# RD Sharma Class 12 Ex 5.1 Solutions Chapter 5 Algebra of Matrices

Here we provide RD Sharma Class 12 Ex 5.1 Solutions Chapter 5 Algebra of Matrices for English medium students, Which will very helpful for every student in their exams. Students can download the RD Sharma Class 12 Ex 5.1 Solutions Chapter 5 Algebra of Matrices book pdf download. Now you will get step-by-step solutions to each question.

## RD Sharma Class 12 Ex 5.1 Solutions Chapter 5 Algebra of Matrices

### Question 1: If a matrix has 8 elements, what are the possible orders it can have? What if it has 5 elements?

Solution:

Part 1:

Let the matrix be of mxn dimension.

Hence, mxn = 8.

Then, all we have to do is, find the divisors of 8, which are : 1, 2, 4, 8.

Thus, the possible orders are: 1×8, 8×1, 2×4 and 4×2.

Part 2:

Following a similar approach:

Since, mxn = 5.

The divisors of 5 are : 1,5.

Thus, the possible orders are: 1×5 and 5×1.

### Question 2(i): If A = [aij] =  and B = [bjj] =  then find a22+b21.

Solution:

We know that every element in a matrix A of dimensions mxn can be addressed as aij, where 1≤ i ≤ m and 1 ≤ j ≤ n.

Thus, a22 is the 2nd element in the 2nd row of A, and b21 is the 1rst element in the 2nd row of B.

That implies, a22 + b21 = 4 + (-3) = 1.

### Question 2(ii): If A = [aij]=  and B = [bij] =  then find a11b11 + a22b22.

Solution:

As seen in the previous question, every element in a matrix A of dimensions mxn can be addressed as aij, where 1≤ i ≤ m and 1 ≤ j ≤ n.

Thus,

a11 = 1rst element in the 1rst row of A = 2.

a22 = 2nd element in the 2nd row of A = 4.

b11 = 1rst element in the 1rst row of B = 2.

b22 = 2nd element in the 2nd row of B = 4.

That implies, a11b11 + a22b22 = (2×2) + (4×4) = 4 + 16 = 20.

### Question 3: Let A be a matrix of order 3×4. If R1 denotes the first row of A and C2 denotes its second column, then determine the orders of matrices R1 and C2.

Solution:

Given, is a matrix of order 3×4.

We know that a matrix having an order of mxn has m rows and n columns.

Thus,  contains, 3 rows, and each row contains 4 elements.

Now, if R1 is a row, it has 4 elements, and thus its order can be written as 1×4,

And similarly, if C2 is a column, it will have 3 rows, each with 1 element, and thus its order is 3×1,

### Question 4(i): Construct a 2×3 matrix A = [aij] whose elements aij are given by : aij = i x j.

Solution:

We know that A is a matrix of order 2×3.

Thus A can be depicted as:

Since each element is a product of its row number and column number (i x j):

a11 = 1    a12 = 2    a13 = 3

a21 = 2    a22 = 4    a23 = 6

Thus, A can be depicted as :

### Question 4(ii): Construct a 2×3 matrix A = [aij] whose elements aij are given by : aij = 2i – j.

Solution:

We know that A is a matrix of order 2×3.

Thus A can be depicted as:

Since each element can be defined as (2 x (row number)) – column number:

a11 = 1    a12 = 0    a13 = -1

a21 = 3    a22 = 2    a23 = 1

Thus, A can be depicted as :

### Question 4(iii): Construct a 2×3 matrix A = [aij] whose elements aij are given by : aij = i + j.

Solution:

We know that A is a matrix of order 2×3.

Thus A can be depicted as:

Since each element can be defined as the sum of its row number and column number:

a11 = 2    a12 = 3    a13 = 4

a21 = 3    a22 = 4    a23 = 5

Thus, A can be depicted as :

### Question 4(iv): Construct a 2×3 matrix A = [aij] whose elements aij are given by : aij = .

Solution:

We know that A is a matrix of order 2×3.

Thus A can be depicted as:

Since each element can be defined as :  ,

a11 = 2      a12 = 4.5    a13 = 8

a21 = 4.5   a22 = 8       a23 = 12.5

Thus, A can be depicted as :

### Question 5(i): Construct a 2×2 matrix A = [aij] whose aij are given by :  .

Solution:

We know that A is a matrix of order 2×2.

Thus A can be depicted as :

Since each element can be defined as :,

a11 = 2      a12 = 4.5

a21 = 4.5   a22 = 8

Thus, A can be depicted as :

### Question 5(ii): Construct a 2×2 matrix A = [aij] whose aij are given by :  .

Solution:

We know that A is a matrix of order 2×2.

Thus A can be depicted as :

Since each element can be defined as :  ,

a11 = 0      a12 = 0.5

a21 = 0.5   a22 = 0

Thus, A can be depicted as :

### Question 5(iii): Construct a 2×2 matrix A = [aij] whose aij are given by :  .

Solution:

We know that A is a matrix of order 2×2.

Thus A can be depicted as :

Since each element can be defined as : ,

a11 = 0.5    a12 = 4.5

a21 = 0       a22 = 2

Thus, A can be depicted as :

### Question 5(iv): Construct a 2×2 matrix A = [aij] whose aij are given by :  .

Solution:

We know that A is a matrix of order 2×2.

Thus A can be depicted as :

Since each element can be defined as :  ,

a11 = 4.5      a12 = 8

a21 = 12.5    a22 = 18

Thus, A can be depicted as :

### Question 5(v): Construct a 2×2 matrix A = [aij] whose aij are given by :   .

Solution:

We know that A is a matrix of order 2×2.

Thus A can be depicted as :  ,

Since each element can be defined as :  ,

a11 = 0.5      a12 = 2

a21 = 0.5      a22 = 1

Thus, A can be depicted as :

### Question 5(vi): Construct a 2×2 matrix A = [aij] whose aij are given by :  .

Solution:

We know that A is a matrix of order 2×2.

Thus A can be depicted as :  ,

Since each element can be defined as : ,

a11 = 1         a12 = 0.5

a21 = 2.5      a22 = 2

Thus, A can be depicted as :

### Question 5(vii) Construct a 2×2 matrix A = [aij] whose aij are given by :  .

Solution:

We know that A is a matrix of order 2×2.

Thus A can be depicted as : ,

Since each element can be defined as : ,

a11 = e2xsinx         a12 = e2xsin2x

a21 = e4xsinx         a22 = e4xsin2x

Thus, A can be depicted as :

### Question 6(i): Construct a 3×4 matrix A = [aij] whose aij are given by : aij = i + j .

Solution:

A is a matrix of order 3×4.

Thus, A can be depicted as : ,

Since each element can be defined as : ( row number + column number ),

a11 = 2    a12 = 3    a13 = 4    a14 = 5

a21 = 3    a22 = 4    a23 = 5    a24 = 6

a31 = 4    a32 = 5    a33 = 6    a34 = 7

Thus, A can be depicted as :

### Question 6(ii): Construct a 3×4 matrix A = [aij] whose aij are given by : aij = i – j.

Solution:

A is a matrix of order 3×4.

Thus, A can be depicted as : ,

Since each element can be defined as : (row number – column number),

a11 = 0    a12 = -1    a13 = -2    a14 = -3

a21 = 1    a22 = 0     a23 = -1    a24 = -2

a31 = 2    a32 = 1     a33 = 0     a34 = -1

Thus, A can be depicted as :

### Question 6(iii): Construct a 3×4 matrix A = [aij] whose aij are given by : aij = 2i .

Solution:

A is a matrix of order 3×4.

Thus, A can be depicted as : ,

Since each element can be defined as : (2 x row number ),

a11 = 2    a12 = 2     a13 = 2    a14 = 2

a21 = 4    a22 = 4     a23 = 4    a24 = 4

a31 = 6    a32 = 6     a33 = 6    a34 = 6

Thus, A can be depicted as :

### Question 6(iv): Construct a 3×4 matrix A = [aij] whose aij are given by : aij = j.

Solution:

A is a matrix of order 3×4.

Thus, A can be depicted as :  ,

Since each element can be defined as : (column number ),

a11 = 1    a12 = 2     a13 = 3    a14 = 4

a21 = 1    a22 = 2     a23 = 3    a24 = 4

a31 = 1    a32 = 2     a33 = 3    a34 = 4

Thus, A can be depicted as :

### Question 6(v): Construct a 3×4 matrix A = [aij] whose aij are given by : aij =  .

Solution:

A is a matrix of order 3×4.

Thus, A can be depicted as :  ,

Since each element can be defined as : ,

a11 = -1       a12 = -1/2     a13 = 0         a14 = 1/2

a21 = -5/2   a22 = -2         a23 = -3/2    a24 = -1

a31 = -4       a32 = -7/2     a33 = -3        a34 = -5/2

Thus, A can be depicted as :

### Question 7(i): Construct a 4×3 matrix A = [aij] whose aij are given by : aij = .

Solution:

A is a matrix of order 4×3.

Thus, A can be depicted as :

Since each element can be defined as : ,

a11 = 3       a12 = 5/2         a13 = 7/3

a21 = 6       a22 = 5            a23 = 14/3

a31 = 9       a32 = 15/2      a33 = 7

a41 = 12     a42 = 10        a43 = 28/3

Thus, A can be depicted as :

### Question 7(ii): Construct a 4×3 matrix A = [aij] whose aij are given by : aij = .

Solution:

A is a matrix of order 4×3.

Thus, A can be depicted as:

Since each element can be defined as : ,

a11 = 0          a12 = -1/3      a13 = -1/2

a21 = 1/3       a22 = 0          a23 = -1/5

a31 = 1/2       a32 = 1/5       a33 = 0

a41 = 3/5       a42 = 1/3       a43 = 1/7

Thus, A can be depicted as :

### Question 7(iii): Construct a 4×3 matrix A = [aij] whose aij are given by : aij = i.

Solution:

A is a matrix of order 4×3.

Thus, A can be depicted as :

Since each element can be defined as : (row number)

a11 = 1     a12 = 1      a13 = 1

a21 = 2     a22 = 2      a23 = 2

a31 = 3     a32 = 3      a33 = 3

a41 = 4     a42 = 4      a43 = 4

Thus, A can be depicted as :

### Question 8: Find x, y, a and b if:

Solution:

We can see that both, the matrix on the Left Hand Side (LHS) and the matrix on the Right Hand Side (RHS are of the dimension 2×3.

Since, the matrix on LHS is equal to the matrix on the RHS, each element on the LHS at the index (i, j) must be equal to each element on the RHS at the index (i, j).

Hence, equating each element on RHS to LHS:

a11: 3x+4y = 2 ………………(eq.1)       a12: 2 = 2                                   a13: x-2y = 4 ………………(eq.2)

a21: a+b = 5    ………………(eq.3)       a22: 2a-b = -5……………..(eq.4)      a23: -1 = -1

Thus, (eq.1) and (eq.2) form one system of equations comprising variables x and y.

Solving (eq.1) and (eq.2): (eq.1) + 2x(eq.2)

=> (3x+2x) + (4y-2(2y)) = 2+ (2(4))

=> 5x = 10

=> x = 2

Substituting (x=2) in (eq.1) :

=> (3(2)) + 4y = 2

=> 4y = 2-6 = -4

=> y=-1

Similarly, (eq.3) and (eq.4) form a system of equations comprising variables a and b.

Solving (eq.3) and (eq.4) : (eq.1) + (eq.2)

=> (a+2a) + (b-b) = 5 – 5

=> 3a = 0

=> a = 0

Substituting (a=0) in (eq.3):

=> 0 + b = 5

=>b = 5

Thus, a=0, b=5, x=2 and y=-1.

### Question 9: Find x, y, a and b if : .

Solution:

We can see that both, the matrix on the Left Hand Side (LHS) and the matrix on the Right Hand Side (RHS are of the dimension 2×3.

Since, the matrix on LHS is equal to the matrix on the RHS, each element on the LHS at the index (i, j) must be equal to each element on theRHS at the index (i, j).

Hence, equating each element on RHS to LHS:

a11 : 2x-3y = 1………………(eq.1)      a12 : a-b = -2………………(eq.2)     a13 : 3 = 3

a21 : 1 = 1                                   a22 : x+4y = 6……………..(eq.3)      a23 : 3a+4b = 29………………(eq.4)

Thus, (eq.1) and (eq.3) form one system of equations comprising of variables x and y.

Solving (eq.1) and (eq.2) : (eq.1) – 2x(eq.2)

=> (2x-2x) + (-3y-2(4y)) = 1- (2(6))

=> -11y = -11

=> y = 1

Substituting (y=1) in (eq.1) :

=> 2x – 3(1) = 1

=> 2x = 3+1 = 4

=> x = 2

Similarly, (eq.2) and (eq.4) form a system of equations comprising of variables a and b.

Solving (eq.2) and (eq.4) : 4x(eq.1) + (eq.2)

=> (4a+3a) + (-4(b)+4b) = 4(-2) + 29

=> 7a = 21

=> a = 3

Substituting (a=3) in (eq.2) :

=> 3 – b = -2

=>b = 5

Thus, a=3, b=5, x=2 and y=1.

### Question 10: Find a, b, c and d if :

Solution:

We can see that both, the matrix on the Left Hand Side (LHS) and the matrix on the Right Hand Side (RHS are of the dimension 2×2.

Since, the matrix on LHS is equal to the matrix on the RHS, each element on the LHS at the index (i, j) must be equal to each element on the RHS at the index (i, j).

Hence, equating each element on RHS to LHS:

a11 : 2a+b = 4 ………….(eq.1)

a12 : a-2b = -3 …………(eq.2)

a21 : 5c-d = 11 …………(eq.3)

a22 : 4c+3d = 24 ……..(eq.4)

Thus, (eq.1) and (eq.2) form one system of equations comprising of variables a and b.

Solving (eq.1) and (eq.2) : (eq.1) – 2x(eq.2)

=> (2a-2a) + (b+4b) = 4 + (-2(-3))

=> 5b = 10

=> b = 2

Substituting (b=2) in (eq.1) :

=> 2a + 2 = 4

=> 2a = 4-2 = 2

=> a=1

Similarly, (eq.3) and (eq.4) form a system of equations comprising of variables c and d.

Solving (eq.3) and (eq.4) : 3x(eq.1) + (eq.2)

=> (15c+4c) + (-3d+3d) = 33 + 24

=> 19c = 57

=> c = 3

Substituting (c=3) in (eq.4) :

=> (4(3)) + 3d = 24

=>3d = 24 – 12 = 12

=>d = 4

Thus, a=1, b=2, c=3 and d=4.

Question 11: Find x, y and z so that A = B, where A= .
Solution:

As it is given that the matrices are equal, every element at a given index on the Left Hand Side (LHS) must be equal to the element at the same index on the Right Hand Side (RHS).
Thus,
a11 : x-2 = y ………….(eq.1)
a12 : z = 3    ………….(eq.2)
a13 : 2z = 6  ………….(eq.3)
a21 : 18z = 6y ……….(eq.4)
a22 : y+2 = x ………..(eq.5)
a23 : 6z = 2y …………(eq.6)
From (eq.2) and (eq.3),
=> z = 3
From (eq.4) and (eq.6),
=> y = 3z
=> y = 3(3)
=> y = 9
Substitute ( y=9 ) in (eq.1),
=> x-2 = 9
=>x = 11
Thus x=11, y=9 and z=3.
Question 12: If , find x, y, z and w.
Solution:
As it is given that the matrices are equal, every element at a given index on the Left Hand Side (LHS) must be equal to the element at the same index on the Right Hand Side (RHS).
Thus,
a11 : x = 3      ………….(eq.1)
a12 : 3x-y = 2 ………….(eq.2)
a21 : 2x+z = 4 ………..(eq.3)
a22 : 3y-w = 7 ……….(eq.4)
From (eq.1) ,
=> x = 3
Substitute (x=3) in (eq.2),
=> 3(3)-y = 2
=> y = 3(3) – 2
=> y = 7
Substitute ( x=3 ) in (eq.3),
=> 2(3)+z = 4
=>z = 4-6
=>z = -2
Substitute (y=7) in (eq.4),
=>3(7) – w = 7
=>w = 21 – 7
=>w = 14
Thus x=3, y=7, z=-2 and w=14.
Question 13: If , find x, y, z and w.
As it is given that the matrices are equal, every element at a given index on the Left Hand Side (LHS) must be equal to the element at the same index on the Right Hand Side (RHS).
Thus,
a11 : x-y = -1 ………….(eq.1)
a12 : z = 4     ………….(eq.2)
a21 : 2x-y = 0 ………..(eq.3)
a22 : w = 5 …………….(eq.4)
From (eq.2),
=> z = 4
And from (eq.4),
=> w = 5
Now, (eq.1) and (eq.3) form a system of equations comprising of variables x and y.
Thus, (eq.1) – (eq.2),
=> (x-2x) +(-y+y) = -1 – 0
=> -x = -1
=> x = 1
Substitute (x=1) in (eq.1),
=> 1-y = -1
=>y = 2
Thus, x=1, y=2, z=4 and w=5.
Question 14: If . Obtain the values of a, b, c, x, y and z.
Solution:
As it is given that the matrices are equal, every element at a given index on the Left Hand Side (LHS) must be equal to the element at the same index on the Right Hand Side (RHS).
Thus,
a11 : x+3 = 0 ……………..(eq.1)
a12 : z+4 = 6    ………….(eq.2)
a13 : 2y-7 = 3y-2  ……..(eq.3)
a21 : 4x+6 = 2x ………….(eq.4)
a22 : a-1 = -3 ……………..(eq.5)
a23 : 0 = 2c+2 …………….(eq.6)
a31 : b-3 = 2b+4…………(eq.7)
a32 : 3b = -21………………(eq.8)
a33 : z+2c = 0………………(eq.9)
From (eq.1) and (eq.4),
=> x = -3
From (eq.2) ,
=> z+4 = 6
=> z = 2
From (eq.3),
=> 2y-7 = 3y-2
=> 3y-2y = -7+2
=> y = -5
From (eq.5),
=> a = -3+1
=> a = -2
From (eq.8),
=> 3b = -21
=> b = -7
Substitute (z=2) in (eq.9),
=> 2+ 2c = 0
=> c = -1
Thus, x=-3, y=-5, z=2, a=-2, b=-7 and c=-1.
Question 15: If , find the value of (x+y).
Solution:
As it is given that the matrices are equal, every element at a given index on the Left Hand Side (LHS) must be equal to the element at the same index on the Right Hand Side (RHS).
Thus,
a11 : 2x+1 = x+3………..(eq.1)
a12 : 5x = 10 ………………(eq.2)
a21 : 0 = 0  …………………(eq.3)
a22 : y2+1 = 26 …………(eq.4)
From (eq.1) and (eq.2),
=> 2x+1 = x+3
=> x=2
From (eq.4),
=> y2+1 = 26
=> y2 = 25
=> y = ± 5
Thus if y = +5,
=> x+y = 7
And if y = -5,
=> x+y = -3
Question 16: If , then find the values of x, y, z and w.
Solution:
As it is given that the matrices are equal, every element at a given index on the Left Hand Side (LHS) must be equal to the element at the same index on the Right Hand Side (RHS).
Thus,
a11 : xy = 8 ……………..(eq.1)
a12 : 4 = w ………………(eq.2)
a21 : z+6 = 0 …………..(eq.3)
a22 : x+y = 6 …………..(eq.4)
From (eq.2),
=> w = 4
And from (eq.3),
=> z = -6
Now, we can see that (eq.1) and (eq.4) form a system of equations comprising of variables x and y.
From (eq.1),
=> x = 8/y
Substitute (x=8/y) in (eq.4):
=> y + (8/y) = 6
=> y2 – 6y +8 = 0
Solving the above equation,
=> y2 – 4y – 2y + 8 = 0
=> y( y-4 ) -2( y-4 ) = 0
=> (y – 2)(y – 4) = 0
=> y = 2 or 4
Substitute in (eq.1):
=> when x=2, y=4 and when x=4, y=2.
Thus, (x,y) = (2,4) or (4,2) and z = -6 and w = 4.
Question 17(i): Give an example of a row matrix which is also a column matrix.
Solution:
We know the order of a row matrix can be written as 1xn (1 row with n elements).
And similarly, the order of a column matrix is mx1.
So, a row matrix which is also a column matrix must be of the order (1×1).
As an example, we can take the matrix :
Question 17(ii): Give an example of a diagonal matrix which is not scalar.
Solution:
In a diagonal matrix, only the diagonal elements possess non-zero values. Thus, for a nxn diagonal matrix, aii ≠ 0, for 1≤ i ≤ n.
And a scalar matrix is a diagonal matrix, such that all the diagonal elements are equal.
Thus, a matrix which is diagonal but not scalar is:

Question 17(iii): Give an example of a triangular matrix.
Solution:
A triangular matrix is a square matrix, and it is filled in such a way that, either the triangle above the main-diagonal is non-zero (upper-triangular) or the triangle below the diagonal is non-zero (lower-triangular).
Thus, an example would be :
Question 18: The sales figure of two car dealers during January 2013 showed that dealer A sold 5 deluxe, 3 premium and 4 standard cars, while dealer B sold 7 deluxe, 2 premium and 3 standard cars. Total sales over the month period of January- February revealed that dealer A sold 8 deluxe, 7 premium and 6 standard cars. In the same 2 month period, dealer B sold 10 deluxe, 5 premium and 7 standard cars. Write 2×3 matrices summarizing sales data for January and 2-month period for each dealer.
Solution:
The above data can be represented in the form of tables:
For January 2013,

Deluxe
Standard
Dealer A
5
3
4
Dealer B
7
2
3
For January to February,

Deluxe
Standard
Dealer A
8
7
6
Dealer B
10
5
7
Thus, the two matrices are :  and
Question 19: For what value of x and y are the following matrices equal?
A=  and B =
Solution:
As it is given that the matrices are equal, every element at a given index on the Left Hand Side (LHS) must be equal to the element at the same index on the Right Hand Side (RHS).
Thus,
a11 : 2x+1 = x+3 …………(eq.1)
a12 : 2y = y2+2 ……………(eq.2)
a21 : 0 = 0 …………………….(eq.3)
a22 : y2-5y = -6 …………..(eq.4)
From (eq.1),
=> 2x-x = 3-1
=> x=2
Taking (eq.2), it can be re-written as,
=> y2-2y+2 = 0
=> y =
=> y = -1 ± i ( No real solutions )
Taking (eq.4), it can be re-written as,
=> y2 – 5y +6 = 0
Solving the equation,
=> y2 – 2y – 3y +6 = 0
=> y( y-2 ) -3( y-2 ) = 0
=> ( y-2 )( y-3 ) = 0
=> y = 2 or 3
As values of y are inconsistent, we can say that the above matrices are not equal for any (x,y) pair.
Question 20. Find the values of x and y if  .
Solution:
As it is given that the matrices are equal, every element at a given index on the Left Hand Side (LHS) must be equal to the element at the same index on the Right Hand Side (RHS).
Thus,
a11 : x+10 = 3x+4 …………(eq.1)
a12 : y2+2y = 3   ……………(eq.2)
a21 : 0 = 0  ……………………..(eq.3)
a22 : y2-5y = -4   …………..(eq.4)
From (eq.1),
=> 2x = 6
=> x = 3
Taking (eq.2), it can be re-written as,
=> y2+2y-3 = 0
=> y+ 3y -y -3 = 0
=> y( y+3 ) -1( y+3 ) = 0
=> ( y+3 )( y-1 ) = 0
=> y = -3 or 1
Taking (eq.3), it can be re-written as,
=> y2-5y+4 = 0
=> y2 -4y -y +4 =0
=> y( y-4 ) -1( y-4 ) = 0
=> ( y-4 )( y-1 ) = 0
=> y = 4 or 1
The value of y that can satisfy both (eq.2) and (eq.3) is 1.
=> y=1
Thus, x=3 and y=1.
Question 21. Find the values of a and b if A = B , where A= , B= .
Solution:
As it is given that the matrices are equal, every element at a given index on the Left Hand Side (LHS) must be equal to the element at the same index on the Right Hand Side (RHS).
Thus,
a11 : a+4 = 2a+2 …………..(eq.1)
a12 : 3b = b2+2   …………..(eq.2)
a21 : 8 = 8  ……………………..(eq.3)
a22 : -6 = b2-5b   ………….(eq.4)
From (eq.1),
=> a = 2
Taking (eq.2), it can be re-written as,
=> b2-3b+2=0
=> b2 – 2b -b +2 = 0
=> b(b-2) -1(b-2) = 0
=> (b-2)(b-1) = 0
=> b=1 or 2
Taking (eq.4) it can be re-written as,
=> b2 -5b +6 = 0
=> b2 – 3b -2b + 6 = 0
=> b( b-3 ) -2( b-3 ) = 0
=> ( b-3 )( b-2 ) = 0
=> b = 2 or 3
Thus, b=2 can satisfy both (eq.2) and (eq.4).
=> b = 2
Thus, a=2 and b=2.

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