# 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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### Access answers to Maths RD Sharma Solutions for Class 12 Chapter 5 – Algebra of Matrices Exercise 5.1

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 = [ai 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 = [aj j] whose elements aj j are given by:

(i) ai j = i × j

(ii) ai j = 2i – j

(iii) ai j = i + j

(iv) ai j = (i + j)2/2

Solution:

(i) Given ai j = i × j

Let A = [ai j]2 × 3

So, the elements in a 2 × 3 matrix are

[a11, a12, a13, a21, a22, a23]

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 ai j = 2i – j

Let A = [ai 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 ai 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 ai j = (i + j)2/2

Let A = [ai j]2×3

So, the elements in a 2 × 3 matrix are

a11, a12, a13, a21, a22, a23

Let A = [ai 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 = [ai j] whose elements ai j are given by:

(i) (i + j)2 /2

(ii) ai j = (i – j)2 /2

(iii) ai j = (i – 2j)2 /2

(iv) ai j = (2i + j)2 /2

(v) ai j = |2i – 3j|/2

(vi) ai j = |-3i + j|/2

(vii) ai j = e2ix sin x j

Solution:

(i) Given (i + j)2 /2

Let A = [ai 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 ai j = (i – j)2 /2

Let A = [ai 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 ai j = (i – 2j)2 /2

Let A = [ai 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 ai j = (2i + j)2 /2

Let A = [ai 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 ai j = |2i – 3j|/2

Let A = [ai 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 ai j = |-3i + j|/2

Let A = [ai 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 ai j = e2ix sin x j

Let A = [ai 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 = [ai j] whose elements ai j are given by:
(i) ai j = i + j

(ii) ai j = i – j

(iii) ai j = 2i

(iv) ai j = j

(v) ai j = ½ |-3i + j|

Solution:

(i) Given ai j = i + j

Let A = [ai j]2×3

So, the elements in a 3 × 4 matrix are

a11, a12, a13, a14, a21, a22, a23, a24, a31, a32, a33, 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

a24 = 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 ai j = i – j

Let A = [ai j]2×3

So, the elements in a 3×4 matrix are

a11, a12, a13, a14, a21, a22, a23, a24, a31, a32, a33, 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

a24 = 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 ai j = 2i

Let A = [ai j]2×3

So, the elements in a 3×4 matrix are

a11, a12, a13, a14, a21, a22, a23, a24, a31, a32, a33, 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

a24 = 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 ai j = j

Let A = [ai j]2×3

So, the elements in a 3×4 matrix are

a11, a12, a13, a14, a21, a22, a23, a24, a31, a32, a33, a34

A =

a11 = 1

a12 = 2

a13 = 3

a14 = 4

a21 = 1

a22 = 2

a23 = 3

a24 = 4

a31 = 1

a32 = 2

a33 = 3

a34 = 4

Substituting these values in matrix A, we get,

A =

(vi) Given ai j = ½ |-3i + j|

Let A = [ai j]2×3

So, the elements in a 3×4 matrix are

a11, a12, a13, a14, a21, a22, a23, a24, a31, a32, a33, a34

A =

a11 =
a12 =
a13 =
a14 =
a21 =

a22 =

a23 =

a24 =
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 = [ai j] whose elements ai j are given by:

(i) ai j = 2i + i/j

(ii) ai j = (i – j)/ (i + j)

(iii) ai j = i

Solution:

(i) Given ai j = 2i + i/j

Let A = [ai j]4×3

So, the elements in a 4 × 3 matrix are

a11, a12, a13, a21, a22, a23, a31, a32, a33, a41, a42, 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 ai j = (i – j)/ (i + j)

Let A = [ai j]4×3

So, the elements in a 4 × 3 matrix are

a11, a12, a13, a21, a22, a23, a31, a32, a33, a41, a42, 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 ai j = i

Let A = [ai j]4×3

So, the elements in a 4 × 3 matrix are

a11, a12, a13, a21, a22, a23, a31, a32, a33, a41, a42, 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