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RD Sharma Solution for Class 12 Science Mathematics Chapter 7 - Adjoint and Inverse of a Matrix

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Our RD Sharma Textbook Solutions are considered extremely helpful for solving the tough questions which are asked in the CBSE Class 12 exam. TopperLearning Textbook Solutions are compiled by subject experts. Herein, you can find all the answers to the textbook questions for Chapter 7 - Adjoint and Inverse of a Matrix.

All our solutions are created in accordance with the latest CBSE syllabus, and they are amended from time to time to provide the most relevant answers. Our free RD Sharma Solutions for CBSE Class 12 Mathematics will strengthen your fundamentals in Mathematics and will help you in your attempts to score more marks in the final examination. CBSE Class 12 students can refer to our solutions any time — while doing their homework and while preparing for the exam.

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Exercise/Page

RD Sharma Solution for Class 12 Science Mathematics Chapter 7 - Adjoint and Inverse of a Matrix Page/Excercise 7.1

Solution 1(i)

Solution 1(ii)

Solution 1(iii)

Solution 1(iv)

Solution 2(i)

Solution 2(ii)

Solution 2(iii)

Solution 2(iv)

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7(i)

Solution 7(ii)

Solution 7(iii)

Solution 7(iv)

Solution 8(i)

Solution 8(ii)

Solution 8(iii)

Solution 8(iv)

Solution 8(v)

Solution 8(vi)

Solution 8(vii)

Solution 9(i)

Solution 9(ii)

Solution 10(i)

Solution 10(ii)

Solution 11

Solution 12

Solution 13

Solution 14

Solution 15

Solution 16(i)

Solution 16(ii)

Solution 16(iii)

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

RD Sharma Solution for Class 12 Science Mathematics Chapter 7 - Adjoint and Inverse of a Matrix Page/Excercise 7.2

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

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

 

 

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.

 

 

RD Sharma Solution for Class 12 Science Mathematics Chapter 7 - Adjoint and Inverse of a Matrix Page/Excercise MCQ

Solution 1

Correct option: (a)

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

Solution 2

Correct option: (c)

  

Solution 3

Correct option:(d)

  

Solution 4

Correct option: (b)

Solution 5

Correct option:(b)

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

Solution 6

Correct option: (a)

Solution 7

Correct option: (c)

Solution 8

Correct option:(a)

Solution 9

Correct option: (c)

Solution 10

Correct option: (c)

Solution 11

Correct option: (d)

Solution 12

Correct option: (d)

Solution 13

Correct option: (a)

Solution 14

Correct option: (b)

Solution 15

Correct option: (c)

Solution 16

Correct option: (d)

Solution 17

Correct option: (d)

Solution 18

Correct option: (b)

Solution 19

Correct option: (d)

Solution 20

Correct option: (c)

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.

Solution 22

Correct option: (b)

Solution 23

Correct option: (a)

Solution 24

Correct option: (b)

Solution 25

Solution 26

Correct option: (d)

Solution 27

Correct option: (b)

Solution 28

Correct option: (d)

Solution 29

Correct option: (b)

Solution 30

Correct option: (a)

Solution 31

Correct option: (a)

RD Sharma Solution for Class 12 Science Mathematics Chapter 7 - Adjoint and Inverse of a Matrix Page/Excercise 7VSAQ

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

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

 

 

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

 

 

TopperLearning provides step-by-step solutions for each question in each chapter in the RD Sharma textbook for class 9. Access the CBSE Class 9 Mathematics Chapter 7 - Adjoint and Inverse of a Matrix for free. The textbook questions have been solved by our subject matter experts to help you understand how to answer them. Our RD Sharma Textbook Solutions will help you to study and revise, and you can easily clear your fundamentals of Chapter 7 - Adjoint and Inverse of a Matrix.

Text Book Solutions

CBSE XII Science - Mathematics

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