CA Foundation Matrices
CA Foundation Matrices
CA Foundation Matrices
LEARNING OBJECTIVES
At the end of this chapter, you will be able to Understand:
Denition of Matrix
Types of Matrix
Addition and Subtraction of Matrix
Multiplication of Matrices
Transpose of Matrix
Inverse of a Square Matrix
Adjoint of a Square Matrix
System of Linear Equations
Cramer’s Rule (Determinant Method)
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 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 Pen
Ram 5 3 2
Sita 10 8 5
Laxman 15 10 2
We can express the above information in the following form:
§ 5 3 2· First Row
¨ ¸ Second Row
¨10 8 5 ¸
¨ 15 10 2 ¸
© ¹3u3 Third Row
Matrix (Denition)
A rectangular array of numbers (real/complex) denoted by:
§ a11 ! a1n ·
¨ ¸
A= ¨ # % # ¸
¨a ¸
© m1 " a mn ¹
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.
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).
§1 2 2·
Consider the matrix A= ¨ 4 6 5 ¸
¨ ¸
¨7 9 8¸
© ¹
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.
§0·
§0 0· ¨ ¸
(m, n) and it is denoted by : Omn. Thus [0, 0]; ¨ ¸ ; ¨ 0 ¸ are all zero matrices, but all of matrices,
but of different orders. ©0 0¹ ¨0¸
© ¹
Square Matrix and Rectangular Matrix: If the number of rows and columns in a matrix are same, such
§ 2 0 0·
§1 0· ¨ ¸
a matrix is called a square matrix; otherwise it is called a rectangular matrix. ¨ ¸ ; ¨ 0 -3 0 ¸ are
© 0 3¹ ¨ 0 0 5¸
© ¹
examples of square matrix are zero except on the leading diagonal, then it is called diagonal matrix.
ª a11 0 0 0 º
«0 a 22 0 0 »»
Diagonal Matrix « Diag > a11 , a 22 ,....a nn @
« . . . . »
« »
¬0 .. .. a nn ¼
Scalar Matrix:
A diagonal matrix whose leading diagonal elements are all equal is called a scalar matrix, for example,
§k 0 0·
¨ ¸ §1 0·
¨ 0 k 0 ¸ ; > 2 @ ; ¨ ¸
¨0 0 k¸ ©0 1¹
© ¹
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,
ª1 0 0 0 º
«0 1 0 0 »
A « » I nun
«. . . .»
« »
¬0 .. .. 1 ¼
A matrix is known as lower triangular matrix if all the elements above the leading diagonal are zero.
For example.
ª1 0 0ª
« 2 3 0« ; ª1 0º
« « « »
«¬ 6 8 2«¬ ¬ 7 5¼
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.
ª 1 3 7º
Let A = « -2 0 3 »
« »
«¬ 6 8 2 »¼
The sub matrix is obtained by deleting second row and the second column from matrix A.
ª1 7º
B= «
¬6 2 »¼
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 3 5 · § 5 3 0 · §7 0 5 ·
Example 1: ¨ ¸¨ ¸ ¨ ¸
© 4 6 0 ¹ ©1 4 2 ¹ © 5 10 2 ¹
§ 2 3 5 · § 5 3 0 · § 3 6 5 ·
¨ ¸ ¨ ¸ ¨ ¸
© 4 6 0 ¹ ©1 4 2 ¹ © 3 2 2 ¹
Negative of a Matrix: If A is any matrix, the negative of A is denoted by
ª-2 1 º
-A= «-5 -4 »
« »
«¬6 0 »¼
Scalar Multiplication
The multiplication of a matrix by scalar k implies the multiplication of every element.
Example 2:
§ 2 3·
Let A = ¨ ¸ k=3
© 4 5¹
§6 9·
Solution: Then k A = ¨ ¸
©12 15 ¹
Multiplication of two matrices.
The product A B of two matrices A and B dened 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 dened.
(iii) Multiplication distributes over addition of Matrices i.e.,
(a) A (B + C) = AB + AC
©
(b)The(AInstitute
+ B) C of
= Chartered
AC + BC Accountants of India
2.40 BUSINESS MATHEMATICS
(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 × null matrix, then AO = O= O - .
(vi) If A is a square matrix and I is a /
matrix of the same order, then AI = IA = -
(vii) Product of the two no-zero matrices is non zero matrix. /
§5 3·
§1 5 3 · ¨ ¸
Example: 3 if A= ¨ ¸ B= ¨ 1 0 ¸ , nd AB
© 4 5 6 ¹ ¨2 8¸
© ¹
§ 1×5+5×1+3×2 1×3+5×0+3×8 ·
Solution: AB= ¨ ¸
© 4×5+5×1+6×2 4×3+5×0+6×8 ¹
§ 5+5+6 3+24 · § 16 27 ·
¨ ¸ =¨ ¸
© 20+5+12 12+48 ¹ © 37 60 ¹
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 Z
I 6,000 9,000 13,000
II 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,
Rs. X Y Z
ª3.50 º X
ª6000 9000 13000 º I
P = ««2.75 »» Y and S = «12000 6000 17000 » II
«¬1.50 »¼ Z ¬ ¼
ª3.50 º
ª6000 9000 13000 º «
Total revenue in markets I and II « » « 2.75»»
¬12000 6000 17000 ¼ «1.50 »
¬ ¼
ª Rs.(21,000+24750+19,500) º
«Rs.(42,000+16,500+25,500)»
¬ ¼
ªRs.65,250 º I
«Rs.84,000 » II
¬ ¼
Transpose of Matrix: The matrix is obtained by interchanging rows and columns of a matrix A is
T
called its transpose. Transpose of a matrix by or A’.
Symbolically, if A=[aij] and A’=[bij]
Then aij=bij
Example:
ª1 3 7º ª1 -2 6º
«
Let A= -2 0 3 » then A = «« 3
» '
0 8 »»
«
«¬ 6 8 2 »¼ «¬ 7 3 2 »¼
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.
ª1 3 7º ª1 3 7º
« » ' « 8 »»
Example 5: Let A= 3
« 0 8» A = «3 0
«¬ 7 8 2 »¼ «¬ 7 8 2 »¼
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
ª1 0 0 º
Example 6: Find A , if A « 0 1 0 »
2
« »
«¬ a b 1»¼
ª1 0 0 º ª1 0 0 º
2 « » «0 1 0 »
Solution: A = A.A= « 0 1 0 » « »
«¬ a b 1»¼ «¬ a b 1»¼
ª1u1 0 u 0 0 u a 1u 0 0 u 1 0 u b 1u 0 0 u 0 0 u 1º
«
= 0 u 1 1u 0 0 u a
»
« 0 u 0 1u 1 0 u b 0 u 0 1u 0 0 u 1 »
«¬ a u1 b u 0 1u a a u 0 b u1 1 u b a u 0 b u 0 1 u 1»¼
ª1 0 0 º
«
= 0 1 0 =
»
« »
«¬0 0 1 »¼
ª1 0 1º
«3 4 5 »» satises the equation:
Example 7: (a) Show that the matrix A «
«¬0 6 7»¼
A 3 2A 2 A 20 , 0 .
Solution:
Now
ªa b º ª 1 0º 2
(b) If A = « » and I = «0 1» , then show that A – (a+d) A = (bc–ad) I.
¬c d¼ ¬ ¼
b) Solution: We may write,
= (bcad) I = R.H.S.
© The
Hence, A2Institute
– (a + d)of A
Chartered Accountants
= (bc–ad) I. of India
MATRICES 2.43
§ 1 1· §a 1 ·
Example 8: If A= ¨ ¸ ; B= ¨ ¸ and (A+B) =A + B , nd the value of a and b
2 2 2
© 2 1¹ © b 1 ¹
Solution: (A+B) =
§1 a 0 · §1 a 0 ·
(A+B)2 = ¨ ¸¨ ¸
© 2 b 2 ¹ © 2 b 2 ¹
§ 1 a 2 0 u 2 b (1 a ) u 0 2 u 0 ·
=¨ ¸
¨ 2 b 1 a 2(2 b) 2 b u 0 4 ¸¹
©
§ 1 a 2 0·
=¨
¨ 2a ab b 2 4 ¸¸
© ¹
§ 1 1· § 1 1· § 1 2 1 1 · § 1 0 ·
A2 = ¨ ¸¨ ¸=¨ ¸ ¨ ¸
© 2 1¹ © 2 1¹ © 2 2 2 1¹ © 0 1¹
§a 1 · § a 1 · § a 2 b a 1·
B2 = ¨ ¸¨ ¸ =¨ ¸
© b 1¹ © b 1¹ © ab b b 1 ¹
§ 1 0 · § a 2 b a 1· § a 2 b 1 a 1·
A +B = ¨ ¸ +¨ ¸ =¨ ¸
2 2
© 0 1¹ © ab b b 1 ¹ © ab b b ¹
Now (A + B)2 = A2 + B2
§ 1 a 2 0 · § a b 1 a 1·
2
¨¨ ¸¸ ¨
= ¸
© 2a ab b 2 4 ¹ © ab b b ¹
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:
§ 5 15 600 ·
¨ ¸
© u ¹
§ 1100 ·
¨ ¸
§ 3 1 1 0 · ¨ 1700 ¸
AB = ¨ ¸
© 5 3 1 1 ¹ ¨ 3000 ¸
¨ ¸
© 2500 ¹
§ 3 u1100 1u1700 1u 3000 0 u 2500 ·
=¨ ¸
© 5 u1100 3 u1700 1u 3000 1u 2500 ¹
§ 8000 ·
=¨ ¸
©16100 ¹
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.
M W C
Aª 4 6 2 º
F =
B «¬ 2 2 4 »¼
And the recommended daily requirement of calories and proteins for each member can be represented
by the 3 × 2 matrix:
Calories Proteins
M ª2400 55 º
5 = W «« 1900 45»»
C «¬1800 33 »¼
The total requirements of calories and proteins for each of the families is given by matrix multiplication.
ª 2400 55 º
ª4 6 2º « Aª 24600 556 º
F.R = « » 1900 45»»
«
¬ 2 2 4 ¼ «1800 33» B «¬15800 332 »¼
¬ ¼
Hence nally A requires 24,600 calories and 556 gm proteins and Family B requires 15,800 calories
and 332 gm proteins.
Example 12: Three 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, nd total amount paid by the contractor to each of these rms, buy using matrix method.
Solution: The amount of stone and sand supplied to a contractor by three rms A, B and C can
represented by using matrix method.
© The Institute of Chartered Accountants of India
2.46 BUSINESS MATHEMATICS
Stone Sand
Aª 40 10 º
X = B «« 35 5 »»
C «¬ 25 8 »¼
The cost per truck load of stone and stand can be represented by the column matrix.
Stone ª1200 º
Y=
Sand «¬ 500 »¼
Thus, the total amount paid by the contractor to each of these rms is given by the matrix product.
Stone ª1200 º
Y=
Sand «¬ 500 »¼
ª 40 10 º ª53,000 º A
« » ª1200 º «
XY= «35 5 » « » = « 44,500 »» B
500 ¼
«¬ 25 8 »¼ ¬ «¬34,000 »¼ C
2.2.4 DETERMINANTS
The 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.
§a b·
The determinant of a (2 × 2) matrix A = ¨ ¸ = ad bc
©c d ¹
§ a1 a 2 a3 ·
The determinant of (3 × 3) matrix, A = ¨ b1 b2 b3 ¸¸
¨
¨c c2 c3 ¸¹
© 1
a1 a 2 a 3
A b1 b2 b3
c1 c2 c3
b2 b3 b1 b3 b1 b2
And its dened as a1 - a2 + a3
c2 c3 c1 c3 c1 c2
= a1 (b2c3-b3c2) – a2(b1c3-b3c1) + a3(b1c2-b2c1)
© The Institute of Chartered Accountants of India
MATRICES 2.47
Singular and Non-Singular Matrices: Any Square Matrix A is singular, if | A | = 0 . The matrix is
non-singular, if | A | 0 .
§ 1 2 ·
Example 13: If A = ¨ ¸ Prove that A is a singular matrix.
© 3 6 ¹
Solution : | A | =
1 2 = -1 × 6 - 2 × -3 = 0
3 6
A is singular matrix.
Example 14: Find the determinant value of the following matrices.
§ 1 2 3·
¨
A= 4 5 6
¸
¨ ¸
¨7 8 9¸
© ¹
Solution by denition
5 6 4 6 4 5
=1 –2 +3
8 9 7 9 7 8
ª 1 4 0º
A = «- 1 2 2 »
« »
«¬ 0 0 2 »¼
Solution:
§ 14 0 ·
¨ ¸
The Co-factors of elements of A = ¨ - 1 2 2 ¸ are calculated below:
¨ 0 02 ¸
2 2 © ¹
A11= (-1) 1+1 =4
0 2
1 2
A12= (-1) 1+2 =2
0 2
1 2
A13= (-1)1+3 =0
0 0
© The Institute of Chartered Accountants of India
MATRICES 2.49
4 0
A21= (-1)2+1 = -8
0 2
10
A22= (-1)2+2 =2
02
14
A23= (-1)2+3 =0
00
4 0
A31= (-1) 3+1 =8
2 2
10
A32= (-1)3+2 = -2
-1 2
14
A33= (-1) 3+3 =6
-1 2
§ 4 -8 8 ·
¨ ¸
Now Adj. A = ¨ 2 2 - 2 ¸
¨0 0 6 ¸
© ¹
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.
§2 8 5· § X· §5 ·
¨ ¸ ¨ ¸ ¨ ¸
Where (A) = ¨1 1 1¸' X ¨ Y ¸ and B ¨ - 2 ¸
¨1 2 1 ¸ ¨Z ¸ ¨2 ¸
© ¹ © ¹ © ¹
285
det (A) = 1 1 1 = 2 (-1–2) – 8 ( -1 -1) + 5(2–1)
1 2 -1
= -6 + 16 + 5
= 15 z 0
Hence, the system has a unique solution as A is non-singular. The solution is given by
X = A-1B
To nd A-1, we nd the cofactors.
A11 = -3; A12 = + 2; A13 = 1, A21 = + 18; A22 =-7; A23
= 4; A31 = 3; A22 = +3; A33 = -6
§3 2 1 ·
¨ ¸
Co-factor of A = ¨18 - 7 4 ¸
¨ 3 3 - 6¸
© ¹
§ 3 18 3 ·
¨ ¸
Adj: A = (co-factor A)T = ¨ 2 -7 3¸
¨ 1 4 - 6¸
© ¹
Solution:
Since A2 – 5A – 2I = 0,
A-1 (A2 – 5A – 2I) = 0
A-5I – 2A-1 = 0
2A-1 = A – 5I
1 ª1 2º 5 ª1 0º
= «3 4 » – « »
2 ¬ ¼ 2 ¬0 1¼
ª1 5 º
« 2 - 2 1- 0 »
= « »
«3 - 0 2 - 5 »
¬« 2 2 ¼»
ª 2 1 º
= « 3 1»
« - »
¬ 2 2¼
ª 2 1º
?A = « 3-1
1»
« - »
¬2 2¼
Example 18:
Fine the inverse of the matrix.
§2 3 ·
¨¨ ¸¸
© 4 -1 ¹
Solution: Hence, solve the system of equations.
2x – 3y = 3
4x – 11y = 11
© The Institute of Chartered Accountants of India
2.52 BUSINESS MATHEMATICS
Hence, Inverse of
? The given equations are consistent, and the solution is given by X =A-1 B.
Let us nd the cofactors:
A11 = 6; A12= -6; A13= 2; A21 = -5; A22 = 8; A23 = -3
A31 = 1; ©AThe
32
= -Institute
2 and Aof33Chartered
=1 Accountants of India
MATRICES 2.53
§ 6 -6 2 · § 6 -5 1·
¨ ¸ ¨ ¸
Then Cof. A = 1 ¨ - 5 8 - 3 ¸; adj. A ¨- 6 8 - 2¸
¨ 1 -2 1 ¸ ¨ 2 -3 1¸
© ¹ © ¹
§ 6 -5 1 ·
1 1¨ ¸
Now, A-1 = adj , A ¨- 6 8 - 2 ¸
A 2¨ ¸
© 2 -3 1 ¹
§ 6 5 1 · §3 · § 4· § 2·
1¨ ¸¨ ¸ 1¨ ¸ ¨ ¸
?X ¨- 6 8 - 2¸ ¨ 4¸ ¨ 2¸ ¨1 ¸,
2¨ ¸¨ ¸ 2¨ ¸ ¨0¸
© 2 - 3 1 ¹ ©6¹ ©0¹ © ¹
Hence, x = –2; y = 1 and z = 0
Notes : 1) If = 0; and x = 0, y = 0 , z= 0; then the given equations will have innite 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
0 + 3y + 2z = 12
3x + 2y + 3z = 16
By using Cramer’s Rule, the solution of the equations are given below:
4 1 1
12 3 2
'x 16 2 3 4(9 4) 1(36 32) 1(24 48)
0= =
' 2 1 1 2(9 4) 1(3 6) 1(2 9)
1 3 2
3 2 3
4 u 5 1u 4 (24) 0
2u5 3 7 0
2 4 1
1 12 2
'y 3 16 3 2(36 32) 4(3 6) 1(16 36)
=
' 0 0
48 20 28 0
0 0
Since = 0; x= 0, y= 0 and z= 0,
There the equations are dependent and will have innite solutions;
Example 21:
x + y + 3z = 6
x – 3y – 3z = –4
5x – 3y + 3z = 8
© The Institute of Chartered Accountants of India
MATRICES 2.55
6 1 3
4 3 3
'x 8 3 3 6(9 9) 1(12 24) 3(12 24)
x= =
' 1 1 3 1(9 9) 1(3 15) 3(3 15)
1 3 3
5 3 3
108 12 u 4 108 12
0 0
'x 38 19
Hence; x = =
' 36 18
'y 96 8
y= = =
' 36 3
'z 130 65
z= =
' 36 18
is a singular matrix?
(a) 5 (b) 6
(c) 7 (d) 8
§ 2i 3i · 2
10) if A= ¨ ¸ (i =-1) then |A| = ?
© 2i -i ¹
(a) 2 (c) 8
(c) 4 (d) 5
11) If
§ 2 3 · §1 5 · §2 5·
Using 12-16 Let A ¨ ¸B ¨ ¸C ¨ ¸
©4 5 ¹ © 6 7 ¹ © 3 4¹
12) Find A + B.
ª3 2º ª 3 2º
(a) « » (b) « »
¬10 2 ¼ ¬ 10 2 ¼
ª 2 3º ª 3 1º
(c) « » (d) « »
¬ 10 2 ¼ ¬ 10 2 ¼
ª1 2º ª 1 8º
(a) « » (b) « »
¬ 2 2 ¼ ¬ 2 12 ¼
ª1 8 º ª 1 8º
(c) « » (d) « »
¬ 2 12 ¼ ¬ 12 2 ¼
14) 3A – C
§ 4 14 · § 4 14 ·
(a) ¨ ¸ (b) ¨ ¸
© 9 11 ¹ © 9 11 ¹
§ 4 14 · § 2 3 ·
(c) ¨ ¸ (d) ¨ ¸
© 9 11 ¹ ©4 5 ¹
15) AB
§ 16 31 · § 16 31 ·
(a) ¨ ¸ (b) ¨ ¸
© 34 15 ¹ © 34 15 ¹
§ 16 31· § 2 3 ·
(c) ¨ ¸ (d) ¨ ¸
© 34 5 ¹ ©4 5 ¹
16) BA
§ 22 22 · § 22 22 ·
(a) ¨ ¸ (b) ¨ ¸
© 16 53 ¹ © 16 53 ¹
§ 22 11· § 22 33 ·
(c) ¨ ¸ (d) ¨ ¸
© 16 53 ¹ © 16 53 ¹
§ a -b · § a b ·
17) ¨ ¸ [ ¨ ¸
© b a ¹ © -b a ¹
ªa 2 b2 0 º ª a 2 b 2 0 º
(a) « » (b) « »
¬ 0 a 2 b2 ¼ ¬ 0 a 2 b2 ¼
ªa 2 b2 0 º ªa 2 b2 0 º
(c) « » (d) « »
¬ 0 a 2 b2 ¼ ¬ 0 a b2 ¼
2
§ a 2 b2 b2 c2 · § 2ab 2bc ·
18) ¨ 2 ¸ ¨ ¸
©a c a 2 b2 ¹ © 2ac 2ab ¹
2
ª a 2 b 2 2ab b 2 c 2 2bc º ª a b 2 b c º
2
(a) « 2 » or « »
«¬ a c »¼
2 2 2 2 2
¬ a c 2 ac a b 2 ab ¼ a b
ª a 2 b 2 2ab b 2 c 2 2bc º ª a b 2 b c º
2
(b) « 2 » or « »
¬ a c 2ac a b 2ab ¼ «¬ a c a b »¼
2 2 2 2 2
ª a 2 b 2 2ab b 2 c 2 2bc º ª a b 2 b c º
2
(c) « 2 » or « »
¬ a c 2ac a b 2ab ¼ «¬ a c a b »¼
2 2 2 2 2
ª a 2 b 2 2ab b 2 c 2 2bc º ª a b 2 b c º
2
(d) « 2 » or « »
¬« a c ¼»
¬ a c 2
2 ac a 2
b 2
2 ab ¼
2
a b
2
§ l m · § -p q ·
19) ¨ ¸ + ¨ ¸
©n o ¹ © r s¹
ªl p m q º ªl p m q º
(a) «
s »¼
(b) «
¬n r ¬n r s »¼
ªl p m q º ªl p m q º
(c) «
s »¼
(d) « »
¬n r ¬n r o s ¼
§ a b· §a b ·
20) ¨ ¸ u ¨ ¸
© b a ¹ © b a ¹
ªa 2 b2 0 º ªa 2 b2 0 º
(a) « » (b) « »
¬ 0 a b 2 ¼
2
¬ 0 a b 2 ¼
2
ª a 2 b 2 0 º ªa 2 b2 0 º
(c) « » (d) « »
¬ 0 a b2 ¼
2
¬ 0 a 2 b2 ¼
§1·
21) ¨ 2 ¸ u 3 4 5 6
¨ ¸
¨5¸
© ¹
ª3 4 5 6º ª3 5 4 6º
« »
(a) « 6 8 10 12 » (b) « 6 8 10 12 »
« »
«¬15 20 25 30 »¼ «¬12 16 20 24 »¼
ª3 4 5 6º ª3 4 5 6º
(c)
« 6 8 10 12 » «
(d) 6 8 10 12
»
« » « »
«¬12 16 20 24 »¼ «¬ 24 16 16 12 »¼
§ x y · §1 2 3 ·
22. ¨ ¸u ¨ ¸
© 2 3 ¹ © x y z¹
ª x 2 xy 3 x y2 3 xyzº ª x xy 2 x y2 3 x yz º
(a) « » (b) « »
¬ 2 3x 4 3 y 6 3z ¼ ¬ 2 3x 4 3 y 6 3z ¼
ª x 2 xy 2 xy y2 12 yzº ª x xy 2 x y2 3 x yzº
(c) « » (d) « »
¬ 2 3x 4 3 y 6 3z ¼ ¬ 2 3x 4 3 y 6 3z ¼
§ 1 2 3 · § 1 3 5 ·
¨ ¸ ¨ ¸
23. ¨ 4 5 6 ¸ u ¨ 0 2 4 ¸
¨7 8 9¸ ¨3 0 5¸
© ¹ © ¹
ª10 1 12 º ª10 1 28 º
« » « 22 2 70 »
(a) « 22 22 70 » (b) « »
«¬34 37 112 »¼ «¬34 5 112 »¼
ª10 1 28 º ª10 1 28 º
«
(c) 22 2 70
» (d)
« 22 2 70 »
« » « »
«¬34 5 112 »¼ «¬34 5 112 »¼
§ 2 3 ·
§ 3 1 3 · ¨ ¸
24. ¨ ¸u¨1 0 ¸
© 1 0 2 ¹ ¨ ¸
©3 1 ¹
ª14 6 º ª2 12 º
(a) « » (b) «4 5 »¼
¬ 4 5¼ ¬
ª14 6 º ª 14 6 º
(c) « » (d) « 4 5 »¼
¬ 4 5 ¼ ¬
§1 2 3 0·
ª3 1 2 º ¨ ¸
25) if A= « » , B= ¨2 3 0 1¸
¬2 0 4¼ ¨3 0 1 2¸
¨ ¸
© ¹
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
§0 3·
¨ ¸
§ 0 2 2 3· ¨1 2¸
26) if A= ¨ ¸ ;B=
©3 2 1 0 ¹ ¨2 1¸
¨ ¸
©3 0¹
§0 i ·
27) If A= ¨ ¸ ; where i
2
1
© i 0¹
ª 1 0 º 3 ª 0 i º ª 1 0 º 3 ª 0 i º
(a) A2 = « » A « i 0 » (b) A2 = « » A « i 0 »
¬ 0 1¼ ¬ ¼ ¬ 0 1¼ ¬ ¼
ª1 0 º 3 ª0 iº ª1 0 º 3 ª 0 i º
(c) A2 = « » A « i 0 » (d) A2 = « » A « i 0 »
¬0 1¼ ¬ ¼ ¬ 0 1 ¼ ¬ ¼
28) Find the elements C23, C32, C31in the product C=AB.
§ 2 3 4· § 1 3 0·
¨ ¸
Where A = 1 2 3 , B = 1 2 1
¨ ¸
¨ ¸ ¨ ¸
¨1 1 2¸ ¨ 0 0 2¸
© ¹ © ¹ ª 1 12 11º
(a) C23= 8, C32= =-1, C22= 7, C31= 5 and AB = « 1 7 8 »
« »
«¬ 0 5 5 »¼
ª 1 12 11º
« »
(b) C23= 8, C32= 5, C22= 7, C31=0 and AB = « 1 7 8 »
«¬ 0 5 5 »¼
© The Institute of Chartered Accountants of India
2.62 BUSINESS MATHEMATICS
ª 1 12 11º
(c) C23= 8, C32= –1, C22 = 7, C31 = 5 and AB = « 1 7 8 »
« »
«¬ 2 5 5 »¼
ª 1 12 11 º
« 8 »»
(d) C23= 8, C32= –1, C22= 7, C31 = 5 and AB = « 1 7
«¬ 0 5 5»¼
29) Using matrix Cramers method
x = 1, y = -1 , Z =1 , = 1 , nd x, y and z values
(a) 0 = 1 , y = –1 and z= -1 (b) 0 = –1 , y = 1 and z = 1
(c)0 = 1 , y = –1 and z= 1 d) 0 = –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) 0 = 1 , y = –1 and z = 1 (b) inconsistent
(c) 0 = 1 , y = –1 and z = 1 (d) none of these
31) x + y = –1; y + z = 1; z + x = 0
(a)0 = –1; y = 0; z = 1 (b) 0 = 1; y = 0; z = 1
(c)0 = 1; y = 0; z = –1 (d) 0 = –1; y = 0; z = –1
ª6 5º
32) If A = « » , nd (A’)’
¬3 9¼
(a) A (b) -A
(c) A2 (d) none of these
ª 0 3 -4 º
34) ««-3 0 -5 »» is a
«¬ 4 4 8 »¼
ª6 10 º
35 ) if A = « »
¬3 5 ¼
(a) Is a singular matrix (b) Non-singular matrix
(c) Identity matrix (d) Symmetric matrix
ANSWERS:
Unit -II Exercise (A)
SUMMARY
In this unit basic applications to matrices and determinates has been studied. Matrix is
dened. Some special types of matrices are mentioned. Operations of matrices dealt with.
Determinants are dened and their properties are discussed. The methods Cramer’s rule.
1) General form matrix of order m × n is
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 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.,
1
8) A1 Adjoint of A
A