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

Algebra of Matrices Exercise MCQ

Solution 46

As A and B are symmetric matrices, we have

AT = A and BT = B … (i)

Consider,

  

Hence, ABT - BAT is a skew-symmetric matrix.

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

  

Solution 10

Solution 11

Solution 12

Solution 13

Solution 14

Solution 15

Solution 16

Solution 17

  

Solution 18

  

Solution 19

Solution 20

Solution 21

Solution 22

  

Solution 23

Solution 24

Solution 25

Solution 26

Solution 27

  

Solution 28

  

Solution 29

Solution 30

Solution 31

Solution 32

Solution 33

Solution 34

Solution 35

Solution 36

Solution 37

Solution 38

Solution 39

   

Solution 40

Solution 41

Solution 42

Solution 43

Solution 44

Solution 45

Correct option: (d)

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

Algebra of Matrices Exercise Ex. 5.1

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

Solution 2

Solution 3

Solution 4

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

Solution 5(vii)

 

Solution 6

 
(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

Solution 7

 

Solution 8

Solution 9

Solution 10

Solution 11

Solution 12

Solution 13

Solution 14

 

 

Solution 15

Solution 16

Solution 17

Solution 18

Solution 19

Solution 20

Solution 21

As A = B,

  

Algebra of Matrices Exercise Ex. 5.2

Solution 1

Solution 2(i)

Solution 2(ii)

Solution 2(iii)

Solution 2(iv)

Solution 3

(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

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

Solution 11

Solution 12

Solution 13

Solution 14

Solution 15(i)

Solution 15(ii)

Solution 15(iii)

Solution 16

Solution 17

Solution 18

(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

 

Solution 19(i)

Solution 19(ii)

Solution 20

Solution 21

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

Solution 22

Let 3x and 4x be the monthly incomes of Aryan and Babban respectively.

Let 5y and 7y be their monthly expenditures respectively.

As each individual saves Rs. 15000 per month, we have

3x - 5y = 15000 … (i)

4x - 7y = 15000 … (ii)

The above equations can be written in matrix form as follows

  

Let AX = B, where

Let's find A-1

  

  

  

Therefore, x = Rs. 30000 and y = Rs. 15000

So, monthly income of Aryan = Rs. 90,000 and monthly income of Babban is Rs. 120,000.

Algebra of Matrices Exercise Ex. 5.3

Solution 1

Solution 2(i)

Solution 2(ii)

Solution 2(iii)

Solution 3(i)

Solution 3(ii)

Solution 3(iii)

Solution 3(iv)

Solution 4(i)

Solution 4(ii)

Solution 5(i)

Solution 5(ii)

Solution 5(iii)

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

Solution 11

Solution 12

Solution 13

Solution 14

Solution 15

Solution 16(i)

Solution 16(ii)

Solution 17(i)

Solution 17(ii)

Solution 18

Solution 19

Solution 20

Solution 21

Solution 22

Solution 23

Solution 24(i)

Solution 24(ii)

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

Solution 25

Solution 26

Solution 27

Solution 28

Solution 29

Solution 31

Solution 32

Solution 33

Solution 34

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

Solution 35

Solution 36

Solution 37

Solution 38

Solution 39

Solution 40(i)

Solution 40(ii)

Solution 40(iii)

Solution 41

Solution 42

Solution 43

Solution 44

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.

Solution 45

Solution 46

Solution 47

Solution 48(i)

Solution 48(ii)


Solution 48(iii)

Solution 48(iv)

Solution 49

Solution 51

 

Solution 52

Solution 53

Solution 54(i)

Solution 54(ii)

Solution 55

Solution 56

Solution 57

Solution 58

Solution 59

Solution 60

Solution 61

Solution 62

Solution 64

Solution 65(i)

Solution 65(ii)

Solution 65(iii)

Solution 65(iv)

Solution 66

Solution 67

Solution 68

 

 

 

 

Solution 69

 

Solution 70

Solution 71

Solution 72

Solution 73

Solution 74


Solution 75

Solution 76

Solution 77

Solution 30

Given: A2 = A

(2 + A)3 - 19A

= 8I + A3 + 12A + 6A2 - 19A

= A3 + 6A2 - 7A + 8I

= A2 + 6A - 7A + 8I

= A - A + 8I

= 8I

Hence, (2 + A)3 - 19A = 8I.

Solution 40(iv)

  

Solution 48(v)

Given:

Let   

  

Hence,

Solution 48(vi)

Given:

Let   

  

Solving equations (i) and (ii), we get, a1 = 1, b1 = -2

Solving (iii) and (iv), we get, a2 = 2, b2 = 0

From equations (v) and (vi), we get, a3 = -5, b3 = 4

Hence,

Solution 50

Given:   and

As A2 + I = kA

  

Hence, k = -4.

Solution 63

To prove

For n = 1,

  

Therefore, it is true for n = 1.

Suppose the result is true for n = k

  

Take n = k + 1

  

Thus,   is true for all n N.

Solution 78

Let 3x and 4x be the monthly incomes of Aryan and Babbar respectively.

Let 5y and 7y be their monthly expenditures respectively.

As each individual saves Rs. 15000 per month, we have

3x - 5y = 15000 … (i)

4x - 7y = 15000 … (ii)

The above equations can be written in matrix form as follows

  

Let AX = B, where

Let's find A-1

  

  

  

Therefore, x = Rs. 30000 and y = Rs. 15000

So, monthly income of Aryan = Rs. 90,000 and monthly income of Babbar is Rs. 120,000.

This encourages us to understand the power of savings and we should save certain money for future.

Solution 79

Let Rs. x and Rs. y is being invested in the first and second bonds respectively.

Let A be the investment matrix and B be the interest matrix.

Therefore,

The annual interest = AB =

  

If the interest had been interchanged, the total interest would be Rs. 100 less.

  

Equations (i) and (ii) can be expressed as

PX = Q, where

Now, |P| = 100 - 144 = -44

So, inverse of P exist.

 

Thus, x = 10000 and y = 15000

Hence, the total amount invested is Rs. 25,000.

Algebra of Matrices Exercise Ex. 5.4

Solution 1(i)

Solution 1(ii)

Solution 1(iii)

Solution 1(iv)

Solution 2

Solution 3(i)

Solution 3(ii)

Solution 3(iii)

Solution 4

Solution 5

Solution 6(i)

Solution 6(ii)

Solution 7

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

Solution 8

Solution 9

Solution 10

Algebra of Matrices Exercise Ex. 5.5

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Given:

  

Consider,

  

Thus, A + AT is a symmetric matrix.

Solution 10

As   is a symmetric matrix, its transpose will be equal to itself.

  

Hence, the value of x is 5.

Algebra of Matrices Exercise Ex. 5VSAQ

Solution 64

As A is a symmetric matrix, we have

AT = A

  

Solution 65

Number of elements in a 2 × 2 matrix = 4

The first element can be 1, 2 or 3.

The second element can be 1, 2 or 3.

Similarly, the remaining two elements can take either of the 3 numbers.

So, for every element we have 3 choices.

Therefore, number of ways of writing 1, 2 or 3 in a 2 × 2 matrix is 34 which is 81.

Thus, the number of all possible matrices of order 2 × 2 with each entry 1, 2 or 3 is 81.

Solution 66

Let

Order of matrix P is 1 × 3

Order of matrix Q is 3 × 3

Order of matrix R is 3 × 1

After multiplying P and Q, we'll get an output matrix B of order 1 × 3.

After multiplying B with R, we'll get an output matrix of order 1 × 1.

Hence, the order of matrix A is 1 × 1.

Solution 67

As P is symmetric, we have

As Q is skew-symmetric, we have

  

  

  

Solution 68

Matrix A is order 3 × 2

Matrix B is of order 2 × 4

Then the product matrix AB will have the order 3 × 4.

Solution 69

As matrix A is skew-symmetric

Therefore, -A = AT

  

Hence, the values of a and b are -2 and 3 respectively.

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

Solution 11

Solution 12

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

Solution 14

Solution 15

Solution 16

Solution 17

Solution 18

Solution 19

Solution 20

Solution 21

Solution 22

Solution 23

Solution 24

Solution 25

Solution 26

Solution 27

Solution 28

Solution 29

Solution 30

Solution 31

Solution 32

Solution 33

Solution 34

Solution 35

Solution 36

Solution 37

Solution 38

Solution 39

Solution 40

Solution 41

Solution 42

Solution 43

Solution 44

Solution 45


Solution 46

Solution 47

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

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.

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.

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

Solution 52

Solution 53

Solution 54

Solution 55

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

Solution 57

Solution 58

Solution 59

Solution 60

Solution 61

Solution 62

Solution 63

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