RD SHARMA Solutions for Class 12-science Maths Chapter 7 - Adjoint and Inverse of a Matrix

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Chapter 7 - Adjoint and Inverse of a Matrix Exercise MCQ

Question 32

If A and B are invertible matrices, then which one of the following is not correct?

  1. adj A = |A|A-1
  2. det(A-1) = [det(A)]-1
  3. (AB)-1 = B-1 A-1
  4. (A + B)-1 = B-1 + A-1
Solution 32

Given: A and B are invertible matrices

Relation between inverse and adjoint of a matrix is given by

  

We know that, AA-1 = I

  

As

  

But

  

Question 33

If   then A-1 exists if

  1. none of these
Solution 33

A matrix is invertible or its inverse exists if determinant is 0.

Now,

Therefore, A-1 exists if |A| = 0

i.e. if

i.e. if

Question 34

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

  1. I
  2. O
  3. I - A
  4. I + A
Solution 34

Given: A2 = A

(I - A)3 + A = (I - A)2(I - A) + A

= (I2 - 2AI + A2)(I - A) + A

= (I- 2A + A)(I - A) + A … (Since A2 = A)

= (I - A)(I - A) + A

= I2 - 2AI + A2 + A

= I - 2A + A + A

= I

Hence, (I - A)3 + A = I.

Question 35

The matrix   is not invertible for

  1.   
Solution 35

The matrix is non-invertible if its determinant is 0.

  

This matrix will not be invertible if

i.e. if

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. 7.1

Question 1(i)
Solution 1(i)
Question 1(ii)

Solution 1(ii)

Question 1(iii)
Solution 1(iii)

Question 1(iv)
Solution 1(iv)

Question 2(i)
Solution 2(i)
Question 2(ii)
Solution 2(ii)
Question 2(iii)
Solution 2(iii)
Question 2(iv)
Solution 2(iv)
Question 3

begin mathsize 12px style For space the space matrix space straight A space equals space open square brackets table row 1 cell negative 1 end cell 1 row 2 3 0 row 18 2 10 end table close square brackets comma space show space that
straight A left parenthesis adj space straight A right parenthesis equals 0 end style

Solution 3

Question 4

begin mathsize 12px style If space straight A equals open square brackets table row cell negative 4 end cell cell negative 3 end cell cell negative 3 end cell row 1 0 1 row 4 4 3 end table close square brackets comma space show space that space adj space straight A equals straight A end style

Solution 4

Question 5

begin mathsize 12px style If space straight A equals open square brackets table row cell negative 1 end cell cell negative 2 end cell cell negative 2 end cell row 2 1 cell negative 2 end cell row 2 cell negative 2 end cell 1 end table close square brackets comma space show space that space adj space straight A equals 3 straight A to the power of straight T. end style 

Solution 5

Question 6

begin mathsize 12px style Find space straight A open parentheses adj space straight A close parentheses space for space the space matrix space straight A equals open square brackets table row 1 cell negative 2 end cell 3 row 0 2 cell negative 1 end cell row cell negative 4 end cell 5 2 end table close square brackets. end style

Solution 6

Question 7(i)

begin mathsize 12px style Find space the space inverse space of space open square brackets table row cosθ sinθ row cell negative sinθ end cell cosθ end table close square brackets end style

Solution 7(i)

Question 7(ii)
Solution 7(ii)
Question 7(iii)
Solution 7(iii)
Question 7(iv)
Solution 7(iv)
Question 8(i)
Solution 8(i)
Question 8(ii)
Solution 8(ii)
Question 8(iii)
Solution 8(iii)
Question 8(iv)
Solution 8(iv)
Question 8(v)
Solution 8(v)
Question 8(vi)
Solution 8(vi)
Question 8(vii)
Solution 8(vii)
Question 9(i)
Solution 9(i)

Question 9(ii)
Solution 9(ii)
Question 10(i)
Solution 10(i)
Question 10(ii)
Solution 10(ii)
Question 11
Solution 11
Question 12
Solution 12
Question 13
Solution 13
Question 14
Solution 14
Question 15

begin mathsize 12px 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 to the power of negative 1 end exponent equals open square brackets table row 1 3 3 row 1 4 3 row 1 3 4 end table close square brackets. space Compute space open parentheses AB close parentheses to the power of negative 1 end exponent. end style

Solution 15

Question 16(i)
Solution 16(i)
Question 16(ii)
Solution 16(ii)
Question 16(iii)
Solution 16(iii)
Question 17

begin mathsize 12px style If space straight A equals open square brackets table row 2 3 row 1 2 end table close square brackets comma space verify space that space straight A squared minus 4 straight A plus straight I equals 0 comma space where space straight I equals open square brackets table row 1 0 row 0 1 end table close square brackets space and space 0 equals open square brackets table row 0 0 row 0 0 end table close square brackets. space Hence space find space straight A to the power of negative 1 end exponent. end style

Solution 17

Question 18

Solution 18


Question 19

begin mathsize 12px style If space straight A equals open square brackets table row 3 1 row cell negative 1 end cell 2 end table close square brackets space show space that space straight A squared minus 5 straight A plus 7 straight I equals 0 comma space Hence space find space straight A to the power of negative 1 end exponent. end style

Solution 19

Question 20
Solution 20
Question 21
Solution 21
Question 22
Solution 22
Question 23

begin mathsize 12px style Show space that space straight A equals open square brackets table row 6 5 row 7 6 end table close square brackets space satisfies space the space equation space straight x squared minus 12 straight x plus 1 equals 0. space Thus space find space straight A to the power of negative 1 end exponent. end style

Solution 23

Question 24

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 24

Question 25
Solution 25
Question 26

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 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

If A is a square matrix of order n; Prove that begin mathsize 12px style open vertical bar AadjA close vertical bar equals open vertical bar straight A close vertical bar to the power of straight n. end style

Solution 35

Question 36

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 36

Question 37

Solution 37

Question 38

   

Solution 38

Question 39

Solution 39

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

Question 1
Solution 1
Question 2
Solution 2
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.

 

 

Question 3

Find the inverse of each of the following matrices by using elementary row transformations:

  

Solution 3

Let

AA-1 = I

  

Applying R2 R2 - 2R1

  

Applying R2 R2/-5

  

Applying R1 R1 - 2R2

  

Question 17

Find the inverse of each of the following matrices by using elementary row transformations:

  

Solution 17

Let

AA-1 = I

  

Applying R2 R2 - 2R1 and R3 R3 + 2R1

  

Applying R1 R1 - 2R2

  

Applying R1 R1 - R3 and R2 R2 - R3

  

Hence,

Question 18

Find the inverse of each of the following matrices by using elementary row transformations:

  

Solution 18

Let

AA-1 = I

  

Applying R3 R1

  

Applying R2 R2 - 3R1 and R3 R3 - 2R1

  

Applying R2 -R2

  

Applying R1 R1 - R2 and R3 R3 + 5R2

  

Applying R3 -R3

  

Applying R2 R2 + 2R3

  

Hence,

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

begin mathsize 12px style If space straight A equals open square brackets table row cosθ cell negative sinθ end cell row cell negative sinθ end cell cosθ end table close square brackets space and space straight A left parenthesis adj space straight A right parenthesis equals open square brackets table row straight k 0 row 0 straight k end table close square brackets comma space then space find space the space value space of space straight k. end style

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

 

 

Question 29

Use elementary column operation C2 C2 + 2C1 in the following matrix equation:

  

Solution 29

Given:   which is in the form X = AB

Now, the column operation is applicable on X and B simultaneously when the equation is X = AB

Therefore, we'll apply the column operation on the first and third matrices.

Applying C2 C2 + 2C1

  

Question 30

In the following matrix equation use elementary operation R2 R2 + R1 and the equation thus obtained:

  

Solution 30

Given:   which is in the form AB = X

Now, the row operation is applicable on A and X simultaneously when the equation is AB = X

Therefore, we'll apply the row operation on the first and third matrices.

Applying R2 R2 + R1

  

Question 31

In A is a square matrix with |A| = 4 then find the value

Solution 31

Given: |A| = 4