This exercise has problems which are solved on Equality of Matrices of two matrices by the subject experts at BYJUâ€™S. Our team of faculty mainly work with the aim of making Mathematics easier for students based on their grasping power. The students can gain a grip on the concepts covered in Maths Class 12 only by regular practise, and while solving 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.

## Download the PDF of RD Sharma Solutions For Class 12 Chapter 5 – Algebra of matrices Exercise 5.1

### Access other exercises of RD Sharma Solutions For Class 12 Chapter 5 – Algebra of Matrices

### 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 Ã— 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 Ã— n be (8 Ã— 1),(1 Ã— 8),(4 Ã— 2),(2 Ã— 4)

Now, if it has 5 elements

Possible orders are (5 Ã— 1), (1 Ã— 5).

**Solution:**

(i)

Now, Comparing with equation (1) and (2)

a_{22}Â = 4 and b_{21}Â = â€“ 3

a_{22}Â + b_{21}Â = 4 + (â€“ 3) = 1

(ii)

Now, Comparing with equation (1) and (2)

a_{11}Â = 2, a_{22}Â = 4, b_{11}Â = 2, b_{22}Â = 4

a_{11}Â b_{11}Â + a_{22}Â b_{22}Â = 2 Ã— 2 + 4 Ã— 4 = 4 + 16 = 20

**3. Let A be a matrix of order 3 Ã— 4. If R _{1}Â denotes the first row of A and C_{2}Â denotes its second column, then determine the orders of matrices R_{1}Â and C_{2}.**

**Solution:**

Given A be a matrix of order 3 Ã— 4.

So, A = [a_{i j}] _{3Ã—4}

R_{1}Â = first row of A = [a_{11}, a_{12}, a_{13}, a_{14}]

So, order of matrix R_{1}Â = 1 Ã— 4

C_{2}Â = second column ofÂ

Therefore order ofÂ C_{2}Â = 3 Ã— 1

**4. Construct a 2 Ã—3 matrix A = [a _{j j}] whose elements a_{j j}Â are given by:**

**(i) a _{i j}Â = i Ã— j**

**(ii) a _{i j }= 2i â€“ j**

**(iii) a _{i j }= i + j**

**(iv) a _{i j} = (i + j)^{2}/2**

**Solution:**

(i) Given a_{i j}Â = i Ã— j

Let A = [a_{i j}]_{2 Ã— 3}

So, the elements in a 2 Ã— 3 matrix are

[a_{11}, a

_{12}, a

_{13}, a

_{21}, a

_{22}, a

_{23}]

a_{11}Â = 1 Ã— 1 = 1

a_{12}Â = 1 Ã— 2 = 2

a_{13}Â = 1 Ã— 3 = 3

a_{21}Â = 2 Ã— 1 = 2

a_{22}Â = 2 Ã— 2 = 4

a_{23}Â = 2 Ã— 3 = 6

Substituting these values in matrix A we get,

(ii) Given a_{i j }= 2i â€“ j

Let A = [a_{i j}]_{2Ã—3}

So, the elements in a 2 Ã— 3 matrix are

a_{11}, a_{12}, a_{13}, a_{21}, a_{22}, a_{23}

a_{11}Â = 2 Ã— 1 â€“ 1 = 2 â€“ 1 = 1

a_{12}Â = 2 Ã— 1 â€“ 2 = 2 â€“ 2 = 0

a_{13}Â = 2 Ã— 1 â€“ 3 = 2 â€“ 3 = â€“ 1

a_{21}Â = 2 Ã— 2 â€“ 1 = 4 â€“ 1 = 3

a_{22}Â = 2 Ã— 2 â€“ 2 = 4 â€“ 2 = 2

a_{23}Â = 2 Ã— 2 â€“ 3 = 4 â€“ 3 = 1

Substituting these values in matrix A we get,

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

Let A = [a _{i j}] _{2Ã—3}

So, the elements in a 2 Ã— 3 matrix are

a_{11}, a_{12}, a_{13}, a_{21}, a_{22}, a_{23}

a_{11}Â = 1 + 1 = 2

a_{12}Â = 1 + 2 = 3

a_{13}Â = 1 + 3 = 4

a_{21}Â = 2 + 1 = 3

a_{22}Â = 2 + 2 = 4

a_{23}Â = 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}]_{2Ã—3}

So, the elements in a 2 Ã— 3 matrix are

a_{11}, a_{12}, a_{13}, a_{21}, a_{22}, a_{23}

Let A = [a_{i j}]_{2Ã—3}

So, the elements in a 2 Ã— 3 matrix are

a_{11}, a_{12}, a_{13}, a_{21}, a_{22}, a_{23}

a_{11}Â =Â

a_{12}Â =Â

a_{13}Â =Â

a_{21}Â =Â

a_{22}Â =Â

a_{23}Â =Â

Substituting these values in matrix A we get,

**5. Construct a 2 Ã— 2 matrix A = [a _{i j}] whose elements a_{i j}Â 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 â€“ 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}]_{2Ã—2}

So, the elements in a 2 Ã— 2 matrix are

a_{11}, a_{12}, a_{21}, a_{22}

a_{11}Â =Â

a_{12}Â =Â

a_{21}Â =Â

a_{22}Â =Â

Substituting these values in matrix A we get,

(ii) Given a_{i j} = (i – j)^{2 }/2

Let A = [a_{i j}]_{2Ã—2}

So, the elements in a 2 Ã— 2 matrix are

a_{11}, a_{12}, a_{21}, a_{22}

a_{11}Â =Â

a_{12}Â =Â

a_{21}Â =Â

a_{22}Â =Â

Substituting these values in matrix A we get,

(iii) Given a_{i j} = (i – 2j)^{2 }/2

Let A = [a_{i j}]_{2Ã—2}

So, the elements in a 2 Ã— 2 matrix are

a_{11}, a_{12}, a_{21}, a_{22}

a_{11}Â =Â

a_{12}Â =Â

a_{21}Â =Â

a_{22}Â =Â

Substituting these values in matrix A we get,

(iv) Given a_{i j} = (2i + j)^{2 }/2

Let A = [a_{i j}]_{2Ã—2}

So, the elements in a 2 Ã— 2 matrix are

a_{11}, a_{12}, a_{21}, a_{22}

a_{11}Â =Â

a_{12}Â =Â

a_{21}Â =Â

a_{22}Â =Â

Substituting these values in matrix A we get,

(v) Given a_{i j} = |2i â€“ 3j|/2

Let A = [a_{i j}]_{2Ã—2}

So, the elements in a 2Ã—2 matrix are

a_{11}, a_{12}, a_{21}, a_{22}

a_{11}Â =Â

a_{12}Â =Â

a_{21}Â =Â

a_{22}Â =Â

Substituting these values in matrix A we get,

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

Let A = [a_{i j}]_{2Ã—2}

So, the elements in a 2 Ã— 2 matrix are

a_{11}, a_{12}, a_{21}, a_{22}

a_{11}Â =Â

a_{12}Â =Â

a_{21}Â =Â

a_{22}Â =Â

Substituting these values in matrix A we get,

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

Let A = [a_{i j}]_{2Ã—2}

So, the elements in a 2 Ã— 2 matrix are

a_{11}, a_{12}, a_{21}, a_{22},

a_{11}Â =Â

a_{12}Â =Â

a_{21}Â =Â

a_{22}Â =Â

Substituting these values in matrix A we get,

**6. Construct a 3Ã—4 matrix A = [a _{i j}] whose elements a_{i j}Â are given by:**

(i) a_{i j}Â = i + j

**(ii) a _{i j}Â = i â€“ j**

**(iii) a _{i j}Â = 2i**

**(iv) a _{i j}Â = j**

**(v) a _{i j}Â = Â½ |-3i + j|**

**Solution:**

(i) Given a_{i j}Â = i + j

Let A = [a_{i j}]_{2Ã—3}

So, the elements in a 3 Ã— 4 matrix are

a_{11}, a_{12}, a_{13}, a_{14,}Â a_{21}, a_{22}, a_{23}, a_{24}, a_{31, }a_{32, }a_{33, }a_{34}

A =Â

a_{11}Â = 1 + 1 = 2

a_{12}Â = 1 + 2 = 3

a_{13}Â = 1 + 3 = 4

a_{14}Â = 1 + 4 = 5

a_{21}Â = 2 + 1 = 3

a_{22}Â = 2 + 2 = 4

a_{23}Â = 2 + 3 = 5

a_{24 =}Â 2 + 4 = 6

a_{31}Â = 3 + 1 = 4

a_{32}Â = 3 + 2 = 5

a_{33}Â = 3 + 3 = 6

a_{34}Â = 3 + 4 = 7

Substituting these values in matrix A we get,

A =Â

(ii) Given a_{i j}Â = i â€“ j

Let A = [a_{i j}]_{2Ã—3}

So, the elements in a 3Ã—4 matrix are

a_{11}, a_{12}, a_{13}, a_{14,}Â a_{21}, a_{22}, a_{23}, a_{24}, a_{31, }a_{32, }a_{33, }a_{34}

A =Â

a_{11}Â = 1 â€“ 1 = 0

a_{12}Â = 1 â€“ 2 = â€“ 1

a_{13}Â = 1 â€“ 3 = â€“ 2

a_{14}Â = 1 â€“ 4 = â€“ 3

a_{21}Â = 2 â€“ 1 = 1

a_{22}Â = 2 â€“ 2 = 0

a_{23}Â = 2 â€“ 3 = â€“ 1

a_{24 =}Â 2 â€“ 4 = â€“ 2

a_{31}Â = 3 â€“ 1 = 2

a_{32}Â = 3 â€“ 2 = 1

a_{33}Â = 3 â€“ 3 = 0

a_{34}Â = 3 â€“ 4 = â€“ 1

Substituting these values in matrix A we get,

A =Â

(iii) Given a_{i j}Â = 2i

Let A = [a_{i j}]_{2Ã—3}

So, the elements in a 3Ã—4 matrix are

a_{11}, a_{12}, a_{13}, a_{14,}Â a_{21}, a_{22}, a_{23}, a_{24}, a_{31, }a_{32, }a_{33, }a_{34}

A =Â

a_{11}Â = 2Ã—1 = 2

a_{12}Â = 2Ã—1 = 2

a_{13}Â = 2Ã—1 = 2

a_{14}Â = 2Ã—1 = 2

a_{21}Â = 2Ã—2 = 4

a_{22}Â = 2Ã—2 = 4

a_{23}Â = 2Ã—2 = 4

a_{24 =}Â 2Ã—2 = 4

a_{31}Â = 2Ã—3 = 6

a_{32}Â = 2Ã—3 = 6

a_{33}Â = 2Ã—3 = 6

a_{34}Â = 2Ã—3 = 6

Substituting these values in matrix A we get,

A =Â

(iv) Given a_{i j}Â = j

Let A = [a_{i j}]_{2Ã—3}

So, the elements in a 3Ã—4 matrix are

_{11}, a_{12}, a_{13}, a_{14,}Â a_{21}, a_{22}, a_{23}, a_{24}, a_{31, }a_{32, }a_{33, }a_{34}

A =Â

a_{11}Â = 1

a_{12}Â = 2

a_{13}Â = 3

a_{14}Â = 4

a_{21}Â = 1

a_{22}Â = 2

a_{23}Â = 3

a_{24 =}Â 4

a_{31}Â = 1

a_{32}Â = 2

a_{33}Â = 3

a_{34}Â = 4

Substituting these values in matrix A we get,

A =Â

(vi) Given a_{i j}Â = Â½ |-3i + j|

Let A = [a_{i j}]_{2Ã—3}

So, the elements in a 3Ã—4 matrix are

_{11}, a_{12}, a_{13}, a_{14,}Â a_{21}, a_{22}, a_{23}, a_{24}, a_{31, }a_{32, }a_{33, }a_{34}

A =Â

a_{11}Â =Â

a_{12}Â =Â

a_{13}Â =Â

a_{14}Â =Â

a_{21}Â =Â

a_{22}Â =Â

a_{23}Â =Â

a_{24 =}Â

a_{31}Â =Â

a_{32}Â =Â

a_{33}Â =Â

a_{34}Â =Â

Substituting these values in matrix A we get,

A =Â

Multiplying by negative sign we get,

**7. Construct a 4 Ã— 3 matrix A = [a _{i j}] whose elements a_{i j}Â are given by:**

**(i) a _{i j}Â = 2i + i/j**

**(ii) a _{i j}Â = (i â€“ j)/ (i + j)**

**(iii) a _{i j}Â = i**

**Solution:**

(i) Given a_{i j}Â = 2i + i/j

Let A = [a_{i j}]_{4Ã—3}

So, the elements in a 4 Ã— 3 matrix are

a_{11}, a_{12}, a_{13}, a_{21}, a_{22}, a_{23}, a_{31, }a_{32, }a_{33, }a_{41,}Â a_{42,}Â a_{43}

A =Â

a_{11}Â =Â

a_{12}Â =Â

a_{13}Â =Â

a_{21}Â =Â

a_{22}Â =Â

a_{23}Â =Â

a_{31}Â =Â

a_{32}Â =Â

a_{33}Â =Â

a_{41}Â =Â

a_{42}Â =Â

a_{43}Â =Â

Substituting these values in matrix A we get,

A =Â

(ii) Given a_{i j}Â = (i â€“ j)/ (i + j)

Let A = [a_{i j}]_{4Ã—3}

So, the elements in a 4 Ã— 3 matrix are

a_{11}, a_{12}, a_{13}, a_{21}, a_{22}, a_{23}, a_{31, }a_{32, }a_{33, }a_{41,}Â a_{42,}Â a_{43}

A =Â

a_{11}Â =Â

a_{12}Â =Â

a_{13}Â =Â

a_{21}Â =Â

a_{22}Â =Â

a_{23}Â =Â

a_{31}Â =Â

a_{32}Â =Â

a_{33}Â =Â

a_{41}Â =Â

a_{42}Â =Â

a_{43}Â =Â

Substituting these values in matrix A we get,

A =Â

(iii) Given a_{i j}Â = i

Let A = [a_{i j}]_{4Ã—3}

So, the elements in a 4 Ã— 3 matrix are

a_{11}, a_{12}, a_{13}, a_{21}, a_{22}, a_{23}, a_{31, }a_{32, }a_{33, }a_{41,}Â a_{42,}Â a_{43}

A =Â

a_{11}Â = 1

a_{12}Â = 1

a_{13}Â = 1

a_{21}Â = 2

a_{22}Â = 2

a_{23}Â = 2

a_{31}Â = 3

a_{32}Â = 3

a_{33}Â = 3

a_{41}Â = 4

a_{42}Â = 4

a_{43}Â = 4

Substituting these values in matrix A we get,

A =Â

**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 matrices are also equal.

Therefore by equating them we get,

3x + 4y = 2Â â€¦â€¦Â (1)

x â€“ 2y = 4Â â€¦â€¦Â (2)

a + b = 5Â â€¦â€¦ (3)

2a â€“ b = â€“ 5Â â€¦â€¦ (4)

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

3x + 4y + 2x â€“ 4y = 2 + 8

â‡’Â 5x = 10

â‡’Â x = 2

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

3 Ã— 2 + 4y = 2

â‡’Â 6 + 4y = 2

â‡’Â 4y = 2 â€“ 6

â‡’Â 4y = â€“ 4

â‡’Â y = â€“ 1

Now by adding equation (3) and (4)

a + b + 2a â€“ b = 5 + (â€“ 5)

â‡’Â 3a = 5 â€“ 5 = 0

â‡’Â a = 0

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

0 + b = 5

â‡’Â b = 5

âˆ´Â a = 0, b = 5, x = 2 and y = â€“ 1

**9. Find x, y, a and b if**

**Solution:**

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

Given that two matrices are equal.

Therefore by equating them we get,

2a + b = 4Â â€¦â€¦ (1)

And a â€“ 2b = â€“ 3Â â€¦â€¦Â (2)

And 5c â€“ d = 11Â â€¦â€¦Â (3)

4c + 3d = 24Â â€¦â€¦Â (4)

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

4a + 2b + a â€“ 2b = 8 â€“ 3

â‡’Â 5a = 5

â‡’Â a = 1

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

2 Ã— 1 + b = 4

â‡’Â 2 + b = 4

â‡’Â b = 4 â€“ 2

â‡’Â b = 2

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

15c â€“ 3d + 4c + 3d = 33 + 24

â‡’Â 19c = 57

â‡’Â c = 3

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

4 Ã— 3 + 3d = 24

â‡’Â 12 + 3d = 24

â‡’Â 3d = 24 â€“ 12

â‡’Â 3d = 12

â‡’Â d = 4

âˆ´Â 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 matrices are also equal.

Given that two matrices are equal.

Therefore by equating them we get,

2a + b = 4Â â€¦â€¦Â (1)

And a â€“ 2b = â€“ 3Â â€¦â€¦ (2)

And 5c â€“ d = 11Â â€¦â€¦Â (3)

4c + 3d = 24Â â€¦â€¦Â (4)

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

4a + 2b + a â€“ 2b = 8 â€“ 3

â‡’Â 5a = 5

â‡’Â a = 1

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

2 Ã— 1 + b = 4

â‡’Â 2 + b = 4

â‡’Â b = 4 â€“ 2

â‡’Â b = 2

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

15c â€“ 3d + 4c + 3d = 33 + 24

â‡’Â 19c = 57

â‡’Â c = 3

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

4 Ã— 3 + 3d = 24

â‡’Â 12 + 3d = 24

â‡’Â 3d = 24 â€“ 12

â‡’Â 3d = 12

â‡’Â d = 4

âˆ´Â a = 1, b = 2, c = 3 and d = 4