagonal matrix .
0 0 3
3x3
8 3 2 1
0 12 3 4
0 0 9 8 is a 4 x 4 upper diagonal matrix .
0 0 0 1 4x4
Lower Diagonal Matrix : If all elments above the main diagonal are zero it is lower
triangular matrix.
13 0 0
7 5 0 is a 3 x 3 lower diagonal matrix .
6 8 12
3x3
8 0 0 0
13 5 0 0
1 17 9 0 is a 4 x 4 lower diagonal matrix .
7 9 5 1 4x4
(6) Scalar Matrix : If in the diagonal matrix D , diagonal elements are same , it behaves
like a scalar matrix .
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2 0 0
0 2 0 is a 3 x 3 scalar matrix .
0 0 2
3x3
11 0 0 0
0 11 0 0 is a 4 x 4 scalar matrix .
0 0 11 0
0 0 0 11
4x4
(7) The Identity Matrix : An identity matrix has 1 for every diagonal element and
zero elsewhere .
1 0 0
0 1 0 is a 3 x 3 identity matrix .
0 0 1
3x3
1 0 0 0
0 1 0 0
0 0 1 0 is a 4 x 4 identity matrix .
0 0 0 1
4x4
(8) Transpose Matrix : Interchanging the rows into columns or columns into rows in any given
matrix , the new transformed matrix is called as transpose matrix.
A = 3 2 5
6 9 0 2x3
3 6
A'(A Transpose , AT
) = 2 9 is transpose of matrix A
5 03x2
10 7 6
4 3 0
B = 3 8 7
6 13 19
4x3
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10 4 3 6
B' (B Transpose ,BT
)
= 7 3 8 13 is transpose of matrix B .6 0 7 19
3x4
(9) Symmetric Matrix :A square matrix M is symmetric ifM = MT
1 2 5 1 2 5
M = 2 8 9 ,MT
= 2 8 9
5 9 4 5 9 4
MATRIX ALGEBRA
MATRIX ADDITION : For matrix addition , order of matrix should be same .
i.e. To add two matrices A and B:
# of rows in A = # of rows in B
# of columns in A = # of columns in B
2 3 5
A = 5 0 6 can be added with 3x3 matrix only .
7 8 1 3x3
1 7 0
B = 3 4 8
12 5 4
3x3
2 3 5 1 7 0 2 + 1 3 + 7 5 + 0
A + B = 5 0 6 + 3 4 8 = 5 + ( -3) 0 + 4 6 + 87 8 1 12 5 4 7 + 12 8 + 5 1 + 4
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3 10 5
= 2 4 14
5 13 5
Subtraction :
To subtract two matrices A and B:
# of rows in A = # of rows in B
# of columns in A = # of columns in B
Multiplication of Matrices
Scalar Multiplication : 47 9 53 6 3 =28 36 2012 24 12 i.e. k[aij] = [kaij]
If any element is multiplied with any matrix, each element inside the matrix will be
multiplied .
This also means that we can take a common factor out of every element.
Multiplication of two matrices : Two matrices can be multiplied together only when the
number ofcolumns in the matrix on the left equals the number of rows in the matrix on
the right.
A B = C
mn pq = mq for n=p
Regular Multiplication : To multiply two matrices A and B ,
# of columns in A = # of rows in B
Multiply: A (m x n ) by B (n by p)
Matrices A and B can be multiplied if : [r x c] and [s x d]
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1
4
2
3
5
8
6
7+ =
6
12
8
10
A B+ = C
1
4
2
3
5
8
6
7=
4
4
4
4
B A = C
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c = s
The resulting matrix will have the dimensions : [r x c] and [s x d]
r x d
2 5 7 7 2 5 1
Q. Multiply A = , B = 1 0 8 2
3 4 9 3 3 4 4
2x3 3x4
Here , matrix multiplication AB is possible as number of columns in first
matrix A is equal to number of rows in second matrix B , i.e. Matrix A and B satisfies
condition required for matrix multiplication.
2 x 7+ 5 x 1+7 x ( 3) 2x2 + 5 x 0 + 7x3 2 x 5 + 5 x 8 + 7 x 4 2 x ( 1) + 5 x 2 + 7 x( 4)
AB =
3 x 7 + 4 x 1+ 9 x ( 3) 3x2 + 4 x 0 + 9x3 3 x 5 + 4 x 8 + 9 x 4 3 x ( 1) + 4 x 2 + 9 x ( 4)
2 x 4
2 25 78 20
=
2 33 83 31
2 x 4 Ans .
2 1
1 0 2 3 0 2
Q. 2 . Multiply A = 2 1 0 3 and B = 1 4
3 2 2 0 2 0
3x4 4x2
Soln : Here , Product AB is possible because number of columns in A = number of rows in B
i.e. A and B satisfies the condition of multiplication and hence AB is possible .
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1 0 2 3 0 2
Q. 2 . AB = 2 1 0 3 1 4
3 2 2 0 2 0
3x4 4x2
1.2 + 0.0 + 2.1 + 3.2 1.1 + 0.2 + 2.4 + 3.0
= 2.2 1.0 + 0.1 + 3.2 2.1 1.2 + 0.4 + 3.0
3.2 + 2.0 2.1 + 0.2 3.1 + 2.2 2.4 + 0.0
3x2
( Here, . is used for ( x ) multiplication )
10 9
= 10 0
4 1
3x2 Ans .
Find the product of following matrices :
2 4 1 2 10 2
(1)
6 8 2 0 Ans . 22 6
(2) 2 1 2
3 2 Ans . 6 9 14
6 3 4
1 2 3 5 2 1 1 1 9
0 2 4 3 1 7 Ans . 6 2 6
1 3 1 0 1 2 4 0 18
3 4
2 2 1 3 0 1 11 0
(3) 1 0 1 0 1 0 Ans . 4 4
0 1 4 0 2 2 4 1
Cofactor Matrix of a Given Matrix :
1 3 3
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Q. 1 . Find cofactor matrix for the given matrix : 1 4 3
1 3 4
1 3 3Soln : Let A = 1 4 3
1 3 4
In order to find cofactor matrix , we need to find cofactors of each element aij in matrix A .
Let us take notation Aij for the cofactors of each aij respectively .
4 3
Cofactor of a11(=1) is , A11 = + = 7
3 4
1 3
Cofactor of a12(=3) is , A12 = = 11 4
1 4
Cofactor of a13(=3) is , A13 = + = 11 3
3 3
Cofactor of a21(=1) is , A21 = = 33 4
1 3
Cofactor of a22(=4) is , A22 = + = ( 4 3 ) = 11 4
1 3
Cofactor of a23(=3) is , A23 = = 01 3
3 3
Cofactor of a31(=1) is , A31 = + = 3
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4 3
1 3
Cofactor of a32(=3) is , A32 = = 01 3
1 3
Cofactor of a33(=4) is , A33 = + = 31 4
Then . cofactor matrix for the given matrix A is ,
A11 A12 A13
B = A21 A22 A23
A31 A32 A33
7 1 1
= 3 1 0
3 0 1
1 4 2
Q. Given matrix X = 1 2 1 , find cofactor matrix.
1 3 2
In order to find cofactor matrix , we need to find cofactors of each element aij in matrix A .
Let us take notation Xij for the cofactors of each xij respectively .
2 1
Cofactor of x11(=1) is , X11 = + = 13 2
1 1
Cofactor of x12(=4) is , X12 = = 01 2
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1 2
Cofactor of x13(=2) is , X13 = + = 5
1 3
4 2
Cofactor of x21(=1) is , X21 = = 23 2
1 2
Cofactor of x22(=2) is , X22 = + = 0
1 2
1 4
Cofactor of x23(=1) is , X23 = = 11 3
4 2
Cofactor of x31(=1) is , X31 = + = 02 1
1 2
Cofactor of x32(=3) is , X32 = = 31 1
1 4
Cofactor of x33(=2) is , A33 = + = 61 2
Then . cofactor matrix for the given matrix X is ,
X11 X12 X13
Y = X21 X22 X23
X31 X32 X33
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1 3 5
= 2 0 1
0 3 6
1 3 3
Q. Find adjoint matrix for the given matrix : 1 4 3
1 3 4
1 3 3
Soln : Let A = 1 4 3
1 3 4
In order to find adjoint matrix for the given matrix , first we need to find cofactor matrix of the
matrix A .
A11 A12 A13
Step 1 : Cofactor matrix B for the given matrix A = A21 A22 A23
A31 A32 A33
7 1 1 See solution of Q .1
on page 8,9 ,and 10
3 1 0
3 0 1
Step 2 : Transpose matrix of cofactor matrix B will give adjoint matrix A .
T
7 1 1
= 3 1 0
3 0 1
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7 3 3
adj (A) = 1 1 0
1 0 1
Which is the required adjoint matrix for the matrix A denoted by adj A .
1 4 2
Q. Given matrix X = 1 2 1 , find adjoint matrix.
1 3 2
In order to find adjoint matrix for the given matrix , first we need to find cofactor matrix of the
matrix A .
X11 X12 X13
Step 1 : Cofactor matrix B for the given matrix A = X21 X22 X23
X31 X32 X33
1 For cofactor matrix,see page no 10 , 11 , 12
1 3 5
= 2 0 1
0 3 6
Step 2 : Transpose matrix of cofactor matrix Y will give adjoint matrix X .
T
1 3 5
= 2 0 1
0 3 6
1 2 0
adj (X) = 3 0 3
5 1 6
Which is the required adjoint matrix for the matrix X denoted by adj X .
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Condition for evaluating inverse of the matrix : The conditions required to calculate
inverse of the matrix is as follows :1. Matrix shuold be a square matrix
2. Determinant of the matrix should be non zero. If determinat of any matrix becomes zero ,
inverse of that matrix can not be calculated .
1 3 3
Q. Find inverse matrix for the given matrix : 1 4 3
1 3 4
1 3 3
Soln : Let A = 1 4 3
1 3 4
In order to find inverse matrix for the given matrix , first we need to find determinant of the
matrix A .
Step 1 : We find determinant of the given matria as :
1 3 3
A = 1 4 3 = 1(4x4 3x3) 3(1 x4 1 x 3)
1 3 4 + 3 ( 1x3 1x4)
= 1(7) 3(4 3) + 3(3 4)
= 1(7) 3 (1) + 3( 1)
= 1 0
Here , Determinant A , inverse A is possible .
Step 2 : To find cofactor matrix B for the given matrix A
Cofactor matrix B = 7 1 1 See solution of Q .1
on page 8,9 ,and 10
3 1 0
3 0 1
Step 3 : To find transpose of matrix B , which is adjoint A
adj(A) = Transpose cofactor matrix B is given as ,
T
7 1 1
adj (A) = 3 1 0
3 0 1
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7 3 3
adj (A) = 1 1 0
1 0 1
adj(A)
Step 3 : Inverse A = =
A
A11 A12 A13
Cofactor matrix B for the given matrix A = A21 A22 A23
A31 A32 A33
7 1 1 See solution of Q .1
on page 8,9 ,and 10= 3 1 0
3 0 1
Step 2 : Transpose matrix of cofactor matrix B will give adjoint matrix A .
Which is the required adjoint matrix for the matrix A denoted by adj A .
1 4 2
Q. Given matrix X = 1 2 1 , find adjoint matrix.
1 3 2
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In order to find adjoint matrix for the given matrix , first we need to find cofactor matrix of the
matrix A .
X11 X12 X13
Step 1 : Cofactor matrix B for the given matrix A = X21 X22 X23
X31 X32 X33
2 For cofactor matrix,see page no 10 , 11 , 12
1 3 5
= 2 0 1
0 3 6
Step 2 : Transpose matrix of cofactor matrix Y will give adjoint matrix X .
T
1 3 5
= 2 0 1
0 3 6
1 2 0
adj (X) = 3 0 3
5 1 6
Which is the required adjoint matrix for the matrix X denoted by adj X .
Application of Matrices in Industry :
Q. 1 . Use matrix method to solve the system of equations :
x y + z = 1 , 2x + y z = 2 , x 2y z = 4
Solution : The given system of equations can be written in matrix form as
1 1 1 x 1
2 1 1 y = 2
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1 2 1 z 4
This can be represented as AX = B
where
1 1 1 x 1
A = 2 1 1 , X = y , C = 2
1 2 1 z 4
Solving x , y and z we require the formula , X = A1
C
Here , method to calculate A1
is already discussed on the above pages .
3 3 0
A1
= 1/9 1 2 3
5 1 3
Now , X = A1
C = 1/9
3 3 0 1 9 1
A1
= 1/9 1 2 3 2 = 1/9 9 = 1
5 1 3 4 9 1
x 1
y = 1
z 1
x = 1 , y = 1 , z = 1 Ans ..
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Q. The price of three commodities X , Y and Z are x , y and z per unit respectively . A
purchases 4 units of Z and sells 3 units of X and 5 units of Y . B purchases 3 units of Y
and sells 2 units of X and 1 units of Z . C purchases 1 unit of X and slls 4 units of Y and
6 units of Z . In process A , B , C , earn Rs. 6 ,000 Rs. , RS . 5,000 and RS . 13, 000respectively . Using matrices , find the prices per unit of the three
( Note that selling the units is positive earnings and buying the units is negative
earnings . )
Solution : We can formulate the above data in the form of simultaneous equations as :
3x + 5y 4z = 6,000 , 2x 3y + z = 5,000 , x + 4y + 6z = 13,000
The above system of equations can be written in the matrix form as :
3 5 4 x 6,000
2 3 1 y = 5,000
1 4 6 z 13,000
We can represent above marix
AX = B
22 46 7 6,000
X = A1
B = 1/ 151 13 14 11 5,000
5 17 19 13,000
x 4,53,000 3,000
y = (1/151) 1,51.000 = 1,000
z 3,02,000 2,000
Hence the prices of three commodities X , Y and Z are RS. 3,000 , Rs. 1,000 and RS.
2,000 per unit . Ans ...
Q. A company is to employ 60 laborers from either of the party X and Y comprising of
persons in different skills are as under :
Category
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Party Skill I Skill II Skill III
X 25 20 15Y 20 30 10
Rate of labor applicable to categories I , II and III are Rs. 500 , Rs . 700 and Rs.
600 respectively . Using matrices , find which party is economically preferable .
Soln : Hint :Let A be the matrix representing the nmber of labourers of three categories I , II
and III of the party X and Y ; and B be the matrix representing the rate of labour applicable to
categories I , II and III .
Then ,
Category
Rate
I II III I 500
A = 25 20 15 X B = II 700
20 30 10 Y III 600
The total labour charges payable to each party are given by the elements of product of the
matrices A and B , i.e. ,
I II III I 500
12500 + 20000 + 9000
X 25 20 15 II 1000 =
AB =
Y 20 30 10 III 600 10000 + 30000 + 6000
X 41500
=
Y 46000
This shows that the company has to pay Rs. 41,500 to party X and RS . 46,000 to party Y as
labour charges . Since Rs. 46000 < 41500 , therefore party X is economicaly preferable .
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Determinant
A scalar quantity obtained by expanding the elements of a sqaure matrix , say A with respect to
the elements of any row or column is called is called a determinat of matrix A and is denoted by
A or det. A .
Determinat of second order : consider the square matrix of order 2 as :
2 3
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A =
5 7
The product of the elements in the principal diagonal is 2x7 and the product of the lements in
the of diagonal is : 3 x 5 . The difference ( 2 x 7) ( 3 x 5) is called the determinat of A , and is
denoted by A or det. A .
2 4
Illustration : = 2( 8) ( 4) 6 = 16 + 24 = 8
6 8
Determinant of order 3 : Consider the sqaure matrix of order 3 as :
2 3 4
0 7 1 7 1 0 A = 1 0 7 = 2 3 + 4
5 6 4 6 4 5
4 5 6
= 2(0 x 6 7 x 5) 3 ( 1 x6 7 x 4 ) + 4 ( 1 x5 0 x 4)
= 2(0 35 ) 3( 6 28) + 4 ( 5 0 )
= 2 ( 35 ) 3 ( 22 ) + 20
= 70 + 66 + 20
= 4 + 20 = 16 Ans .. ..
What is the relation between matrix and determiants Every determiant is a scalar quantity
( real number) obtained by expanding the elemnts ofsquare matrix .
Properties of matrix :
1. The value of the determinat remains unchanged, if the rows and the columns of
the determiant are interchanged , That is , A = AT
1 2 3 1 4 7 example :
4 5 6 = 2 5 8
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7 8 9 3 6 9
2. If any two adjacent rows( or columns) of the determiant are interchanged , its valueremain unchanged , but its sign is changed , That is , A = A
Changing first two columns ,
1 2 3 2 1 3
4 5 6 = 5 4 6
7 8 9 8 7 9
3 . If two rows( or columns) of a determiant are identical , then the value of the determiant is zero .
That is
1 2 3
A = 1 2 3 = 0 , as first two rows are identical .
4 5 6
1 2 2
B = 4 3 3 = 0 , as last two columns are identica .
5 6 6
4. If any scalar k is multiplied with any determinant , then every element in any one
row ( or column) of the determient gets multiplied by the scalar i.e. ,
1 2 3 k1 2 3
k 4 5 6 = k4 5 6
7 8 9 k7 8 9
5. If to each element of any row ( or column) of a determinant are added ( or
subtracted ) equi multiples of the corresponding elemnts of one or more rows ( or
columns) , the value of the determiant remains same .
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a1 a2 a3 a1 ma3 a2 a3
b1 b2 b3 = b1 mb3 b2 b3
c1 c2 c3 c1 mc3 c2 c3
Cramer's rule for solving unknowns ( variables ) in two / three linear equations :
Let us consider the following linear equations :
ax + by = c
ux + vy = d
Solve for x , y .
We can write the given equatons as follows :
a b x c
u v y d
Let D be the determiant of the coefficients of the variables x and y such that
a b
D = ,
u v
c b
Further , let Dx = be the determiant obtained from D
d v by repalcing the first column by the
elements c , d .
and by repalicng the second column by the elemnts
a c
Dy = be the determinant obtained from D by replacing the second column by
u d the elements c , d .
Thus the values of x and y can be expressed in the form of determiant as
x y 1
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= = , provided D 0
Dx Dy D
1
x = x Dx , provided D 0
D
1
y = X Dy
D
Solve the following linear equations using cramers rule .
Q . Solve the following system of euations using Cramer's Method ,
x + y + z = 6 ; x y + z = 2 ; 2x + y z = 1
Solution : The determinant of the equations of x , y and z is given by ,
1 1 1
D = 1 1 1 = 1(1 1) 1(1 2) + 1(1 + 2 )
2 1 1
= 6 ( 0 )
The solution is given by
x y z 1
= = = , provided D 0
Dx Dy Dz D
6 1 1 1 1 2 1 2 1
Dx = 2 1 1 = 6 1 + 1
1 1 1 1 1 1 1 1 1
= 6 ( 1 1) 1( 2 1) + 1 ( 2 + 1)
= 6 x 0 1 ( 3) + 1 ( 3) = 0 + 3 + 3 = 6 Ans ..
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1 6 1
2 1 1 1 1 2
Similarly , Dy = 1 2 1 = 1 6 +1
1 1 2 1 2 1
2 1 1
= 1( 2 1) 6( 1 2) + 1(1 4)
= 1( 3) 6( 3) + 1( 3)
= 3 + 18 3
= 6 + 18 = 12 Ans .
1 1 6 1 2 1 2 1 1
Also , Dz = 1 1 2 = 1 1 + 6
2 1 1 1 1 2 1 2 1
= 18
Thus using cramer's rule we have ,
x / 6 = y / 12 = z / 18 = 1 / 6
x = 1 , y = 2 , z = 3 Ans . ....sss.
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