Chapter 7 : Adjoint and Inverse of a Matrix - Rd Sharma Solutions for Class 12-science Maths CBSE

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 7 - Adjoint and Inverse of a Matrix Exercise Ex. 7.1

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
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

begin mathsize 14px style Given space straight A equals open square brackets table row 5 0 4 row 2 3 2 row 1 2 1 end table close square brackets comma space straight B minus 1 equals open square brackets table row 1 3 3 row 1 4 3 row 1 3 4 end table close square brackets. Compute space open parentheses AB close parentheses to the power of negative 1 end exponent. end style

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

begin mathsize 14px style For space the space matrix space straight A equals open square brackets table row 1 1 1 row 1 2 cell negative 3 end cell row 2 cell negative 1 end cell 3 end table close square brackets. end style Show that A3 - 6A2 + 5A + 11I3 = Ο. Hence, find A-1.

Solution 43

Question 44
Solution 44
Question 45

begin mathsize 14px style If space straight A equals open square brackets table row 2 cell negative 1 end cell 1 row cell negative 1 end cell 2 cell negative 1 end cell row 1 cell negative 1 end cell 2 end table close square brackets. end style Verify that A3 - 6A2 + 9A - 4I = Ο and hence find A-1.

Solution 45

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

begin mathsize 14px style IfA to the power of negative 1 end exponent equals open square brackets table row 3 cell negative 1 end cell 1 row cell negative 15 end cell 6 cell negative 5 end cell row 5 cell negative 2 end cell 2 end table close square brackets and space space straight B equals open square brackets table row 1 2 cell negative 2 end cell row cell negative 1 end cell 3 0 row 0 cell negative 2 end cell 1 end table close square brackets comma space find open parentheses AB close parentheses to the power of negative 1 end exponent. end style

Solution 55

Question 56

Solution 56

Question 57

   

Solution 57

Question 58

Solution 58

Chapter 7 - Adjoint and Inverse of a Matrix Exercise Ex. 7.2

Question 1
Solution 1
Question 2
Solution 2
Question 3
Solution 3
Question 4
Solution 4
Question 5
Solution 5
Question 6
Solution 6
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
Solution 14

Question 15

F i n d space t h e space i n v e r s e space o f space t h e space f o l l o w i n g space m a t r i x space b y space u sin g space e l e m e n t a r y space t r a n s f o r m a t i o n s :
open square brackets table row 1 3 cell minus 2 end cell row cell minus 3 end cell 0 1 row 2 1 0 end table close square brackets

Solution 15

C o n s i d e r space t h e space g i v e n space m a t r i x :
L e t space A equals open square brackets table row 1 3 cell minus 2 end cell row cell minus 3 end cell 0 1 row 2 1 0 end table close square brackets
W e space k n o w space t h a t space A equals I A
T h u s comma space w e space h a v e comma
open square brackets table row 1 3 cell minus 2 end cell row cell minus 3 end cell 0 1 row 2 1 0 end table close square brackets equals open square brackets table row 1 0 0 row 0 1 0 row 0 0 1 end table close square brackets A
A p p l y i n g space R subscript 2 rightwards arrow 3 R subscript 1 plus R subscript 2 space a n d space R subscript 3 rightwards arrow R subscript 3 minus 2 R subscript 1 comma space w e space h a v e comma
open square brackets table row 1 3 cell minus 2 end cell row 0 9 cell minus 5 end cell row 0 cell minus 5 end cell 4 end table close square brackets equals open square brackets table row 1 0 0 row 3 1 0 row cell minus 2 end cell 0 1 end table close square brackets A
A p p l y i n g space R subscript 1 rightwards arrow R subscript 1 minus 3 R subscript 2 space a n d space R subscript 3 rightwards arrow R subscript 3 plus 5 R subscript 2 comma space w e space h a v e comma
open square brackets table row 1 0 cell fraction numerator minus 1 over denominator 3 end fraction end cell row 0 1 cell fraction numerator minus 5 over denominator 9 end fraction end cell row 0 0 cell 11 over 9 end cell end table close square brackets equals open square brackets table row 0 cell fraction numerator minus 1 over denominator 3 end fraction end cell 0 row cell 1 third end cell cell 1 over 9 end cell 0 row cell fraction numerator minus 1 over denominator 3 end fraction end cell cell 5 over 9 end cell 1 end table close square brackets A
A p p l y i n g space R subscript 1 rightwards arrow fraction numerator R subscript 1 over denominator 11 over 9 end fraction space w e space h a v e comma
open square brackets table row 1 0 cell fraction numerator minus 1 over denominator 3 end fraction end cell row 0 1 cell fraction numerator minus 5 over denominator 9 end fraction end cell row 0 0 1 end table close square brackets equals open square brackets table row 0 cell fraction numerator minus 1 over denominator 3 end fraction end cell 0 row cell 1 third end cell cell 1 over 9 end cell 0 row cell fraction numerator minus 3 over denominator 11 end fraction end cell cell 5 over 11 end cell cell 9 over 11 end cell end table close square brackets A
A p p l y i n g space R subscript 2 rightwards arrow R subscript 2 plus 5 over 9 R subscript 3 space a n d space R subscript 1 rightwards arrow R subscript 1 plus 1 third R subscript 3 comma space w e space h a v e comma
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 cell fraction numerator minus 1 over denominator 11 end fraction end cell cell fraction numerator minus 2 over denominator 11 end fraction end cell cell 3 over 11 end cell row cell 2 over 11 end cell cell 4 over 11 end cell cell 5 over 11 end cell row cell fraction numerator minus 3 over denominator 11 end fraction end cell cell 5 over 11 end cell cell 9 over 11 end cell end table close square brackets A
rightwards double arrow I n v e r s e space o f space t h e space g i v e n space m a t r i x space i s space open square brackets table row cell fraction numerator minus 1 over denominator 11 end fraction end cell cell fraction numerator minus 2 over denominator 11 end fraction end cell cell 3 over 11 end cell row cell 2 over 11 end cell cell 4 over 11 end cell cell 5 over 11 end cell row cell fraction numerator minus 3 over denominator 11 end fraction end cell cell 5 over 11 end cell cell 9 over 11 end cell end table close square brackets

 

 

Question 16

F i n d space t h e space i n v e r s e space o f space t h e space f o l l o w i n g space m a t r i x space b y space u sin g space e l e m e n t a r y space t r a n s f o r m a t i o n s :
open square brackets table row cell minus 1 end cell 1 2 row 1 2 3 row 3 1 1 end table close square brackets

Solution 16

C o n s i d e r space t h e space g i v e n space m a t r i x space open square brackets table row cell minus 1 end cell 1 2 row 1 2 3 row 3 1 1 end table close square brackets
L e t space A equals open square brackets table row cell minus 1 end cell 1 2 row 1 2 3 row 3 1 1 end table close square brackets
W e space k n o w space t h a t space A equals I A
rightwards double arrow open square brackets table row cell minus 1 end cell 1 2 row 1 2 3 row 3 1 1 end table close square brackets equals open square brackets table row 1 0 0 row 0 1 0 row 0 0 1 end table close square brackets A
A p p l y i n g space R subscript 1 rightwards arrow open parentheses minus 1 close parentheses R subscript 1 comma space w e space h a v e
open square brackets table row 1 cell minus 1 end cell cell minus 2 end cell row 1 2 3 row 3 1 1 end table close square brackets equals open square brackets table row cell minus 1 end cell 0 0 row 0 1 0 row 0 0 1 end table close square brackets A
A p p l y i n g space R subscript 2 rightwards arrow R subscript 2 minus R subscript 1 comma space R subscript 3 rightwards arrow R subscript 3 minus 3 R subscript 1 comma space w e space h a v e
open square brackets table row 1 cell minus 1 end cell cell minus 2 end cell row 0 3 5 row 0 4 7 end table close square brackets equals open square brackets table row cell minus 1 end cell 0 0 row 1 1 0 row 3 0 1 end table close square brackets A
A p p l y i n g space R subscript 2 rightwards arrow R subscript 2 over 3 comma space w e space h a v e comma
open square brackets table row 1 cell minus 1 end cell cell minus 2 end cell row 0 1 cell 5 over 3 end cell row 0 4 7 end table close square brackets equals open square brackets table row cell minus 1 end cell 0 0 row cell 1 third end cell cell 1 third end cell 0 row 3 0 1 end table close square brackets A
A p p l y i n g space R subscript 1 rightwards arrow R subscript 1 plus R subscript 2 space a n d space R subscript 3 rightwards arrow R subscript 3 minus 4 R subscript 2 comma space w e space h a v e
open square brackets table row 1 0 cell minus 1 third end cell row 0 1 cell 5 over 3 end cell row 0 0 cell 1 third end cell end table close square brackets equals open square brackets table row cell minus 2 over 3 end cell cell 1 third end cell 0 row cell 1 third end cell cell 1 third end cell 0 row cell 5 over 3 end cell cell fraction numerator minus 4 over denominator 3 end fraction end cell 1 end table close square brackets A
A p p l y i n g space R subscript 3 rightwards arrow space R subscript 3 over 3 comma space w e space h a v e
open square brackets table row 1 0 cell minus 1 third end cell row 0 1 cell 5 over 3 end cell row 0 0 1 end table close square brackets equals open square brackets table row cell minus 2 over 3 end cell cell 1 third end cell 0 row cell 1 third end cell cell 1 third end cell 0 row 5 cell minus 4 end cell 3 end table close square brackets A
A p p l y i n g space R subscript 1 rightwards arrow R subscript 1 plus 1 third R subscript 3 comma space R subscript 2 rightwards arrow R subscript 2 minus 5 over 3 R subscript 3 comma space w e space h a v e comma
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 1 cell minus 1 end cell 1 row cell minus 8 end cell 7 cell minus 5 end cell row 5 cell minus 4 end cell 3 end table close square brackets A
T h u s comma space t h e space i n v e r s e space o f space t h e space g i v e n space m a t r i x space i s space open square brackets table row 1 cell minus 1 end cell 1 row cell minus 8 end cell 7 cell minus 5 end cell row 5 cell minus 4 end cell 3 end table close square brackets.

 

 

Chapter 7 - Adjoint and Inverse of a Matrix Exercise MCQ

Question 1

If A is an invertible matrix, then which of the following is not true

 

a. (A2)-1 = (A-1)2

b. |A-1| = |A|-1

c. (AT)-1 = (A-1)T

d. |A| 0

Solution 1

Correct option: (a)

|A-1| = |A|-1, (AT)-1 = (A-1)T, |A| 0 are properties of an invertible matrix.

Question 2

If A is an invertible matrix of order 3, then which of the following is not true

 

a. |adj A| = |A|2

b. (A-1)-1 = A

c. If BA = CA, then B C, where B and C are square matrices of order 3

d. (AB)-1 = B-1 A-1, where B = [bij]3×3 and |B| 0

Solution 2

Correct option: (c)

  

Question 3

 

a. is a skew-symmetric matrix

b. A-1 + B-1

c. Does not exist

d. None of these

Solution 3

Correct option:(d)

  

Question 4

 

a. 

b. 

c. 

d. 

Solution 4

Correct option: (b)

Question 5

If A is singular matrix, then adj A is

 

a. non-singular

b. singular

c. symmetric

d. not defined

Solution 5

Correct option:(b)

If A is singular matrix then adjoint of A is also singular.

Question 6

If A, B are two n × n non - singular matrices, then

 

a. AB is non-singular

b. AB is singular

c. (AB)-1 = A-1B-1

d. (AB)-1 does not exist

Solution 6

Correct option: (a)

Question 7

 

a. a27

b. a9

c. a6

d. a2

Solution 7

Correct option: (c)

Question 8

 

a. 144

b. 143

c. 142

d. 14

Solution 8

Correct option:(a)

Question 9

If B is non-singular matrix and A is a square matrix, then det (B-1AB) is equal to

 

 

a. Det (A-1)

b. Det (B-1)

c. Det (A)

d. Det (B)

Solution 9

Correct option: (c)

Question 10

 

 

a. 20

b. 100

c. 10

d. 0

Solution 10

Correct option: (c)

Question 11

If A5 = O such that An I for 1 n 4, then (I - A)-1 equals

 

a. A4

b. A3

c. I + A

d. none of these

Solution 11

Correct option: (d)

Question 12

If A satisfies the equation x3 - 5x2 + 4x + λ = 0, then A-1 exists if

 

a. λ  1

b. λ  2

c. λ  -1

d. λ  0

Solution 12

Correct option: (d)

Question 13

If for the matrix A, A3 = I, then A-1 =

 

a. A2

b. A3

c. A

d. none of these

Solution 13

Correct option: (a)

Question 14

If A and B are square matrices such that B = - A- 1 BA, then (A + B)2 =

 

 

a. O

b. A2 + B2

c. A2 + 2AB + B2

d. A + B

Solution 14

Correct option: (b)

Question 15

 

 

a. 5A

b. 10A

c. 16A

d. 32A

Solution 15

Correct option: (c)

Question 16

For non-singular square matrix A,B and C of the same order (AB-1C)-1 =

 

a. A-1BC-1

b. C-1 B-1A-1

c. CBA-1

d. C-1BA-1

Solution 16

Correct option: (d)

Question 17

 

 

a. -3

b. 3

c. 0

d. Non-existent

Solution 17

Correct option: (d)

Question 18

If d is the determinant of a square matrix A of order n, then the determinant of its adjoint is

 

a. dn

b. dn-1

c. dn+1

d. d

Solution 18

Correct option: (b)

Question 19

If A is a matrix of order 3 and |A| = 8, then |adj A| =

 

a. 1

b. 2

c. 23

d. 26

Solution 19

Correct option: (d)

Question 20

If A2 - A + I = O, then the inverse of A is

 

a. A-2

b. A + I

c. I - A

d. A - I

Solution 20

Correct option: (c)

Question 21

If A and B are invertible matrices, which of the following statement is not correct.

 

a. Adj A = |A| A-1

b. Det(A-1) = (det A)-1

c. (A + B)-1 = A-1 + B-1

d. (AB)-1 = B-1A-1

Solution 21

Correct option: (c)

Adj A = |A| A-1, Det(A-1) = (det A)-1, (AB)-1 = B-1A-1 are all the properties of invertible matrix.

Question 22

If A is a square matrix such that A2 = I, then A-1 is equal to

 

a. A + I

b. A

c. 0

d. 2A

Solution 22

Correct option: (b)

Question 23

 

a. 

b. 

c. 

d. none of these

Solution 23

Correct option: (a)

Question 24

 

 

a. 19

b. 1/19

c. -19

d. -1/19

Solution 24

Correct option: (b)

Question 25

 

 

a. 3

b. 0

c. -3

d. 1

Solution 25

Question 26

 

a. A

b. -A

c. ab A

d. none of these

Solution 26

Correct option: (d)

Question 27

 

 

a. a = 1, b = 1

b. a = cos 2θ, b = sin 2θ

c. a = sin 2θ, b = cos 2θ

d. none of these

Solution 27

Correct option: (b)

Question 28

If a matrix A is such that 3A3 + 2A2 + 5A + I = 0, then A-1 is equal to

 

 

a. - (3A2 + 2A + 5)

b. 3A2 + 2A + 5

c. 3A2 - 2A - 5

d. none of these

Solution 28

Correct option: (d)

Question 29

If A is an invertible matrix, then det (A-1) is equal to

 

a. det (A)

b. 

c. 1

d. none of these

Solution 29

Correct option: (b)

Question 30

 

 

a. 

b. 

c. 

d. none of these

Solution 30

Correct option: (a)

Question 31

If x, y, z are non-zero real numbers, then the inverse of the matrix

 

 

a. 

b. 

c. 

d. 

Solution 31

Correct option: (a)

Chapter 7 - Adjoint AND Inverse of a Matrix Exercise Ex. 7VSAQ

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

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

I f space A equals open square brackets table row 2 3 row 5 cell minus 2 end cell end table close square brackets comma space w r i t e space A to the power of minus 1 end exponent space i n space t e r m s space o f space A.

Solution 27

G i v e n space t h a t space space A equals open square brackets table row 2 3 row 5 cell minus 2 end cell end table close square brackets
W e space k n o w space t h a t space A equals I A
rightwards double arrow open square brackets table row 2 3 row 5 cell minus 2 end cell end table close square brackets equals open square brackets table row 1 0 row 0 1 end table close square brackets A
A p p l y i n g space R subscript 1 rightwards arrow R subscript 1 over 2 comma space w e space h a v e comma
open square brackets table row 1 cell 3 over 2 end cell row 5 cell minus 2 end cell end table close square brackets equals open square brackets table row cell 1 half end cell 0 row 0 1 end table close square brackets A
A p p l y i n g space R subscript 2 rightwards arrow R subscript 2 minus 5 R subscript 1 comma space w e space h a v e comma
open square brackets table row 1 cell 3 over 2 end cell row 0 cell minus 19 over 2 end cell end table close square brackets equals open square brackets table row cell 1 half end cell 0 row cell fraction numerator minus 5 over denominator 2 end fraction end cell 1 end table close square brackets A
A p p l y i n g space R subscript 2 rightwards arrow fraction numerator R subscript 2 over denominator minus 19 over 2 end fraction comma space w e space h a v e comma
open square brackets table row 1 cell 3 over 2 end cell row 0 1 end table close square brackets equals open square brackets table row cell 1 half end cell 0 row cell 5 over 19 end cell cell fraction numerator minus 2 over denominator 19 end fraction end cell end table close square brackets A
A p p l y i n g space R subscript 1 rightwards arrow R subscript 1 minus 3 over 2 R subscript 2 comma space w e space h a v e comma
open square brackets table row 1 0 row 0 1 end table close square brackets equals open square brackets table row cell 2 over 19 end cell cell 3 over 19 end cell row cell 5 over 19 end cell cell fraction numerator minus 2 over denominator 19 end fraction end cell end table close square brackets A
rightwards double arrow open square brackets table row 1 0 row 0 1 end table close square brackets equals 1 over 19 open square brackets table row 2 3 row 5 cell minus 2 end cell end table close square brackets A
rightwards double arrow I n v e r s e space o f space A space i s space A to the power of minus 1 end exponent equals 1 over 19 open square brackets table row 2 3 row 5 cell minus 2 end cell end table close square brackets
rightwards double arrow I n v e r s e space o f space A space i s space A to the power of minus 1 end exponent equals 1 over 19 A

 

 

Question 28

W r i t e space A to the power of minus 1 end exponent space f o r space A equals open square brackets table row 2 5 row 1 3 end table close square brackets

Solution 28

C o n s i d e r space t h e space g i v e n space m a t r i x.
A equals open square brackets table row 2 5 row 1 3 end table close square brackets
W e space k n o w space t h a t space A equals I A
open square brackets table row 2 5 row 1 3 end table close square brackets equals open square brackets table row 1 0 row 0 1 end table close square brackets A
A p p l y i n g space R subscript 1 rightwards arrow R subscript 1 over 2 comma space w e space h a v e
open square brackets table row 1 cell 5 over 2 end cell row 1 3 end table close square brackets equals open square brackets table row cell 1 half end cell 0 row 0 1 end table close square brackets A
A p p l y i n g space R subscript 2 rightwards arrow R subscript 1 minus R subscript 2 comma space w e space h a v e
open square brackets table row 1 cell 5 over 2 end cell row 0 cell minus 1 half end cell end table close square brackets equals open square brackets table row cell 1 half end cell 0 row cell 1 half end cell cell minus 1 end cell end table close square brackets A
A p p l y i n g space space R subscript 2 rightwards arrow fraction numerator R subscript 2 over denominator minus 1 half end fraction comma space w e space h a v e
open square brackets table row 1 cell 5 over 2 end cell row 0 1 end table close square brackets equals open square brackets table row cell 1 half end cell 0 row cell minus 1 end cell 2 end table close square brackets A
A p p l y i n g space R subscript 1 rightwards arrow R subscript 1 minus 5 over 2 R subscript 2 comma space w e space h a v e comma
open square brackets table row 1 0 row 0 1 end table close square brackets equals open square brackets table row 3 cell minus 5 end cell row cell minus 1 end cell 2 end table close square brackets A
T h u s space t h e space i n v e r s e space o f space A space i s space A to the power of minus 1 end exponent equals open square brackets table row 3 cell minus 5 end cell row cell minus 1 end cell 2 end table close square brackets

 

 

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