unit 2 : matrices - ca-foundation.in · 2018-07-09 · 2.2.1 introduction matrices applications are...
TRANSCRIPT
2.2.1 INTRODUCTION Matrices applications are used in Business, Finance and Economics. Matrices applications are helpful to solve the linear equations with the help of this cost estimation, sales projection, etc., can be predicted. In this chapter, we shall fi nd it interesting fundamental applications of matrices.
Matrix:
Ram, Sita and Laxman are three friends. Ram has 5 books, 3 pencils and 2 pens. Sita has 10 books, 8 pencils and 5 pens. Laxman has 15 books, 10pencils and 2 pens. The above information about three friends can be represented in the following form:
Books Pencils PenRam 5 3 2Sita 10 8 5Laxman 15 10 2
We can express the above information in the following form:
3 3
5 3 2
10 8 5
15 10 2
An arrangement or display of information in the above form is called a matrix.
At the end of this chapter, you will be able to:
w Defi nition of Matrix
w Types of Matrix
w Addition and Subtraction of Matrix
w Multiplication of Matrices
w Transpose of Matrix
w Inverse of a Square Matrix
w Adjoint of a Square Matrix
w System of Linear Equations
w Cramer’s Rule (Determinant Method)
LEARNING OBJECTIVES
UNIT 2 : MATRICES
First Row
Second Row
Third Row
First Column
Second Column
Third Column
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Matrix (Defi nition)
A rectangular array of numbers (real/complex) denoted by:
11 1
1
A=n
m mn
a a
a a
A is rectangular matrix with m rows and n columns. The numbers aij, i = 1,2 ……..m; j = 1,2,…..n of this array are called its elements aij, is associated. We shall denote a matrix either using by using brackets [ ]; or ( ).
Notes:
1. It is to be noted that a matrix is just an arrangement of elements without any value in rows and columns.
2. The plural matrix is matrices.
3. It is to be noted that a matrix is just an arrangement of elements without any value in rows and columns.
Order of a Matrix: A matrix A with m rows and n columns is called a matrix of order (m, n) or m × n (read as m by n).
Consider the matrix
1 2 2
A= 4 6 5
7 9 8
It is a matrix of order 3 × 3. Here the 9 occurs in the third row and second column. The elements 5 occurs in the second row and third column. Thus in notations we may write: a32 = 9 and a23 = 5.
2.2.2 TYPES OF MATRICESRow Matrix:
A matrix which has only one row is called a row matrix or row vector.
The matrices of the type [a1, a2, a3 ……..,an]; [1, 2, 5] are examples of row matrices.
Column Matrix:
A matrix which has only one column is called a column matrix or a column vector.
The matrices are of the types
12
2 25 ;
7 36
4
and
are examples of column.
Zero Matrix or Null Matrix:
If every element of a m × n matrix is zero, the matrix is called zero matrix or null matrix of order
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2.37MATRICES
(m, n) and it is denoted by : Omn. Thus [0, 0]; 0 0
0 0
;
0
0
0
are all zero matrices, but all of matrices, but of different orders.
Square Matrix and Rectangular Matrix: If the number of rows and columns in a matrix are same, such
a matrix is called a square matrix; otherwise it is called a rectangular matrix.
2 0 01 0
; 0 -3 00 3
0 0 5
are
examples of square matrix are zero except on the leading diagonal, then it is called diagonal matrix.
Thus, if A = 11 12 1
1 2
n
ij n n
n n nn
a a a
a
a a a
Is an n × n matrix, then the diagonal matrix obtained from it will be following type:
Diagonal Matrix 11
2211 22
0 0 0
0 0 0, ,....
. . . .
0 .. ..
nn
nn
a
aDiag a a a
a
Scalar Matrix:
A diagonal matrix whose leading diagonal elements are all equal is called a scalar matrix, for example,
k 0 0
1 00 k 0 ; 2 ;
0 10 0 k
Unit Matrix:
A scalar matrix whose diagonal elements are equal to unity is called unit matrix and it is denoted by In×n, if it is order of order n .For example,
A
1 0 0 0
0 1 0 0
. . . .
0 .. .. 1
n nI
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Upper triangle matrix:A matrix is known as upper triangular matrix if all the elements below the leading diagonal are zero. For example.
1 6 5 1 4
0 3 4 ; 08
0 0 2 00
;
1 2 3 4
3 6 7 8
2 5 4 6
5 0 7 8
Lower Triangular Matrix:
A matrix is known as lower triangular matrix if all the elements above the leading diagonal are zero. For example.
1 0 0 1 0
2 3 0 ; 7 6
6 8 2 9 2
Sub Matrix:
The matrix obtained by deleting one or more rows or columns or both of a matrix is called its sub matrix. For example.
Let A =
1 3 7
-2 0 3
6 8 2
The sub matrix is obtained by deleting second row and the second column from matrix A.
B = 1 7
6 2
Equal Matrices:
Two matrices A=[aij] and B=[bij] are said to be equal if they satisfy the following two conditions.
(i) The order of both the matrices is same;
(ii) Corresponding elements in both the matrices are equal i.e.,
aij = bij (i = 1,2,…..m and j=1,2…….n)
2.2.3 ALGEBRA OF MATRICESAddition and Subtraction of matrices: Let A and B be two matrices of the same order. Then the addition of A and B, denoted by A+B, is the matrix obtained by adding corresponding entries of A and similarly to subtract two matrices we just subtract their corresponding elements.
Thus, if A = (aij) m×n and B = (bij) m×n, then
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A+B= (aij + bij) m×n
Remark: We can add two matrices of the same order. If they are of the same order, we say they are comfortable for addition. Also, the order of the matrices is the same as that of the two original matrices.
Property: If A, B, C are matrices of same order, then
(i) A + B = B + A (Commutative Law)
(ii) (A + B) + C = A + (B + C) (Associative Law)
(iii) K (A + B) = k.A + m.B, where m is constant
Example 1: 2 3 5 5 3 0
4 6 0 1 4 2
7 0 5
5 10 2
2 3 5 5 3 0
4 6 0 1 4 2
3 6 5
3 2 2
Negative of a Matrix: If A is any matrix, the negative of A is denoted by
-2 1
-A= -5 -4
6 0
Scalar Multiplication
The multiplication of a matrix by scalar k implies the multiplication of every element.
Example 2:
Let A = 2 3
4 5
k = 3
Solution: Then k A = 6 9
12 15
Multiplication of two matrices.
The product A B of two matrices A and B defi ned only if the number of columns in Matrix A is equal to the number of rows in Matrix B.
Properties of matrix Multiplication:
(i) Matrix multiplication is not commutative in general, i.e. AB ≠ BA.
(ii) Matrix multiplication is associative (AB) C = A(BC), where both sides are defi ned.
(iii) Multiplication distributes over addition of Matrices i.e.,
(a) A (B + C) = AB + AC
(b) (A + B) C = AC + BC
MATRICES
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(iv) If A, B and C are three matrices such that AB = AC , then the general B ≠ C.
(v) If A is m×n matrix and O is an n × p null matrix, then AO = O, A= O
(vi) If A is a square matrix and I is a unit matrix of the same order, then AI = IA = O
(vii) Product of the two no-zero matrices is non zero matrix.
Example: 3 if
5 31 5 3
A= B= 1 04 5 6
2 8
, fi nd AB
Solution: AB= 1×5+5×1+3×2 1×3+5×0+3×84×5+5×1+6×2 4×3+5×0+6×8
5+5+6 3+2420+5+12 12+48
=16 27
37 60
Example 4: The annual sale volume of three products X , Y , Z whose sale prices per unit are ` 3.50 `2.75 , ` 1.50 respectively. In two different market I and II are shown below.
Market Product
X Y ZI 6,000 9,000 13,000II 12,000 6,000 17,000
Find the total revenue in each market with the help of matrices.
Solution: Let P denotes the column matrix of prices and S denote the rectangular matrix of sale of volumes of three different commodities at three different markets. Then,
6000 9000 13000
12000 6000 17000
Rs. X Y Z3.50 X
P = 2.75 Y and S = 1.50 Z
I
II
3.506000 9000 13000
2.7512000 6000 17000
1.50
Rs.(21,000+24750+19,500)Rs.(42,000+16,500+25,500)Rs.65,250 IRs.84
Total revenue in m
,000 II
arkets I and II
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Transpose of Matrix: The matrix is obtained by interchanging rows and columns of a matrix A is called its transpose. Transpose of a matrix by AT or A’.
Symbolically, if A=[aij] and A’=[bij]
Then aij=bij
Example:
Let
1 3 7
A= -2 0 3
6 8 2
then '
1 -2 6
A = 3 0 8
7 3 2
Properties of transpose of a Matrix:
(1) A matrix is transpose of its matrix i.e. A = (A’)’.
(2) The transpose of the sum of the two matrices is the sum of their transpose matrices, i.e.(A + B)’= A’ + B’
(3) Transpose of a multiplication of a matrix and constant number is equal to the multiplication of the constant number by the transpose of matrix, i.e. (KA)’ = K.A’
(4) The transpose of the two matrices are equal to the product of their transpose in reverse order, i.e., (AB)’ = B’. A’
Symmetric Matrix: In any matrix if A is called symmetric then A’ = A.
Example 5: Let
1 3 7
A= 3 0 8
7 8 2
'
1 3 7
A = 3 0 8
7 8 2
Here A’= A, A is called symmetric matrix.
Solution: Skew Symmetric matrix: Any matrix A is called skew symmetric. If A’=-A,
for a skew symmetric matric A = [aij], aij = – aij
Example 6: Find A2 , if
1 0 0
0 1 0
1
A
a b
Solution: A2 = A.A=
1 0 0 1 0 0
0 1 0 0 1 0
1 1
a b a b
=
1 1 0 0 0 1 0 0 1 0 1 0 0 0 0 1
0 1 1 0 0 0 0 1 1 0 0 0 1 0 0 1
1 0 1 0 1 1 0 0 1 1
a b
a b
a b a a b b a b
MATRICES
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=
1 0 0
0 1 0
0 0 1
= I
Example 7: (a) Show that the matrix
760543101
A satisfi es the equation:
3 2A 2A A 20 I 0 .
Solution:
Now
(b) If A =
dcba
and I =
1001
, then show that – (a+d) A = (bc–ad) I.
b) Solution: We may write,
L.H.S. of the given equation = A2−(a+d)A
= (bc−ad) I = R.H.S.
Hence, A2 – (a + d) A = (bc–ad) I.
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A2
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Example 8: If A= 1 1
2 1
; B= 1
1
a
b
and (A+B)2 =A2 + B2, fi nd the value of a and b
Solution: (A+B) =
(A+B)2 = 1 0
2 2
a
b
1 0
2 2
a
b
=
21 0 2 (1 ) 0 2 0
2 1 2(2 ) 2 0 4
a b a
b a b b
= 21 0
2 2 4
a
a ab b
A2 = 1 1
2 1
1 1
2 1
= 1 2 1 1 1 0
2 2 2 1 0 1
B2 = 1
1
a
b
1
1
a
b
= 2 1
1
a b a
ab b b
A2 + B2 = 1 0
0 1
+2 1
1
a b a
ab b b
=
2 1 1a b a
ab b b
Now (A + B)2 = A2 + B2
21 0
2 2 4
a
a ab b
= 2 1 1a b a
ab b b
a-1 = 0 a = 1
b = 4
Example 9: A company employs 60 labourers from either of party A and B, comprising of persons in different age groups as under:
Category I (20-25 years) II (26-30 years) III (31-40 years) Party A 25 25 15Party B 20 30 10
Category RatesI 1,200II 1,000III 600
Rate of Labour applicable to categories I, II and III are ` 1,200, ` 1,000 and ` 600 respectively. Using matrices, fi nd which party is economically preferable over the other.
MATRICES
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Solution :
Category I (20-25 years) II (26-30 years) III (31-40 years) Party A 25 25 15Party B 20 30 10
Rates : Category RatesI 1,200II 1,000III 600
Labour charges payable ton each party can be given by
25 1200 20 1000 15 6000
20 1200 30 1000 10 600
1,40,00060,000( )
Therefore, Party B is more economical as compared to Party A.
Example 10: In a certain city there are 5 colleges and 20 schools. Each school has 3 peons, 1 clerk and 1 head-clerk, where a college has 5 peons, 3 clerks, 1 head. clerk and additional staff of a caretaker.The monthly salary of a employee is as follows.
Peon `1100
Clerk `1700
Head-Clerk `3000
Caretaker `2500
Using matrix method, fi nd the total monthly bill of each college.
Solution: Let us put the information in tabular form:
Peon Clerk Head Clerk Caretaker
School 3 1 1 0
College 5 3 1 1
Let A =
1100
3 1 1 0 1700
5 3 1 1 3000
2500
= B
Therefore, the monthly bill for a school and a college is given as:
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AB =
1100
3 1 1 0 1700
5 3 1 1 3000
2500
= 3 1100 1 1700 1 3000 0 2500
5 1100 3 1700 1 3000 1 2500
= 8000
16100
Therefore, the monthly bill for a school is ` 8,000 and for a college is ` 16,100
Example 11: There are two families A and B .There are 4 men, 6 women and 2 Children in a Family A and 2 men, 2 women, and 4 children in Family B .The recommended requirement of calories in Man: 2400, Woman : 1900, Child : 1800 and for proteins in Man: 55 gm, Woman: 45 gm and Child: 33 gm.
Solution: Represent the above information by matrices in using matrix multiplication method
Solution: The members of the two families can be represented by the 2 × 3 matrix.
4 6 2
2 2 4
F =
M W C
A
B
And the recommended daily requirement of calories and proteins for each member can be represented by the 3 × 2 matrix:
Calories Proteins M 2400 55
F = W 1900 45C 1800 33
The total requirements of calories and proteins for each of the families is given by matrix multiplication.
FR =
2400 554 6 2 24600 556
1900 452 2 4 15800 332
1800 33
A
B
Hence fi nally A requires 24,600 calories and 556 gm proteins and Family B requires 15,800 calories and 332 gm proteins.
Example 12: Three fi rms A, B and C supplied 40, 35 and 25 truckloads of stones and 5, 8 truckloads of sand respectively to a contractor. If the costs of stone and sand are ` 1200 and `500 per truck load respectively, fi nd total amount paid by the contractor to each of these fi rms, buy using matrix method.
Solution: The amount of stone and sand supplied to a contractor by three fi rms A, B and C can represented by using matrix method.
MATRICES
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40 10
35 5
25 8
Stone Sand
X = A
B
C
The cost per truck load of stone and stand can be represented by the column matrix.
Stone 1200Y=
Sand 500
Thus, the total amount paid by the contractor to each of these fi rms is given by the matrix product.
Stone 1200Y=
Sand 500
40 10 53,000 A1200
XY= 35 5 = 44,500 B500
25 8 34,000 C
Hence the amount paid to fi rms A = ` 53,000; B = ` 44,500 and C = ` 34,000
2.2.4 DETERMINANTSThe determinant of a square matrix is a number which is associated with the square matrix. This number may be positive, negative or zero > the determinant of a square matrix A commonly denoted by det A or | A| or ∆. The matrices which are not square do not have determinants.
Determinants are quite useful to solving a system of linear equations. They are also equations. They are also helpful in expressing certain formulas.
The determinant of a (2 × 2) matrix A = =a b
ad bcc d
The determinant of (3 × 3) matrix, A = 1 2 3
1 2 3
1 2 3
a a a
b b b
c c c
1 2 3
1 2 3
1 2 3
a a a
A b b b
c c c
And its defi ned as a1
2 3
2 3
b b
c c- a2
1 3
1 3
b b
c c+ a3
1 2
1 2
b b
c c
= a1 (b2c3-b3c2) – a2(b1c3-b3c1) + a3(b1c2-b2c1)
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Minors and Cofactors of a Determinant:
Minor of the element of a determinant is the determinant of Mij by deleting ith row and jth column in which element is existing.
Cofactor of matrix A = (-1) i+j Mij , Where Mij = Minor of the element ith row and jth column.
Minors: Let us consider the determinant 1 2 3
1 2 3
1 2 3
a a a
b b b
c c c
M11 = 2 3
2 3
b b
c c= b2c3 – b3c2 ;
1 3
1 312M
b b
c c = b1c3 – b3c1 ;
1 2
1 213
b bM
c c = b1c2 – b2c1
2 321
2 3
Ma a
c c = a2c3 – a3c2;
1 322
1 3
Ma a
c c = a1c3 – a3c1;
1 222
1 2
Ma a
c c = a1c2 – a2c1;
2 331
2 3
Ma a
b b = a2b3 – a3b1;
1 322
1 3
Ma a
b b = a1b3 – a3b1;
1 322
1 3
Ma a
c c = a1b2 – a2b1;
Cofactors
C11= (-1)1+1M11= M11, C12= (-1) 1+2M12= -M21 ; C13= (-1) 1+3M13= M13
C21= (-1)2+1M21= - M21 , C22= (-1) 2+2M22= M22 ; C23= (-1) 2+3M23= -M23
C31= (-1) 3+1M31= M31 , C32= (-1) 3+2M32= -M32 ; C33= (-1) 3+3M33= M33
The value of determinant can be defi ned in terms of cofactor matrix as
∆ = a 11 c 11 + a 12 c 12 + a13 c13
or
∆ = a 11 c 11 + a 21 c 12 + a31 c13
Properties of Determinants:
1) The value of determinant remains unaltered interchanged if its rows or columns interchanged.
2) The value of determinant change signs if any two rows (or columns) interchanges.
3) The value of determinant is zero if any two rows (any columns) then value of determinant is equal to zero.
4) The value determinant becomes k times (where k is constant) if any row or columns multiplied by k the value of determinant also multiplied by k.
5) The value of determinant is zero if any two rows (or column) are proportional then the value of determinant is equal to zero.
6) If each element of any row (or column) is a sum of two numbers, the determinant can be expressed as the sum of the determinants.
7) The value of determinant remains same if to any (or column) multiple of row (or column) is added or subtracted.
MATRICES
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Singular and Non-Singular Matrices: Any Square Matrix A is singular, if | A | = 0 . The matrix is non-singular, if | A | ≠ 0 .
Example 13: If A = 1 2
3 6 Prove that A is a non singular matrix.
Solution : | A | = 1 2
3 6
= -1 × 6 - 2 × -3 = 0
A is singular matrix.
Example 14: Find the determinant value of the following matrices.
A =
1 2 3
4 5 6
7 8 9
Solution by defi nition
∆ = 15 6
8 9– 2
4 6
7 9+ 3
4 5
7 8
= 1(45 – 48) –2(36 – 42) + 3(32 – 35)
= -3 + 12 - 9 = 0
Adjoint Matrix: Adjoint of A Matrix is the transpose of the Cofactor Matrix
Example 15: Find the Adjoint of the Matrix.
A =
0 0
2 2 1-
0 4 1
Solution:
The Co-factors of elements of A =
0 0
2 2 1 -
0 4 1
are calculated below:
A11= (-1) 1+1
2 0
2 2= 4
A12= (-1) 1+2
2 0
2 1= 2
A13= (-1)1+3
0 0
2 1= 0
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A21= (-1)2+1
2 0
0 4= -8
A22= (-1)2+2
2 0
0 1= 2
A23= (-1)2+34 1
= 0
A31= (-1) 3+1
2 2
0 4= 8
A32= (-1)3+2
2 1-
0 1 = -2
A33= (-1) 3+3
2 1-
4 1 = 6
Now Adj. A =
6 0 0
2- 2 2
8 8- 4
2.2.5 INVERSE OF A MATRIX
If A is a Square matrix and A ≠ 0 then
1 1 Adjoint of AAA
Example 16:
Solve the following system of equations by matrix inversion method :
2x + 8y + 5z = 5
X + y + z = (-2)
X + 2y – z = -2
Solution: The given system of equations can be written in the form, AX = B.
Where (A) =
2
2-
5
B '
1 2 1
1 1 1
5 8 2
and
Z
Y
X
X
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det (A) =
1- 2 1
1 1 1
5 8 2= 2 (-1–2) – 8 ( -1 -1) + 5(2–1)
= -6 + 16 + 5
= 15 0
Hence, the system has a unique solution as A is non-singular. The solution is given by
X = A-1B
To fi nd A-1, we fi nd the cofactors.
A11 = -3; A12 = + 2; A13 = 1, A21 = + 18; A22 =-7; A23 + 4; A31 = 3; A22 = +3; A33 = -6
Co-factor of A =
6- 3 3
4 7- 18
1 2 3
Adj: A = (co-factor A)T =
6- 4 1
3 7- 2
3 18 3
Hence, x = -3; y = 2 and z = -1
Example 17: Show that the matrix A =
4 3
2 1satisfi es the equation
A2 –5A –2I = 0
Hence, deduce the value of A-1
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Solution:
Since A2 – 5A – 2I = 0,
A-1 (A2 – 5A – 2I) = 0
A-5I – 2A-1 = 0
2A-1 = A – 5I
= 2
1
4 3
2 1–
2
5
1 0
0 1
=
2
5 - 2 0 -
2
3
0-1 2
5 -
2
1
=
2
1-
2
3
1 2
A- 1 =
2
1-
2
3
1 2
Example 18:
Fine the inverse of the matrix.
11- 4
3 2
Solution: Hence, solve the system of equations.
2x – 3y = 3
4x – 11y = 11
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Hence, Inverse of
The given system of equations cab be written as AX = B
Where A =
11
3 B and
y
x ,
11- 4
3- 2X
Therefore, X = 1
2
5
11 3 - 10 10
1 - 5
xA B
y
=
1
0
Hence x = 0, y = -1
Example 19: Using the inverse of the coeffi cient matrix, solve the following system of equations:
x + y + z = 3
x + 2y + 3z =4
x + 4y + 9z = 6
Solution: The given equation can be written as AX = B. Where A =
1 1 11 2 31 4 9
, X =
z
y
x
and B =
6
4
3
Since,
The given equations are consistent, and the solution is given by X =A-1 B.
Let us fi nd the cofactors:
A11 = 6; A12= -6; A13= 2; A21 = -5; A22 = 8; A23 = -3
A31 = 1; A32 = - 2 and A33 = 1
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Then Cof. A = 1
1 3- 2
2- 8 6-
1 5- 6
A adj. ;
1 2- 1
3- 8 5-
2 6- 6
Now, A-1 =
1 3- 2
2- 8 6-
1 5- 6
2
1 ,
1Aadj
A
,
0
1
2
0
2
4
2
1
6
4
3
1 3- 2
2- 8 6-
1 5 6
2
1 X
Hence, x = –2; y = 1 and z = 0
2.2.6 SOLUTION OF LINEAR EQUATIONS IN THREE VARIABLES (CRAMER’S RULE)The solution of equations
a11 + a12y + a13z = d1
a21 + a22y + a23z = d2
a31 + a32y + a33z = d3
is given by x =x
; y = y
; z =z
provided ∆≠0, and
∆ = 11 12 13
21 22 23
31 32 33
a a a
a a a
a a a
∆x = 1 12 13
2 22 23
3 32 33
d a a
d a a
d a a
∆y = 11 1 13
21 2 23
31 3 33
a d a
a d a
a d a
∆z = 11 12 1
21 22 2
31 32 3
a a d
a a d
a a d
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x
x
x
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Notes : 1) If ∆ = 0; and ∆x = 0, ∆y = 0 , ∆z= 0; then the given equations will have infi nite solutions and equations will be dependent.
2) If ∆ = 0; and at least ∆x , ∆y = 0 , ∆z; is not zero then the equations will have solution and the equations have no solution and the equations are said to be inconsistent.
Example 20: Solve the equations:
1) 2x – y + z = 4
X + 3y + 2z = 12
3x + 2y + 3z = 16
Solution: Considering the equations:
2x – y + z = 4
X + 3y + 2z = 12
3x + 2y + 3z = 16
By using Cramer’s Rule, the solution of the equations are given below:
X = x
=
4 1 1
12 3 2
16 2 3 4(9 4) 1(36 32) 1(24 48)2 1 1 2(9 4) 1(3 6) 1(2 9)
1 3 2
3 2 3
4 5 1 4 ( 24) 0
2 5 3 7 0
Y =
2 4 1
1 12 2
3 16 3 2(36 32) 4(3 6) 1(16 36)
0 048 20 28 0
0 0
y
Since ∆ = 0; ∆x= 0, ∆y= 0 and ∆z= 0,
There the equations are dependent and will have infi nite solutions;
Example 21:
x + y + 3z = 6
x – 3y – 3z = –4
5x – 3y + 3z = 8
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Solution: By applying Cramer’s rule, we get
x=x
=
6 1 3
4 3 3
8 3 3 6( 9 9) 1(12 24) 3(12 24)1 1 3 1( 9 9) 1(3 15) 3( 3 15)
1 3 3
5 3 3
108 12 4 108 12
0 0
Since ∆x ≠ 0 and ∆ = 0, therefore the system of equations are inconsistent.
Example 22: The given equations are:
x + y + z = –2
3x + 2y + 3z =13
2x + 7y + 4z = 31
Solution:
Here ∆ =
1 1 1
3 2 3
2 7 4
; ∆x =
2 1 1
13 2 3
31 7 4
; ∆y =
1 2 1
3 13 3
2 31 4
; ∆z =
1 1 2
3 2 13
2 7 31
∆ = 1 (8 – 21) –1 (12 – 6) –1(21 – 4) = –13–6–17= –36
∆x = –2 (8 – 21) –1 (52 – 93) –1(91 – 62) = 26 + 41 – 29 = 38
∆y = 1 (52 – 93) +2 (12 – 6) –1(93 – 26) =–14 + 12 – 67= –96
∆z = 1 (62 – 91) –1 (93 – 26) –1(21 – 4) = –13 – 6 – 17 = –36
Hence; x = x
= 38 19
36 18
y=y
=96
36
=8
3
z=z
=130 65
36 18
=
8
3
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UNIT -II EXERCISE: AChoose the most appropriate option (a), (b), (c), or (d)
1) If a matrix has 16 elements; what are the possible orders it can have
(a) 2 × 8 ; 8 × 1; 4 × 4; 1 × 16; 16 × 1 (b) 2 × 8 ; 8 × 2; 4 × 4; 1 × 16; 16 × 1
(c) 2 × 8 ; 8 × 2; 4 × 1; 1 × 16; 16 × 1 (d) 2 × 4 ; 8 × 2; 4 × 4; 1 × 16; 16 × 1
2) Transpose of a rectangular matrix is a
(a) rectangular matrix (b) diagonal matrix
(c) square matrix (d) scaler matrix
3) Transpose of a row matrix is
(a) zero matrix (b) diagonal matrix
(c) column matrix (d) row matrix
4) Two matrices A and B are multiplied to get AB if
(a) both are rectangular
(b) both have same order
(c) no. of columns of A is equal to rows of B
(d) no. of rows of A is equal to no. of columns of B
5) If |A| = 0, then A is
(a) zero matrix (b) singular matrix
(c) non-singular matrix (d) 0
6) If A is a symmetric matrix, then At =
(a) A (b) |A|
(c) 0 (d) diagonal matrix
7) If the order of matrix A is m × p. And the order of B is p × n. Then the order of matrix AB is?
(a) m × n (b) n × m
(c) n × p (d) m × p
8) If A and B are matrices, then which from the following is true?
(a) A + B ≠ B + A (b) (At)t ≠ A
(c) AB ≠ BA (d) all are true
9) What is a, if
2 3
4A
a
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is a singular matrix?
(a) 5 (b) 6
(c) 7 (d) 8
2i 3i10) if A=
2i -i
(i2=-1) then |A| = ?
(a) 2 (c) 8
(c) 4 (d) 5
11) If
then order of matrix A = ?
(a) 2 x 2 (b) 2 x 3
(c) 3 x 2 (d) 3 x 3
Using 12-16 Let 2 3
4 5A
1 5
6 7B
2 5
3 4C
12) Find A + B.
(a) 3 2
10 2
(b) 3 2
10 2
(c) 2 3
10 2
(d) 3 1
10 2
13) Find A–B.
(a) 1 2
2 2
(b) 1 8
2 12
(c) 1 8
2 12
(d) 1 8
12 2
MATRICES
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14) 3A – C
(a) 4 14
9 11
(b) 4 14
9 11
(c) 4 14
9 11
(d) 2 3
4 5
15) AB
(a) 16 31
34 15
(b) 16 31
34 15
(c) 16 31
34 5
(d) 2 3
4 5
16) BA
(a) 22 22
16 53
(b) 22 22
16 53
(c) 22 11
16 53
(d) 22 33
16 53
17) a -b a b
+ b a -b a
(a) 2 2
2 2
0
0
a b
a b
(b) 2 2
2 2
0
0
a b
a b
(c) 2 2
2 2
0
0
a b
a b
(d) 2 2
2 2
0
0
a b
a b
18) 2 2 2 2
2 2 2 2
2 2
2 2
ab bca b b cac aba c a b
(a)
2 22 2 2 2
2 2 2 2 2 2
2 2
2 2 or
a b b ca b ab b c bc
a c ac a b ab a c a b
(b)
2 22 2 2 2
2 2 2 2 2 2
2 2
2 2 or
a b b ca b ab b c bc
a c ac a b ab a c a b
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(c)
2 22 2 2 2
2 2 2 2 2 2
2 2
2 2 or
a b b ca b ab b c bc
a c ac a b ab a c a b
(d)
2 22 2 2 2
2 2 2 2 2 2
2 2
2 2 or
a b b ca b ab b c bc
a c ac a b ab a c a b
19) l m -p q
+ n o r s
(a) l p m q
n r s
(b) l p m q
n r s
(c) l p m q
n r s
(d) l p m q
n r s
20) a b a b
b a b a
(a) 2 2
2 2
0
0
a b
a b
(b) 2 2
2 2
0
0
a b
a b
(c) 2 2
2 2
0
0
a b
a b
(d) 2 2
2 2
0
0
a b
a b
21) 1
2 3 4 5 6
5
(a)
3 4 5 6
6 8 10 12
15 20 25 30
(b)
3 5 4 6
6 8 10 12
12 16 20 24
(c)
3 4 5 6
6 8 10 12
12 16 20 24
(d)
3 4 5 6
6 8 10 12
24 16 16 12
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22. 1 2 3
2 3
x y
x y z
(a) 22 3 3
2 3 4 3 6 3
x xy x y xyz
x y z
(b) 22 3
2 3 4 3 6 3
x xy x y x yz
x y z
(c) 22 2 12
2 3 4 3 6 3
x xy xy y yz
x y z
(d) 22 3
2 3 4 3 6 3
x xy x y x yz
x y z
23.
1 2 3 1 3 5
4 5 6 0 2 4
7 8 9 3 0 5
(a)
10 1 12
22 22 70
34 37 112
(b)
10 1 28
22 2 70
34 5 112
(c)
10 1 28
22 2 70
34 5 112
(d)
10 1 28
22 2 70
34 5 112
24.
2 33 1 3
1 01 0 2
3 1
(a) 14 6
4 5
(b) 14 6
4 5
(c) 14 6
4 5
(d) 14 6
4 5
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25) if A= 3 1 2
2 0 4
, B=
1 2 3 0
2 3 0 1
3 0 1 2
Find AB. Does BA exist?
(a) AB exists but BA not Exists (b) AB not exists BA Exists
(c) Both Ab and BA not Exists (d) None of these
26) if A=
0 3
0 2 2 3 1 2
3 2 1 0 2 1
3 0
;B=
(a) AB ≠ BA (b) AB= BA
(c) AB exists BA not exists (d) AB not exists BA exists
27) If A= 201
0 ; where i
i
i
Find A2, A3 etc.
(a) A =1 0 0
0 1 03 A
i
i
(b) A =1 0 0
0 1 03 A
i
i
(c) A =1 0 0
0 1 03 A
i
i
(d) A =
1 0 0
0 1 03 A
i
i
28) Find the elements C23, C32, C31in the product C=AB.
Where A =
2 3 4 1 3 0
1 2 3 1 2 1
1 1 2 0 0 2
, B =
(a) C23= 8, C32= =-1, C22= 7, C31= 5 and AB = 1 12 11
1 7 8
0 5 5
(b) C23= 8, C32= =5, C22= 7, C31=0 and AB =
1 12 11
1 7 8
0 5 5
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2
2
2
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(c) C23= 8, C32= –1, C22 = 7, C31 = 5 and AB = 1 12 11
1 7 8
2 5 5
(d) C23= 8, C32= –1, C22= 7, C31 = 5 and AB =
1 12 11
1 7 8
0 5 5
29) Using matrix Cramers method
∆x = 1, ∆y = -1 , ∆Z =1 , ∆ = 1 , fi nd x, y and z values
(a) X = 1 , y = –1 and z= -1 (b) X = –1 , y = 1 and z = 1
(c) X = 1 , y = –1 and z= 1 (d) X = –1 , y = –1 and z = 1
30) 4x – 5y – 2z= 0; 2x + 2y + z = 2; 2x + 2y + 8z = –1 then the values of x, y, z using crammers rule
(a) X = 1 , y = –1 and z = 1 (b) inconsistent
(c) X = 1 , y = –1 and z = 1 (d) none of these
31) x + y = –1; y + z = 1; z + x = 0
(a) X = –1; y = 0; z = 1 (b) X = 1; y = 0; z = 1
(c) X = 1; y = 0; z = –1 (d) X = –1; y = 0; z = –1
32) If A = 6 5
3 9
, fi nd (A’)’
(a) A (b) -A
(c) A2 (d) none of these
33) Chose the correct alternative;
if 2 2 3
1 0 -9 =18
x y
z p
(a) X=18 ; z= 9/2 (b) x= 0 , z= -9/2
(c) X= 0 ; z= 9/2 (d) None of these
0 3 -434) -3 0 -5 is a
4 4 8
(a) Symmetric matrix (b) Null matrix
(c) Skew - symmetric matrix (d) None of these
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35 ) if A = 6 10
3 5
(a) Is a singular matrix (b) Non-singular matrix
(c) Identity matrix (d) Symmetric matrix
ANSWERS:Unit -II Exercise (A)
1 (b) 2 (a) 3 (c) 4 (c) 5 (b)
6 (a) 7 (a) 8 (c) 9 (b) 10 (b)
11 (d) 12 (a) 13 (b) 14 (c) 15 (a)
16 (a) 17 (a) 18 (c) 19 (d) 20 (a)
21 (a) 22 (b) 23 (a) 24 (b) 25 (a)
26 (a) 27 (b) 28 (b) 29 (c) 30 (b)
31 (a) 32 (a) 33 (c) 34 (d) 35 (a)
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SUMMARY
In this unit basic applications to matrices and determinates has been studied. Matrix is defi ned. Some special types of matrices are mentioned. Operations of matrices dealt with. Determinants are defi ned and their properties are discussed. The methods Cramer’s rule.
1) General form matrix of order m × n is
1
11 12 13 1
21 22 23 2
2 3
......
......
... ... ... ...
....
.....
m
n
n
m m mn
a a a a
a a a a
a a a a
2) Only square matrices have determinates. A Determinant of n rows and n columns is called determinant of order n . General form of determinant of order n is
1
11 12 13 1
21 22 23 2
2 3
......
......
... ... ... ...
....
.....
n
n
n
n n nn
a a a a
a a a a
a a a a
3) Only matrices of the same order can be added or subtracted. To add (or subtract) two matrices, we add (or subtract) their corresponding elements.
4) To multiply a matrix with a number, we multiply every element of the matrix with that number whereas to multiply a determinant with a number we multiply only one row (or column) of the determinant with that number.
5) Two matrices can be multiplied only if the number of columns of the fi rst is the same as the number of rows of the second, E.g. , a 2 × 3 matrix can be multiplied by a 3 × 4 matrix. The order of resulting matrix will be 3 × 4.
6) Transpose of a matrix A is the matrix obtained by interchanging rows and columns of the matrix A. It is denoted by A’or AT .
7) Adjoint of matrix A is transpose of the co-factor matrix of A, e.g.,
If 11 12 13
21 22 23
31 32 33
a a a
A a a a
a a a
Let C11 be co-factor of a11 and so on
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Then Co-factor matrix of 11 21 31
12 22 32
13 23 33
C C C
A C C C
C C C
8) 1 1 Adjoint of AAA
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