RD SHARMA Solutions for Class 12-science Maths Chapter 5 - Algebra of Matrices

Your CBSE Class 12 syllabus for Maths consists of topics such as linear programming, vector quantities, determinants, etc. which lay the foundation for further education in science, engineering, management, etc. Studying differential equations will be useful for exploring subjects such as Physics, Biology, Chemistry, etc. where your knowledge can be applied for scientific investigations.

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Chapter 5 - Algebra of Matrices Exercise Ex. 5.1

Question 1

If a matrix has 8 elements, what are the possible orders it can have? What if it has 5 elements?

Solution 1

We know that if a matrix is of the order m cross times n, it has mn elements. Thus, to find all the possible orders of a matrix having 8 elements, we have to find all the ordered pairs of natural numbers whose products is 8.

The ordered pairs are: open parentheses 1 cross times 8 close parentheses comma space open parentheses 8 cross times 1 close parentheses comma space open parentheses 2 cross times 4 close parentheses comma space open parentheses 4 cross times 2 close parentheses

Error: the service is unavailable. are the ordered pairs of natural numbers whose product is 5.

Hence, the possible orders of a matrix having 5 elements are 1 cross times 5 space a n d space 5 cross times 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Construct a 2 × 2 matrix A = [aij] whose elements aij are given by:

begin mathsize 14px style open parentheses straight i close parentheses open parentheses straight i plus straight j close parentheses squared over 2 space space space space space space space space space space space space space space space space space space space space space space space open parentheses ii close parentheses straight a subscript ij equals open parentheses straight i minus straight j close parentheses squared over 2
open parentheses iii close parentheses straight a subscript ij equals open parentheses straight i minus 2 straight j close parentheses squared over 2 space space space space space space space space space space open parentheses iv close parentheses straight a subscript ij equals open parentheses 2 straight i plus straight j close parentheses squared over 2
open parentheses straight v close parentheses straight a subscript ij equals fraction numerator open vertical bar 2 straight i minus 3 straight j close vertical bar over denominator 2 end fraction space space space space space space space space space space open parentheses vi close parentheses straight a subscript ij equals fraction numerator open vertical bar negative 3 straight i plus straight j close vertical bar over denominator 2 end fraction end style

Solution 5

 
begin mathsize 14px style open parentheses vi close parentheses straight a subscript ij equals fraction numerator open vertical bar negative 3 straight i plus straight j close vertical bar over denominator 2 end fraction
Thus comma space straight a subscript 11 equals fraction numerator open vertical bar negative 3 cross times 1 plus 1 close vertical bar over denominator 2 end fraction equals fraction numerator open vertical bar negative 3 plus 1 close vertical bar over denominator 2 end fraction equals 1
space straight a subscript 12 equals fraction numerator open vertical bar negative 3 cross times 1 plus 2 close vertical bar over denominator 2 end fraction equals fraction numerator open vertical bar negative 3 plus 2 close vertical bar over denominator 2 end fraction equals 1 half
space straight a subscript 21 equals fraction numerator open vertical bar negative 3 cross times 2 plus 1 close vertical bar over denominator 2 end fraction equals fraction numerator open vertical bar negative 6 plus 1 close vertical bar over denominator 2 end fraction equals 5 over 2
straight a subscript 22 equals fraction numerator open vertical bar negative 3 cross times 2 plus 2 close vertical bar over denominator 2 end fraction equals fraction numerator open vertical bar negative 6 plus 2 close vertical bar over denominator 2 end fraction equals 2
Thus comma space straight A equals open square brackets table row 1 cell 1 half end cell row cell 5 over 2 end cell 2 end table close square brackets end style
Question 6

Construct a 2 × 2 matrix A = [aij] whose elements aij are given by:

 

Aij = e2ix sin xj

Solution 6

 

Question 7

Construct a 3 × 4 matrix A = [aij] whose elements aij are given by :

(i) aij = i + j

(ii) aij = i - j

(iii) aij = 2i

(iv) aij = j

begin mathsize 14px style open parentheses straight v close parentheses space aij space equals space 1 half open vertical bar negative 3 straight i plus straight j close vertical bar end style

Solution 7

 
(v)
 
a subscript i j end subscript equals fraction numerator open vertical bar minus 3 i plus j close vertical bar over denominator 2 end fraction
T h u s comma space a subscript 11 equals fraction numerator open vertical bar minus 3 cross times 1 plus 1 close vertical bar over denominator 2 end fraction equals fraction numerator open vertical bar minus 3 plus 1 close vertical bar over denominator 2 end fraction equals 1
space a subscript 12 equals fraction numerator open vertical bar minus 3 cross times 1 plus 2 close vertical bar over denominator 2 end fraction equals fraction numerator open vertical bar minus 3 plus 2 close vertical bar over denominator 2 end fraction equals 1 half
a subscript 13 equals fraction numerator open vertical bar minus 3 cross times 1 plus 3 close vertical bar over denominator 2 end fraction equals fraction numerator open vertical bar 0 close vertical bar over denominator 2 end fraction equals 0
a subscript 14 equals fraction numerator open vertical bar minus 3 cross times 1 plus 4 close vertical bar over denominator 2 end fraction equals 1 half
space a subscript 21 equals fraction numerator open vertical bar minus 3 cross times 2 plus 1 close vertical bar over denominator 2 end fraction equals fraction numerator open vertical bar minus 6 plus 1 close vertical bar over denominator 2 end fraction equals 5 over 2
a subscript 22 equals fraction numerator open vertical bar minus 3 cross times 2 plus 2 close vertical bar over denominator 2 end fraction equals fraction numerator open vertical bar minus 6 plus 2 close vertical bar over denominator 2 end fraction equals 2
a subscript 23 equals fraction numerator open vertical bar minus 3 cross times 2 plus 3 close vertical bar over denominator 2 end fraction equals fraction numerator open vertical bar minus 3 close vertical bar over denominator 2 end fraction equals 3 over 2
a subscript 24 equals fraction numerator open vertical bar minus 3 cross times 2 plus 4 close vertical bar over denominator 2 end fraction equals 1
space a subscript 31 equals fraction numerator open vertical bar minus 3 cross times 3 plus 1 close vertical bar over denominator 2 end fraction equals fraction numerator open vertical bar minus 9 plus 1 close vertical bar over denominator 2 end fraction equals 4
a subscript 32 equals fraction numerator open vertical bar minus 3 cross times 3 plus 2 close vertical bar over denominator 2 end fraction equals fraction numerator open vertical bar minus 9 plus 2 close vertical bar over denominator 2 end fraction equals 7 over 2
a subscript 33 equals fraction numerator open vertical bar minus 3 cross times 3 plus 3 close vertical bar over denominator 2 end fraction equals fraction numerator open vertical bar minus 6 close vertical bar over denominator 2 end fraction equals 3
a subscript 34 equals fraction numerator open vertical bar minus 3 cross times 3 plus 4 close vertical bar over denominator 2 end fraction equals 5 over 2

T h u s comma space A equals open square brackets table row 1 cell 1 half end cell 0 cell 1 half end cell row cell 5 over 2 end cell 2 cell 3 over 2 end cell 1 row 4 cell 7 over 2 end cell 3 cell 5 over 2 end cell end table close square brackets
Question 8

Solution 8

 

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

 

 

Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

The sales figure of two car dealer during january 2013 showed that dealer A sold 5 deluxe, 3 premium and 4 standard cars, while dealer B sold 7 deluxe, 2 premium and 3 standard cars. Total sales over the 2 month period of january - february revealed that dealer A sold 8 deluxe 7 premium and 6 standard cars. in the same 2 month period, dealer b sold 10 deluxe, 5 premium and 7 standard cars. Write 2 × 3 matrices summarizing for january and 2 - month period for each dealer.

Solution 19

Question 20

Solution 20

Question 21

Solution 21

Question 22

Find the values of a and b if A = B, where

 

Solution 22

Chapter 5 - Algebra of Matrices Exercise Ex. 5.2

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

(ii)

W e space n e e d space t o space f i n d space 2 B plus 3 A space a n d space 3 C minus 4 B
T h u s s comma space 2 B plus 3 A space d o e s space n o t space e x i s t space a s space t h e space o r d e r space o f space A space a n d space B space a r e space d i f f e r e n t.
L e t space u s space f i n d space 3 C minus 4 B equals 3 open square brackets table row cell minus 1 end cell 2 3 row 2 1 0 end table close square brackets minus 4 open square brackets table row cell minus 1 end cell 0 2 row 3 4 1 end table close square brackets
equals open square brackets table row cell minus 3 end cell 6 9 row 6 3 0 end table close square brackets minus open square brackets table row cell minus 4 end cell 0 8 row 12 16 4 end table close square brackets
equals open square brackets table row cell minus 3 end cell 6 9 row 6 3 0 end table close square brackets plus open square brackets table row 4 0 cell minus 8 end cell row cell minus 12 end cell cell minus 16 end cell cell minus 4 end cell end table close square brackets
equals open square brackets table row 1 6 1 row cell minus 6 end cell cell minus 13 end cell cell minus 4 end cell end table close square brackets

Question 7

Solution 7

Question 8

begin mathsize 14px style if space straight A space equals space diag space open parentheses table row 2 cell negative 5 end cell 9 end table close parentheses comma space straight B space equals space diag space open parentheses table row 1 1 cell negative 4 end cell end table close parentheses space and space straight c space equals open parentheses table row cell negative 6 end cell 3 4 end table close parentheses space find
open parentheses straight i close parentheses space straight A space minus space 2 straight B space space space open parentheses ii close parentheses space straight B space plus space straight C space minus space 2 straight A space space open parentheses iii close parentheses space 2 straight A space plus space 3 straight B space minus space 5 straight C end style

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

Find x, y satisfying the matrix equations

 

 

Solution 20

Question 21

Solution 21

Question 22

Solution 22

Question 23

(i)

(ii)

I f space A equals open square brackets table row 8 0 row 4 cell minus 2 end cell row 3 6 end table close square brackets space a n d space B equals open square brackets table row 2 cell minus 2 end cell row 4 2 row cell minus 5 end cell 1 end table close square brackets comma space t h e n space f i n d space t h e space m a t r i x space X space o f space o r d e r space 3 cross times 2
s u c h space t h a t space 2 A plus 3 X equals 5 B.

Solution 23

(i)

(ii)

G i v e n comma space A equals open square brackets table row 8 0 row 4 cell minus 2 end cell row 3 6 end table close square brackets space a n d space open square brackets table row 2 cell minus 2 end cell row 4 2 row cell minus 5 end cell 1 end table close square brackets
A l s o space w e space h a v e space 2 A plus 3 X equals 5 B
T h u s comma space w e space h a v e comma space 3 X equals 5 B minus 2 A
rightwards double arrow 3 x equals 5 open square brackets table row 2 cell minus 2 end cell row 4 2 row cell minus 5 end cell 1 end table close square brackets minus 2 open square brackets table row 8 0 row 4 cell minus 2 end cell row 3 6 end table close square brackets
rightwards double arrow 3 x equals open square brackets table row 10 cell minus 10 end cell row 20 10 row cell minus 25 end cell 5 end table close square brackets minus open square brackets table row 16 0 row 8 cell minus 4 end cell row 6 12 end table close square brackets
rightwards double arrow 3 x equals open square brackets table row cell 10 minus 16 end cell cell minus 10 minus 0 end cell row cell 20 minus 8 end cell cell 10 minus open parentheses minus 4 close parentheses end cell row cell minus 25 minus 6 end cell cell 5 minus 12 end cell end table close square brackets
rightwards double arrow 3 x equals open square brackets table row cell minus 6 end cell cell minus 10 end cell row 12 14 row cell minus 31 end cell cell minus 7 end cell end table close square brackets
rightwards double arrow x equals open square brackets table row cell minus 2 end cell cell fraction numerator minus 10 over denominator 3 end fraction end cell row 4 cell 14 over 3 end cell row cell fraction numerator minus 31 over denominator 3 end fraction end cell cell fraction numerator minus 7 over denominator 3 end fraction end cell end table close square brackets

 

Question 24

Solution 24

Question 25

Solution 25

Question 26

If X and Y are 2 × 2 matrices, then solve the following matrix equations for X and Y.

 

Solution 26

Question 27

Solution 27

begin mathsize 12px style Let space straight A space represent space the space post space allocation space matrix space for space straight a space collage comma space So
straight A equals open square brackets table row 15 row 6 row 1 row 1 end table close square brackets space table row Peons row Clerks row Typist row cell Section space officer end cell end table
The space total space number space of space posts space of space each space kind space in space 30 space colleges space in space given space by colon
equals space 30 straight A
equals space 30 open square brackets table row 15 row 6 row 1 row 1 end table close square brackets
30 space straight A equals open square brackets table row 450 row 180 row 30 row 30 end table close square brackets space table row Peons row Clerks row Typist row cell Section space Officers end cell end table end style

Chapter 5 - Algebra of Matrices Exercise Ex. 5.3

Question 1

Solution 1
Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

begin mathsize 14px style If space straight A equals open square brackets table row 1 0 row 0 1 end table close square brackets comma straight B equals open square brackets table row 1 0 row 0 cell negative 1 end cell end table close square brackets space space space and space straight C equals open square brackets table row 0 1 row 1 0 end table close square brackets comma
then space show space that space straight A squared space equals space straight B squared space equals space straight C squared space equals space straight I subscript 2. end style

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

Solution 20

Question 21

Solution 21

Question 22

Solution 22

Question 23

Solution 23

Question 24

Solution 24

Question 25

Solution 25

Question 26

Solution 26

Question 27

Solution 27

Question 28

Solution 28

Question 29

Solution 29

Question 30

Solution 30

Question 31

Solution 31

Question 32

Solution 32

Question 33

Solution 33

Question 34

begin mathsize 14px style If space open square brackets table row 1 1 straight x end table close square brackets open square brackets table row 1 0 2 row 0 2 1 row 2 1 0 end table close square brackets open square brackets table row 1 row 1 row 1 end table close square brackets equals 0 comma find space space space straight x. end style

Solution 34

Question 35

I f space open square brackets table row 2 3 row 5 7 end table close square brackets open square brackets table row 1 cell minus 3 end cell row cell minus 2 end cell 4 end table close square brackets equals open square brackets table row cell minus 4 end cell 6 row cell minus 9 end cell x end table close square brackets space f i n d space x

Solution 35

G i v e n space t h a t space open square brackets table row 2 3 row 5 7 end table close square brackets open square brackets table row 1 cell minus 3 end cell row cell minus 2 end cell 4 end table close square brackets equals open square brackets table row cell minus 4 end cell 6 row cell minus 9 end cell x end table close square brackets
B y space m u l t i p l i c a t i o n space o f space m a t r i c e s comma space w e space h a v e comma
open square brackets table row cell 2 cross times 1 plus 3 cross times open parentheses minus 2 close parentheses end cell cell 2 cross times open parentheses minus 3 close parentheses plus 3 cross times 4 end cell row cell 5 cross times 1 plus 7 cross times open parentheses minus 2 close parentheses end cell cell 5 cross times open parentheses minus 3 close parentheses plus 7 cross times 4 end cell end table close square brackets equals open square brackets table row cell minus 4 end cell 6 row cell minus 9 end cell x end table close square brackets
rightwards double arrow open square brackets table row cell minus 4 end cell 6 row cell minus 9 end cell 13 end table close square brackets equals open square brackets table row cell minus 4 end cell 6 row cell minus 9 end cell x end table close square brackets
rightwards double arrow x equals 13

Question 36

Solution 36

Question 37

Solution 37

Question 38

Solution 38

Question 39

Solution 39

Question 40

Solution 40

Question 41

Solution 41

Question 42

Solution 42

Question 43

Solution 43

Question 44

Solution 44

Question 45

Use this to find A4

Solution 45

S i n c e space A squared minus 5 A plus 7 I equals O comma space w e space h a v e
space A squared equals 5 A minus 7 I
T h e r e f o r e comma space A to the power of 4 equals A squared cross times A squared equals open parentheses 5 A minus 7 I close parentheses open parentheses 5 A minus 7 I close parentheses
rightwards double arrow A to the power of 4 equals 25 A squared minus 35 A I minus 35 I A plus 49 I
rightwards double arrow A to the power of 4 equals 25 A squared minus 70 A plus 49 I
rightwards double arrow A to the power of 4 equals 25 open parentheses 5 A minus 7 I close parentheses minus 70 A plus 49 I
rightwards double arrow A to the power of 4 equals 125 A minus 175 I minus 70 A plus 49 I
rightwards double arrow A to the power of 4 equals 55 A minus 126 I
rightwards double arrow A to the power of 4 equals 55 open square brackets table row 3 1 row cell minus 1 end cell 2 end table close square brackets minus 126 open square brackets table row 1 0 row 0 1 end table close square brackets
rightwards double arrow A to the power of 4 equals open square brackets table row 165 55 row cell minus 55 end cell 110 end table close square brackets minus open square brackets table row 126 0 row 0 126 end table close square brackets
rightwards double arrow A to the power of 4 equals open square brackets table row cell 165 minus 126 end cell cell 55 minus 0 end cell row cell minus 55 minus 0 end cell cell 110 minus 126 end cell end table close square brackets
rightwards double arrow A to the power of 4 equals open square brackets table row 39 55 row cell minus 55 end cell cell minus 16 end cell end table close square brackets

Question 46

Solution 46

Question 47

Solution 47

Question 48

Solution 48

Question 49

Solution 49

Question 50

Solution 50

Question 51

Solution 51

Question 52

Solution 52

Question 53

Solution 53

Question 54

Solution 54

Question 55

Solution 55

Question 56

Solution 56

Question 57

I f space A equals open square brackets table row 1 0 2 row 0 2 1 row 2 0 3 end table close square brackets space t h e n space s h o w space t h a t space A space i s space a space r o o t space o f space t h e space p o l y n o m i a l space
f open parentheses x close parentheses equals x cubed minus 6 x squared plus 7 x plus 2

Solution 57

G i v e n space t h a t comma space A equals open square brackets table row 1 0 2 row 0 2 1 row 2 0 3 end table close square brackets space a n d space f open parentheses x close parentheses equals x cubed minus 6 x squared plus 7 x plus 2
T h e r e f o r e comma space f open parentheses A close parentheses equals A cubed minus 6 A squared plus 7 A plus 2 I subscript 3
F i r s t space f i n d space A squared :
A squared equals A cross times A equals open square brackets table row 1 0 2 row 0 2 1 row 2 0 3 end table close square brackets cross times open square brackets table row 1 0 2 row 0 2 1 row 2 0 3 end table close square brackets equals open square brackets table row 5 0 8 row 2 4 5 row 8 0 13 end table close square brackets
N o w comma space L e t space u s space f i n d space A cubed :
A cubed equals A squared cross times A equals open square brackets table row 5 0 8 row 2 4 5 row 8 0 13 end table close square brackets cross times open square brackets table row 1 0 2 row 0 2 1 row 2 0 3 end table close square brackets equals open square brackets table row 21 0 34 row 12 8 23 row 34 0 55 end table close square brackets
T h u s comma space
f open parentheses A close parentheses equals A cubed minus 6 A squared plus 7 A plus 2 I subscript 3
equals open square brackets table row 21 0 34 row 12 8 23 row 34 0 55 end table close square brackets minus 6 open square brackets table row 5 0 8 row 2 4 5 row 8 0 13 end table close square brackets plus 7 open square brackets table row 1 0 2 row 0 2 1 row 2 0 3 end table close square brackets space plus 2 open square brackets table row 1 0 0 row 0 1 0 row 0 0 1 end table close square brackets
equals open square brackets table row 21 0 34 row 12 8 23 row 34 0 55 end table close square brackets minus open square brackets table row 30 0 48 row 12 24 30 row 48 0 78 end table close square brackets plus open square brackets table row 7 0 14 row 0 14 7 row 14 0 21 end table close square brackets space plus open square brackets table row 2 0 0 row 0 2 0 row 0 0 2 end table close square brackets
equals open square brackets table row cell 21 minus 30 plus 7 plus 2 end cell 0 cell 34 minus 48 plus 14 plus 0 end cell row cell 12 minus 12 plus 0 end cell cell 8 minus 24 plus 14 plus 2 end cell cell 23 minus 30 plus 7 plus 0 end cell row cell 34 minus 48 plus 14 plus 0 end cell 0 cell 55 minus 78 plus 21 plus 2 end cell end table close square brackets
equals open square brackets table row 0 0 0 row 0 0 0 row 0 0 0 end table close square brackets equals O
T h u s comma space A space i s space a space r o o t space o f space t h e space p o l y n o m i a l.

Question 58

Solution 58

Question 59

Solution 59

Question 60

Solution 60

Question 61

Solution 61

Question 62

Solution 62


Question 63

Find the matrix A such that

 

 

Solution 63

Question 64

Find the matrix A such that

 

 

Solution 64

Question 65

Solution 65

Question 66

Solution 66

So,

A16 is null matrix.

Question 67

Then show that (A + B)2 = A2 + B2.

 

Solution 67

 

Question 68

 

 

Solution 68

Question 69

 

 

Solution 69

Question 70

Solution 70

Question 71

Solution 71

Question 72

Solution 72

Question 73

Solution 73

Question 74

Solution 74

Question 75

Solution 75

Question 76

Solution 76

Question 77

Error converting from MathML to accessible text.

Solution 77

Question 78

Solution 78

Question 79

Solution 79

Question 80

Solution 80

Question 81

Solution 81

Question 82

Solution 82

Question 83

Solution 83

Question 84

Solution 84

Question 85

Solution 85

Question 86

Solution 86

Question 87

Let A and B be square matrices of the order 3 × 3.

Is (AB)2 = A2 B2? Give reasons.

Solution 87

 

 

 

 

Question 88

If A and B be square matrices of the same order such that AB = BA, then show that (A + B)2 = A2 + 2AB + B2.

 

Solution 88

 

Question 89

Solution 89

Question 90

Solution 90

Question 91

Solution 91

Question 92

Solution 92

Question 93

Solution 93


Question 94

To promote making of toilets for women, an organization tried to generate awareness through (i) house calls (ii) letters and (iii) announcements. The cost for each mode per attempt is given below:

 i. Rs. 50

 ii. Rs. 20

 iii. Rs. 40

The number of attempts made in three villages X, Y, and Z are given below:

 

 

(i)

(ii)

(iii)

X

400

300

100

Y

300

250

75

z

500

400

150

 

Find the total cost incurred by the organization for three villages separately, using matrices.

Solution 94

Question 95

There are 2 families A and B. There are 4 men, 6 women and 2 children in family A, and 2 men, 2 women and 4 children in family B. The recommend daily amount of calories is 2400 for men, 1900 for women, 1800 for children and 45 grams of proteins for men, 55 grams for women and 33 grams for children. Represent the above information using matrix. Using matrix multiplication, Calculate the total requirement of calories and proteins for each of the families. What awareness can you create among people about the planned diet from this question?

Solution 95

Question 96

In a parliament election, a political party hired a public relations firm to promote its candidates in three ways - telephone, house calls and letters. The cost per contact (in paisa) is given in matrix A as

 

 

The number of contacts of each type made in two cities X and Y is given in the matrix B as

 

Find the total amount spent by the party in the two cities.

What should one consider before casting his/her vote - party's promotional activity or their social activities?

Solution 96

Chapter 5 - Algebra of Matrices Exercise Ex. 5.4

Question 1
Solution 1
Question 2
Solution 2
Question 3
Solution 3
Question 4
Solution 4
Question 5
Solution 5
Question 6
Solution 6
Question 7
Solution 7
Question 8
Solution 8
Question 9
Solution 9
Question 10
Solution 10
Question 11
Solution 11
Question 12
Solution 12
Question 13

I f space A to the power of T equals open square brackets table row 3 4 row cell minus 1 end cell 2 row 0 1 end table close square brackets space a n d space B equals open square brackets table row cell minus 1 end cell 2 1 row 1 2 3 end table close square brackets comma space f i n d space A to the power of T minus B to the power of T.

Solution 13

G i v e n space t h a t space A to the power of T equals open square brackets table row 3 4 row cell minus 1 end cell 2 row 0 1 end table close square brackets space a n d space B equals open square brackets table row cell minus 1 end cell 2 1 row 1 2 3 end table close square brackets.
W e space n e e d space t o space f i n d space A to the power of T minus B to the power of T.
G i v e n space t h a t space comma space B equals open square brackets table row cell minus 1 end cell 2 1 row 1 2 3 end table close square brackets
rightwards double arrow B to the power of T equals open square brackets table row cell minus 1 end cell 2 1 row 1 2 3 end table close square brackets to the power of T equals open square brackets table row cell minus 1 end cell 1 row 2 2 row 1 3 end table close square brackets
L e t space u s space f i n d space A to the power of T minus B to the power of T :
A to the power of T minus B to the power of T equals open square brackets table row 3 4 row cell minus 1 end cell 2 row 0 1 end table close square brackets minus open square brackets table row cell minus 1 end cell 1 row 2 2 row 1 3 end table close square brackets
rightwards double arrow A to the power of T minus B to the power of T equals open square brackets table row cell 3 plus 1 end cell cell 4 minus 1 end cell row cell minus 1 minus 2 end cell cell 2 minus 2 end cell row cell 0 minus 1 end cell cell 1 minus 3 end cell end table close square brackets
rightwards double arrow A to the power of T minus B to the power of T equals open square brackets table row 4 3 row cell minus 3 end cell 0 row cell minus 1 end cell cell minus 2 end cell end table close square brackets

Question 14

begin mathsize 14px style If space space straight A equals open square brackets table row cosα sinα row cell negative sinα end cell cosα end table close square brackets comma space then space verify space that space straight A apostrophe straight A equals straight I end style

Solution 14

Question 15

 

begin mathsize 14px style straight A space equals open square brackets table row sinα cosα row cell negative cosα end cell sinα end table close square brackets comma space then space verify space that space straight A apostrophe straight A space equals space straight I end style

Solution 15

Question 16

If li, mi, ni ; i = 1, 2, 3 denote the direction cosines of three mutually perpendicular vectors in space, prove that AAT = I,

begin mathsize 14px style Where space straight A space equals space open square brackets table row cell straight l subscript 1 end cell cell straight m subscript 1 end cell cell straight n subscript 1 end cell row cell straight l subscript 2 end cell cell straight m subscript 2 end cell cell straight n subscript 2 end cell row cell straight l subscript 3 end cell cell straight m subscript 3 end cell cell straight n subscript 3 end cell end table close square brackets. end style

Solution 16

Chapter 5 - Algebra of Matrices Exercise Ex. 5.5

Question 1
Solution 1
Question 2
Solution 2
Question 3
Solution 3
Question 4
Solution 4
Question 5
Solution 5
Question 6
Solution 6
Question 7
Solution 7
Question 8
Solution 8

Chapter 5 - Algebra of Matrices Exercise Ex. 5VSAQ

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

T h u s comma space w e space c a n space s a y space t h a t comma space A squared equals minus open square brackets table row cell minus 1 end cell 0 0 row 0 cell minus 1 end cell 0 row 0 0 cell minus 1 end cell end table close square brackets equals minus A equals I subscript 3

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

Solution 20

Question 21

Solution 21

Question 22

Solution 22

Question 23

Solution 23

Question 24

Solution 24

Question 25

Solution 25

Question 26

Solution 26

Question 27

Solution 27

Question 28

Solution 28

Question 29

Solution 29

Question 30

Solution 30

Question 31

Solution 31

Question 32

Solution 32

Question 33

Solution 33

Question 34

Solution 34

Question 35

Solution 35

Question 36

Solution 36

Question 37

Solution 37

Question 38

Solution 38

Question 39

Solution 39

Question 40

Solution 40

Question 41

Solution 41

Question 42

Solution 42

Question 43

Solution 43

Question 44

Solution 44

Question 45

Solution 45


Question 46

Solution 46

Question 47

Solution 47

Question 48

I f space open square brackets table row x cell x minus y end cell row cell 2 x plus y end cell 7 end table close square brackets equals open square brackets table row 3 1 row 8 7 end table close square brackets comma space t h e n space f i n d space t h e space v a l u e space o f space y.

Solution 48

G i v e n space t h a t space open square brackets table row x cell x minus y end cell row cell 2 x plus y end cell 7 end table close square brackets equals open square brackets table row 3 1 row 8 7 end table close square brackets
W e space n e e d space t o space f i n d space t h e space v a l u e space o f space y.
c o n s i d e r space t h e space g i v e n space e q u a t i o n comma
open square brackets table row x cell x minus y end cell row cell 2 x plus y end cell 7 end table close square brackets equals open square brackets table row 3 1 row 8 7 end table close square brackets
rightwards double arrow x equals 3 space a n d space x minus y equals 1
rightwards double arrow 3 minus y equals 1
rightwards double arrow y equals 2

Question 49

If a matrix has 5 elements, write all possible orders it can have.

Solution 49

If a matrix is of order m cross times n, then the number of elements in the matrix is the product m n.

Given that the required matrix is having 5 elements and 5 is a prime number.

Hence the prime factorization of 5 is either 5 cross times 1 space o r space 1 cross times 5.

Thus, the order of the matrix is either 5 cross times 1 space o r space 1 cross times 5.

Question 50

F o r space a space 2 cross times 2 space m a t r i x space A equals open square brackets a subscript i j end subscript close square brackets space w h o s e space e l e m e n t s space a r e space g i v e n space b y space a subscript i j end subscript equals i over j comma space w r i t e space t h e space
v a l u e space o f space a subscript 12.

Solution 50

G i v e n space t h a t space space a space 2 cross times 2 space m a t r i x space A equals open square brackets a subscript i j end subscript close square brackets space w h o s e space e l e m e n t s space a r e space g i v e n space b y space a subscript i j end subscript equals i over j.
W e space n e e d space t o space f i n d space t h e space v a l u e space o f space a subscript 12.
T h u s comma space a subscript 12 equals 1 half.

Question 51

I f space x open square brackets table row 2 row 3 end table close square brackets plus y open square brackets table row cell minus 1 end cell row 1 end table close square brackets equals open square brackets table row 10 row 5 end table close square brackets comma space f i n d space t h e space v a l u e space o f space x.

Solution 51

G i v e n space t h a t space x open square brackets table row 2 row 3 end table close square brackets plus y open square brackets table row cell minus 1 end cell row 1 end table close square brackets equals open square brackets table row 10 row 5 end table close square brackets.... left parenthesis 1 right parenthesis
W e space n e e d space t o space f i n d space t h e space v a l u e space o f space x.
R e w r i t i n g space e q u a t i o n space left parenthesis 1 right parenthesis comma space w e space h a v e comma
open square brackets table row cell 2 x end cell row cell 3 x end cell end table close square brackets plus open square brackets table row cell minus y end cell row y end table close square brackets equals open square brackets table row 10 row 5 end table close square brackets
rightwards double arrow open square brackets table row cell 2 x minus y end cell row cell 3 x plus y end cell end table close square brackets equals open square brackets table row 10 row 5 end table close square brackets
T h u s comma space t h e space c o r r e s p o n d i n g space e l e m e n t s space a r e space e q u a l.
rightwards double arrow 2 x minus y equals 10... left parenthesis 2 right parenthesis
a n d
3 x plus y equals 5... left parenthesis 3 right parenthesis
A d d i n g space e q u a t i o n s space left parenthesis 2 right parenthesis space a n d space left parenthesis 3 right parenthesis comma space w e space h a v e
5 x equals 15
rightwards double arrow x equals 15 over 5 equals 3
S u b s t i t u t i n g space t h e space v a l u e space o f space x space i n space e q u a t i o n space left parenthesis 2 right parenthesis comma space w e space h a v e
2 open parentheses 3 close parentheses minus y equals 10
rightwards double arrow 6 minus y equals 10
rightwards double arrow 6 minus 10 equals y
rightwards double arrow y equals minus 4

Question 52

Solution 52

Question 53

Solution 53

Question 54

Solution 54

Question 55

Solution 55

Question 56

If A is a square matrix such that A2=A, then write the value of 7 A-(I+A)3, where I is the identity matrix.

Solution 56

A2 = A

A3 = A2 = A

7A - (I + A)3

= 7A - (I3 + A3 + 3A2I + 3AI2)

= 7A - (I + A + 3A + 3A)

= 7A - (I + 7A)

= -I

Question 57

Solution 57

Question 58

Solution 58

Question 59

Solution 59

Question 60

Write 2×2 matrix which is both symmetric and skew-symmetric.

Solution 60

Question 61

Solution 61

Question 62

Construct a 2 × 2 matrix A = [aij] whose elements aij are given by

 

 

Solution 62

Question 63

Solution 63

Chapter 5 - Algebra of Matrices Exercise MCQ

Question 1

 

a. a null matrix

b. a unit matrix

c. -A

d. A

Solution 1

Question 2

 

a. 

b. 

c. 

d. 

Solution 2

Question 3

If A and B are two matrices such that AB = A and BA = B, then B2 is equal to

 

a. B

b. A

c. 1

d. 0

Solution 3

Question 4

If AB = A and BA = B, where A and B are square matrices, then

 

a. B2 = B and A2 = A

b. B2 B and A2 = A

c. A2 A,B2 = B

d. A2A, B2 B

Solution 4

Question 5

If A and B are two matrices such that AB = B and BA =A, then A2 + B2 is equal to

 

a. 2 AB

b. 2 BA

c. A + B

d. AB

Solution 5

Question 6

a. 3

b. 4

c. 6

d. 7

Solution 6

Question 7

If the matrix AB is zero , then

 

a. It is not necessary that either A = 0 or B = 0

b. A = 0 or B = 0

c. A = O and B = 0

d. All the above statements are wrong

Solution 7

Question 8

 

a. 

b. 

c. 

d. 

Solution 8

Question 9

If A, B are square matrices of order 3, A is non-singular and AB = 0, then B is a

 

a. Null matrix

b. Singular matrix

c. Unit matrix

d. Non-singular matrix

Solution 9

  

Question 10

 

a. B

b. nB

c. Bn

d. A + B

Solution 10

Question 11

 

a. 

b. 

c. 

d. 

Solution 11

Question 12

a. 0

b. -1

c. 2

d. None of these

Solution 12

Question 13

 

a. a = 4, b = 1

b. a = 1, b = 4

c. a = 0, b = 4

d. a = 2, b = 4

Solution 13

Question 14

 

a. 1 + α2 + βγ = 0

b. 1 - α2 + βγ = 0

c. 1 - α2 - βγ = 0

d. 1 + α2 - βγ = 0

Solution 14

Question 15

If S = [sij] is a scalar matrix such that sii = k and A is a square matrix of the same order, then AS = SA = ?

 

a. Ak

b. k + A

c. kA

d. kS

Solution 15

Question 16

If A is a square matrix such that A2 = A, then (I + A)3 - 7A is equal to

 

a. A

b. I - A

c. I

d. 3A

Solution 16

Question 17

If a matrix A is both symmetric and skew-symmetric, then

 

a. A is a diagonal matrix

b. A is a zero matrix

c. A is a scalar matrix

d. A is a square matrix

Solution 17

  

Question 18

 

a. A skew-symmetric matrix

b. A symmetric matrix

c. A diagonal matrix

d. An upper triangular matrix

Solution 18

  

Question 19

If A is a square matrix, then AA is a

 

a. Skew-symmetric matrix

b. Symmetric matrix

c. Diagonal matrix

d. None of these

Solution 19

Question 20

If A and B are symmetric matrices, then ABA is

 

a. Symmetric matrix

b. Skew-symmetric matrix

c. Diagonal matrix

d. Scalar matrix

Solution 20

Question 21

 

a. X = 0, y = 0

b. X + y = 5

c. X = y

d. None of these

Solution 21

Question 22

If A is 3 × 4 matrix and B is a matrix such that ATB and BAT are both defined. Then, B is of the type

 

a. 3 × 4

b. 3 × 3

c. 4 × 4

d. 4 × 3

 

Solution 22

  

Question 23

If A = [aij] is a square matrix of even order such that aij = i2 - j2, then

 

a. A is a skew - symmetric matrix and |A| = 0

b. A is symmetric matrix and |A| is a square

c. A is symmetric matrix and |A| = 0

d. None of these

Solution 23

Question 24

 

a. 

b. 

c. 

d. none of these

Solution 24

Question 25

a. 

b. 

c. 

d. 

Solution 25

Question 26

Out of the following matrices, choose that matrix which is a scalar matrix:

 

a. 

b. 

c. 

d. 

Solution 26

Question 27

The number of all possible matrices of order 3 × 3 with each entry 0 or 1 is

 

a. 27

b. 18

c. 81

d. 512

Solution 27

  

Question 28

Which of the given values of x and y make the following pairs of matrices equal?

 

a. 

b. 

c. 

d. Not possible to find

 

Solution 28

  

Question 29

 

a. -6, -12, -18

b. -6, 4, 9

c. -6, -4, -9

d. -6, 12, 18

Solution 29

Question 30

 

a. I cos  θ + J sin θ

b. I sin θ + J cos θ

c. I cos θ - J sin θ

d. - I cos θ + J sin θ

Solution 30

Question 31

 

a. 17

b. 25

c. 3

d. 12

Solution 31

Question 32

If A = [aij] is a scalar matrix of order n × n such that aii = k for all I, then trace of A is equal to

 

a. nk

b. n + k

c. 

d. none of these

Solution 32

Question 33

 

a. square matrix

b. diagonal matrix

c. unit matrix

d. none of these

Solution 33

Question 34

The number of possible matrices of order 3 × 3 with each entry 2 or 0 is

 

a. 9

b. 27

c. 81

d. none of these

Solution 34

Question 35

 

a. x = 3, y = 1

b. x = 2, y = 3

c. x = 2, y = 3

d. x = 3, y = 3

Solution 35

Question 36

If A is a square matrix such that A2 = I, then (A - I)3 + (A + I)3 - 7A is equal to

 

a. A

b. I - A

c. I + A

d. 3A

Solution 36

Question 37

If A and B are two matrix of order 3 × m and 3 × n respectively and m = n, then the order of 5A - 2B is

 

a. m × n

b. 3 × 3

c. m × n

d. 3 × n

Solution 37

Question 38

If A is a matrix of order m × n and B is a matrix such that ABT and BTA are both defined, then the order of matrix B is

 

a. m × n

b. n × n

c. n × m

d. 3 × n

Solution 38

Question 39

If A and B are matrices of the same order, then (ABT-BAT)T is a

 

a. skew-symmetric matrix

b. null matrix

c. unit matrix

d. symmetric matrix

Solution 39

   

Question 40

 

a. I

b. A

c. O

d. -I  

Solution 40

Question 41

 

a. I

b. 0

c. 2I

d. 

Solution 41

Question 42

If A and B are square matrices of the same order, then (A + B) (A - B) is equal to

 

a. A2 - B2

b. A2 - BA - AB - B2

c. A2 - B2 + BA - AB

d. A2 - BA + B2 + AB

Solution 42

Question 43

 

a. Only AB is defined

b. Only BA is defined

c. AB ad BA both are defined

d. AB and BA both are not defined

Solution 43

Question 44

 

a. Diagonal matrix

b. Symmetric matrix

c. Skew-symmetric matrix

d. Scalar matrix

Solution 44

Question 45

 

a. Identity matrix

b. Symmetric matrix

c. Skew-symmetric matrix

d. Diagonal matrix

Solution 45

Correct option: (d)

A matrix is called Diagonal matrix if all the elements, except those in the leading diagonal, are zero.

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