RD Sharma Solutions for Class 12 Maths Exercise 5.1 Chapter 5 Algebra of Matrices

RD Sharma Solutions for Class 12 Maths Exercise 5.1 Chapter 5 Algebra of Matrices are provided here for students to study and excel in their board exams. This exercise has problems which are solved on the Equality of Matrices of two matrices by the subject experts at BYJU’S. Our team of faculty work with the aim of making Mathematics easier for students based on their grasping abilities.

The students can gain a grip on the concepts covered in Maths Class 12 only by regular practice, and while solving, they can make use of the PDF as a reference. Solving the main problems will become easier if the students first solve the examples which are present before each exercise. RD Sharma Solutions for Class 12 Maths Chapter 5 Algebra of Matrices Exercise 5.1 are provided here.

RD Sharma Solutions for Class 12 Chapter 5 – Algebra of Matrices Exercise 5.1

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

Solution:

If a matrix is of order m × n elements, it has m n elements. So, if the matrix has 8 elements, we will find the ordered pairs m and n.

m n = 8

Then, ordered pairs m and n will be

m × n be (8 × 1),(1 × 8),(4 × 2),(2 × 4)

Now, if it has 5 elements

Possible orders are (5 × 1), (1 × 5).

Solution:

(i)

Now, comparing with equations (1) and (2),

a₂₂ = 4 and b₂₁ = – 3

a₂₂ + b₂₁ = 4 + (– 3) = 1

(ii)

Now, comparing with equations (1) and (2),

a₁₁ = 2, a₂₂ = 4, b₁₁ = 2, b₂₂ = 4

a₁₁ b₁₁ + a₂₂ b₂₂ = 2 × 2 + 4 × 4 = 4 + 16 = 20

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, A be a matrix of order 3 × 4.

So, A = [a_{i j}] _3×4

R₁ = first row of A = [a₁₁, a₁₂, a₁₃, a₁₄]

So, the order of matrix R₁ = 1 × 4

C₂ = second column of

Therefore, order of C₂ = 3 × 1

4. Construct a 2 ×3 matrix A = [a_{j j}] whose elements a_{j j} are given by:

(i) a_{i j} = i × j

(ii) a_{i j} = 2i – j

(iii) a_{i j} = i + j

(iv) a_{i j} = (i + j)²/2

Solution:

(i) Given a_{i j} = i × j

Let A = [a_{i j}]_{2 × 3}

So, the elements in a 2 × 3 matrix are

a₁₁ = 1 × 1 = 1

a₁₂ = 1 × 2 = 2

a₁₃ = 1 × 3 = 3

a₂₁ = 2 × 1 = 2

a₂₂ = 2 × 2 = 4

a₂₃ = 2 × 3 = 6

Substituting these values in matrix A, we get,

(ii) Given a_{i j} = 2i – j

Let A = [a_{i j}]_2×3

So, the elements in a 2 × 3 matrix are

a₁₁, a₁₂, a₁₃, a₂₁, a₂₂, a₂₃

a₁₁ = 2 × 1 – 1 = 2 – 1 = 1

a₁₂ = 2 × 1 – 2 = 2 – 2 = 0

a₁₃ = 2 × 1 – 3 = 2 – 3 = – 1

a₂₁ = 2 × 2 – 1 = 4 – 1 = 3

a₂₂ = 2 × 2 – 2 = 4 – 2 = 2

a₂₃ = 2 × 2 – 3 = 4 – 3 = 1

Substituting these values in matrix A, we get,

(iii) Given a_{i j} = i + j

Let A = [a _{i j}] _2×3

So, the elements in a 2 × 3 matrix are

a₁₁, a₁₂, a₁₃, a₂₁, a₂₂, a₂₃

a₁₁ = 1 + 1 = 2

a₁₂ = 1 + 2 = 3

a₁₃ = 1 + 3 = 4

a₂₁ = 2 + 1 = 3

a₂₂ = 2 + 2 = 4

a₂₃ = 2 + 3 = 5

Substituting these values in matrix A, we get,

(iv) Given a_{i j} = (i + j)²/2

Let A = [a_{i j}]_2×3

So, the elements in a 2 × 3 matrix are

a₁₁, a₁₂, a₁₃, a₂₁, a₂₂, a₂₃

Let A = [a_{i j}]_2×3

So, the elements in a 2 × 3 matrix are

a₁₁, a₁₂, a₁₃, a₂₁, a₂₂, a₂₃

a₁₁ =

a₁₂ =

a₁₃ =

a₂₁ =

a₂₂ =

a₂₃ =

Substituting these values in matrix A, we get,

5. Construct a 2 × 2 matrix A = [a_{i j}] whose elements a_{i j} are given by:

(i) (i + j)² /2

(ii) a_{i j} = (i – j)² /2

(iii) a_{i j} = (i – 2j)² /2

(iv) a_{i j} = (2i + j)² /2

(v) a_{i j} = |2i – 3j|/2

(vi) a_{i j} = |-3i + j|/2

(vii) a_{i j} = e^2ix sin x j

Solution:

(i) Given (i + j)² /2

Let A = [a_{i j}]_2×2

So, the elements in a 2 × 2 matrix are

a₁₁, a₁₂, a₂₁, a₂₂

a₁₁ =

a₁₂ =

a₂₁ =

a₂₂ =

Substituting these values in matrix A, we get,

(ii) Given a_{i j} = (i – j)² /2

Let A = [a_{i j}]_2×2

So, the elements in a 2 × 2 matrix are

a₁₁, a₁₂, a₂₁, a₂₂

a₁₁ =

a₁₂ =

a₂₁ =

a₂₂ =

Substituting these values in matrix A, we get,

(iii) Given a_{i j} = (i – 2j)² /2

Let A = [a_{i j}]_2×2

So, the elements in a 2 × 2 matrix are

a₁₁, a₁₂, a₂₁, a₂₂

a₁₁ =

a₁₂ =

a₂₁ =

a₂₂ =

Substituting these values in matrix A, we get,

(iv) Given a_{i j} = (2i + j)² /2

Let A = [a_{i j}]_2×2

So, the elements in a 2 × 2 matrix are

a₁₁, a₁₂, a₂₁, a₂₂

a₁₁ =

a₁₂ =

a₂₁ =

a₂₂ =

Substituting these values in matrix A, we get,

(v) Given a_{i j} = |2i – 3j|/2

Let A = [a_{i j}]_2×2

So, the elements in a 2×2 matrix are

a₁₁, a₁₂, a₂₁, a₂₂

a₁₁ =

a₁₂ =

a₂₁ =

a₂₂ =

Substituting these values in matrix A, we get,

(vi) Given a_{i j} = |-3i + j|/2

Let A = [a_{i j}]_2×2

So, the elements in a 2 × 2 matrix are

a₁₁, a₁₂, a₂₁, a₂₂

a₁₁ =

a₁₂ =

a₂₁ =

a₂₂ =

Substituting these values in matrix A, we get,

(vii) Given a_{i j} = e^2ix sin x j

Let A = [a_{i j}]_2×2

So, the elements in a 2 × 2 matrix are

a₁₁, a₁₂, a₂₁, a₂₂,

a₁₁ =

a₁₂ =

a₂₁ =

a₂₂ =

Substituting these values in matrix A, we get,

6. Construct a 3×4 matrix A = [a_{i j}] whose elements a_{i j} are given by:
(i) a_{i j} = i + j

(ii) a_{i j} = i – j

(iii) a_{i j} = 2i

(iv) a_{i j} = j

(v) a_{i j} = ½ |-3i + j|

Solution:

(i) Given a_{i j} = i + j

Let A = [a_{i j}]_2×3

So, the elements in a 3 × 4 matrix are

a₁₁, a₁₂, a₁₃, a_14, a₂₁, a₂₂, a₂₃, a₂₄, a_31, a_32, a_33, a₃₄

A =

a₁₁ = 1 + 1 = 2

a₁₂ = 1 + 2 = 3

a₁₃ = 1 + 3 = 4

a₁₄ = 1 + 4 = 5

a₂₁ = 2 + 1 = 3

a₂₂ = 2 + 2 = 4

a₂₃ = 2 + 3 = 5

a_{24 =} 2 + 4 = 6

a₃₁ = 3 + 1 = 4

a₃₂ = 3 + 2 = 5

a₃₃ = 3 + 3 = 6

a₃₄ = 3 + 4 = 7

Substituting these values in matrix A, we get,

A =

(ii) Given a_{i j} = i – j

Let A = [a_{i j}]_2×3

So, the elements in a 3×4 matrix are

a₁₁, a₁₂, a₁₃, a_14, a₂₁, a₂₂, a₂₃, a₂₄, a_31, a_32, a_33, a₃₄

A =

a₁₁ = 1 – 1 = 0

a₁₂ = 1 – 2 = – 1

a₁₃ = 1 – 3 = – 2

a₁₄ = 1 – 4 = – 3

a₂₁ = 2 – 1 = 1

a₂₂ = 2 – 2 = 0

a₂₃ = 2 – 3 = – 1

a_{24 =} 2 – 4 = – 2

a₃₁ = 3 – 1 = 2

a₃₂ = 3 – 2 = 1

a₃₃ = 3 – 3 = 0

a₃₄ = 3 – 4 = – 1

Substituting these values in matrix A, we get,

A =

(iii) Given a_{i j} = 2i

Let A = [a_{i j}]_2×3

So, the elements in a 3×4 matrix are

a₁₁, a₁₂, a₁₃, a_14, a₂₁, a₂₂, a₂₃, a₂₄, a_31, a_32, a_33, a₃₄

A =

a₁₁ = 2×1 = 2

a₁₂ = 2×1 = 2

a₁₃ = 2×1 = 2

a₁₄ = 2×1 = 2

a₂₁ = 2×2 = 4

a₂₂ = 2×2 = 4

a₂₃ = 2×2 = 4

a_{24 =} 2×2 = 4

a₃₁ = 2×3 = 6

a₃₂ = 2×3 = 6

a₃₃ = 2×3 = 6

a₃₄ = 2×3 = 6

Substituting these values in matrix A, we get,

A =

(iv) Given a_{i j} = j

Let A = [a_{i j}]_2×3

So, the elements in a 3×4 matrix are

a₁₁, a₁₂, a₁₃, a_14, a₂₁, a₂₂, a₂₃, a₂₄, a_31, a_32, a_33, a₃₄

A =

a₁₁ = 1

a₁₂ = 2

a₁₃ = 3

a₁₄ = 4

a₂₁ = 1

a₂₂ = 2

a₂₃ = 3

a_{24 =} 4

a₃₁ = 1

a₃₂ = 2

a₃₃ = 3

a₃₄ = 4

Substituting these values in matrix A, we get,

A =

(vi) Given a_{i j} = ½ |-3i + j|

Let A = [a_{i j}]_2×3

So, the elements in a 3×4 matrix are

a₁₁, a₁₂, a₁₃, a_14, a₂₁, a₂₂, a₂₃, a₂₄, a_31, a_32, a_33, a₃₄

A =

a₁₁ =
a₁₂ =
a₁₃ =
a₁₄ =
a₂₁ =

a₂₂ =

a₂₃ =

a_{24 =}
a₃₁ =
a₃₂ =
a₃₃ =
a₃₄ =

Substituting these values in matrix A we get,

A =

Multiplying by a negative sign, we get,

7. Construct a 4 × 3 matrix A = [a_{i j}] whose elements a_{i j} are given by:

(i) a_{i j} = 2i + i/j

(ii) a_{i j} = (i – j)/ (i + j)

(iii) a_{i j} = i

Solution:

(i) Given a_{i j} = 2i + i/j

Let A = [a_{i j}]_4×3

So, the elements in a 4 × 3 matrix are

a₁₁, a₁₂, a₁₃, a₂₁, a₂₂, a₂₃, a_31, a_32, a_33, a_41, a_42, a₄₃

A =

a₁₁ =

a₁₂ =

a₁₃ =

a₂₁ =

a₂₂ =

a₂₃ =

a₃₁ =

a₃₂ =

a₃₃ =

a₄₁ =

a₄₂ =

a₄₃ =

Substituting these values in matrix A, we get,

A =

(ii) Given a_{i j} = (i – j)/ (i + j)

Let A = [a_{i j}]_4×3

So, the elements in a 4 × 3 matrix are

a₁₁, a₁₂, a₁₃, a₂₁, a₂₂, a₂₃, a_31, a_32, a_33, a_41, a_42, a₄₃

A =

a₁₁ =

a₁₂ =

a₁₃ =

a₂₁ =

a₂₂ =

a₂₃ =

a₃₁ =

a₃₂ =

a₃₃ =

a₄₁ =

a₄₂ =

a₄₃ =

Substituting these values in matrix A, we get,

A =

(iii) Given a_{i j} = i

Let A = [a_{i j}]_4×3

So, the elements in a 4 × 3 matrix are

a₁₁, a₁₂, a₁₃, a₂₁, a₂₂, a₂₃, a_31, a_32, a_33, a_41, a_42, a₄₃

A =

a₁₁ = 1

a₁₂ = 1

a₁₃ = 1

a₂₁ = 2

a₂₂ = 2

a₂₃ = 2

a₃₁ = 3

a₃₂ = 3

a₃₃ = 3

a₄₁ = 4

a₄₂ = 4

a₄₃ = 4

Substituting these values in matrix A, we get,

A =

8. Find x, y, a and b if

Solution:

Given

Given that two matrices are equal.

We know that if two matrices are equal, then the elements of each matrix are also equal.

Thereforem by equating them, we get,

3x + 4y = 2 …… (1)

x – 2y = 4 …… (2)

a + b = 5 …… (3)

2a – b = – 5 …… (4)

Multiplying equation (2) by 2 and adding to equation (1), we get

3x + 4y + 2x – 4y = 2 + 8

⇒ 5x = 10

⇒ x = 2

Now, substituting the value of x in equation (1)m

3 × 2 + 4y = 2

⇒ 6 + 4y = 2

⇒ 4y = 2 – 6

⇒ 4y = – 4

⇒ y = – 1

Now by adding equations (3) and (4)

a + b + 2a – b = 5 + (– 5)

⇒ 3a = 5 – 5 = 0

⇒ a = 0

Now, again by substituting the value of a in equation (3), we get

0 + b = 5

⇒ b = 5

∴ a = 0, b = 5, x = 2 and y = – 1

9. Find x, y, a and b if

Solution:

We know that if two matrices are equal, then the elements of each matrix are also equal.

Given that two matrices are equal.

Therefore by equating them, we get,

2a + b = 4 …… (1)

And a – 2b = – 3 …… (2)

And 5c – d = 11 …… (3)

4c + 3d = 24 …… (4)

Multiplying equation (1) by 2 and adding to equation (2)

4a + 2b + a – 2b = 8 – 3

⇒ 5a = 5

⇒ a = 1

Now, substituting the value of a in equation (1)

2 × 1 + b = 4

⇒ 2 + b = 4

⇒ b = 4 – 2

⇒ b = 2

Multiplying equation (3) by 3 and adding to equation (4)

15c – 3d + 4c + 3d = 33 + 24

⇒ 19c = 57

⇒ c = 3

Now, substituting the value of c in equation (4)

4 × 3 + 3d = 24

⇒ 12 + 3d = 24

⇒ 3d = 24 – 12

⇒ 3d = 12

⇒ d = 4

∴ a = 1, b = 2, c = 3 and d = 4

10. Find the values of a, b, c and d from the following equations:

Solution:

Given

We know that if two matrices are equal, then the elements of each matrix are also equal.

Given that two matrices are equal.

Therefore by equating them, we get,

2a + b = 4 …… (1)

And a – 2b = – 3 …… (2)

And 5c – d = 11 …… (3)

4c + 3d = 24 …… (4)

Multiplying equation (1) by 2 and adding to equation (2)

4a + 2b + a – 2b = 8 – 3

⇒ 5a = 5

⇒ a = 1

Now, substituting the value of a in equation (1)

2 × 1 + b = 4

⇒ 2 + b = 4

⇒ b = 4 – 2

⇒ b = 2

Multiplying equation (3) by 3 and adding to equation (4)

15c – 3d + 4c + 3d = 33 + 24

⇒ 19c = 57

⇒ c = 3

Now, substituting the value of c in equation (4)

4 × 3 + 3d = 24

⇒ 12 + 3d = 24

⇒ 3d = 24 – 12

⇒ 3d = 12

⇒ d = 4

∴ a = 1, b = 2, c = 3 and d = 4