Class 12 RD Sharma Solutions - Chapter 5 Algebra of Matrices - Exercise 5.1 | Set 1

Chapter 5 of RD Sharma's Class 12 Mathematics textbook focuses on the Algebra of Matrices, with Exercise 5.1 Set 1 introducing fundamental concepts and operations of matrices. This exercise covers matrix notation, types of matrices, and basic matrix operations. Students will learn to represent data in matrix form, identify different matrix types, and perform elementary matrix calculations, laying the groundwork for more advanced matrix algebra concepts.

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: 1x8, 8x1, 2x4 and 4x2.

Part 2:

Following a similar approach:

Since, mxn = 5.

The divisors of 5 are : 1,5.

Thus, the possible orders are: 1x5 and 5x1.

Question 2(i): If A = [a_ij] =  and B = [b_jj] =  then find a₂₂+b₂₁.

Solution:

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

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

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

Question 2(ii): If A = [a_ij]=  and B = [b_ij] =  then find a₁₁b₁₁ + a₂₂b₂₂.

Solution:

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

Thus,

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

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

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

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

That implies, a₁₁b₁₁ + a₂₂b₂₂ = (2x2) + (4x4) = 4 + 16 = 20.

Question 3: Let A be a matrix of order 3x4. 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, A is a matrix of order 3x4. 

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

Thus, A  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 1x4, 

And similarly, if C2 is a column, it will have 3 rows, each with 1 element, and thus its order is 3x1, 

Question 4(i): Construct a 2x3 matrix A = [a_ij] whose elements a_ij are given by : a_ij = i x j.

Solution:

We know that A is a matrix of order 2x3.

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 2x3 matrix A = [a_ij] whose elements a_ij are given by : a_ij = 2i - j.

Solution:

We know that A is a matrix of order 2x3.

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 2x3 matrix A = [a_ij] whose elements a_ij are given by : a_ij = i + j.

Solution:

We know that A is a matrix of order 2x3.

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 2x3 matrix A = [a_ij] whose elements a_ij are given by : a_ij = .

Solution:

We know that A is a matrix of order 2x3.

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 2x2 matrix A = [a_ij] whose a_ij are given by :  .

Solution:

We know that A is a matrix of order 2x2.

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 2x2 matrix A = [a_ij] whose a_ij are given by :  .

Solution:

We know that A is a matrix of order 2x2.

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 2x2 matrix A = [a_ij] whose a_ij are given by :  .

Solution:

We know that A is a matrix of order 2x2.

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 2x2 matrix A = [a_ij] whose a_ij are given by :  .

Solution: 

We know that A is a matrix of order 2x2.

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 2x2 matrix A = [a_ij] whose a_ij are given by :   .

Solution:

We know that A is a matrix of order 2x2.

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 2x2 matrix A = [a_ij] whose a_ij are given by :  .

Solution:

We know that A is a matrix of order 2x2.

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 2x2 matrix A = [a_ij] whose a_ij are given by :  .

Solution:

We know that A is a matrix of order 2x2.

Thus A can be depicted as : ,

Since each element can be defined as : ,

a₁₁ = e^2xsinx         a₁₂ = e^2xsin2x

a₂₁ = e^4xsinx         a₂₂ = e^4xsin2x

Thus, A can be depicted as : 

Question 6(i): Construct a 3x4 matrix A = [a_ij] whose a_ij are given by : a_ij = i + j .

Solution:

A is a matrix of order 3x4.

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 3x4 matrix A = [a_ij] whose a_ij are given by : a_ij = i - j.

Solution:

A is a matrix of order 3x4.

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 3x4 matrix A = [a_ij] whose a_ij are given by : a_ij = 2i .

Solution:

A is a matrix of order 3x4.

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 3x4 matrix A = [a_ij] whose a_ij are given by : a_ij = j.

Solution:

A is a matrix of order 3x4.

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 3x4 matrix A = [a_ij] whose a_ij are given by : a_ij =  .

Solution:

A is a matrix of order 3x4.

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 4x3 matrix A = [a_ij] whose a_ij are given by : a_ij = .

Solution:

A is a matrix of order 4x3.

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 4x3 matrix A = [a_ij] whose a_ij are given by : a_ij = .

Solution:

A is a matrix of order 4x3.

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 4x3 matrix A = [a_ij] whose a_ij are given by : a_ij = i.

Solution:

A is a matrix of order 4x3.

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 2x3.

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 2x3.

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 2x2.

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.

Summary

Exercise 5.1 Set 1 in Chapter 5 introduces students to the basics of matrix algebra. It covers matrix representation, identification of matrix types (such as square, rectangular, symmetric), and fundamental matrix operations like addition, subtraction, and scalar multiplication. This exercise builds a strong foundation for understanding more complex matrix operations and applications in later sections.