## RD Sharma Solutions for Class 12 Maths Chapter 5 – Free PDF Download Updated for (2022-23)

**RD Sharma Solutions for Class 12 Maths Chapter 5 – Algebra of Matrices** is provided here. In order to have a good academic score in Mathematics, the important thing to be done by the students is to solve the questions of each and every exercise. RD Sharma Solutions for Class 12 is prepared by a team of experts who work with their full potential to help the students excel in their exams. Students who practice these solutions on a regular basis undoubtedly enhance their skills which are essential from an exam point of view.

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Exercise 5.1 Page No: 5.6

**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 equation (1) and (2)

a_{22} = 4 and b_{21} = – 3

a_{22} + b_{21} = 4 + (– 3) = 1

(ii)

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

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

a_{11} b_{11} + a_{22} b_{22} = 2 × 2 + 4 × 4 = 4 + 16 = 20

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

**Solution:**

Given A be a matrix of order 3 × 4.

So, A = [a_{i j}] _{3×4}

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

So, order of matrix R_{1} = 1 × 4

C_{2} = second column of

Therefore order of C_{2} = 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

[a_{11}, a

_{12}, a

_{13}, a

_{21}, a

_{22}, a

_{23}]

a_{11} = 1 × 1 = 1

a_{12} = 1 × 2 = 2

a_{13} = 1 × 3 = 3

a_{21} = 2 × 1 = 2

a_{22} = 2 × 2 = 4

a_{23} = 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

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

a_{11} = 2 × 1 – 1 = 2 – 1 = 1

a_{12} = 2 × 1 – 2 = 2 – 2 = 0

a_{13} = 2 × 1 – 3 = 2 – 3 = – 1

a_{21} = 2 × 2 – 1 = 4 – 1 = 3

a_{22} = 2 × 2 – 2 = 4 – 2 = 2

a_{23} = 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

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

a_{11} = 1 + 1 = 2

a_{12} = 1 + 2 = 3

a_{13} = 1 + 3 = 4

a_{21} = 2 + 1 = 3

a_{22} = 2 + 2 = 4

a_{23} = 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

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

Let A = [a_{i j}]_{2×3}

So, the elements in a 2 × 3 matrix are

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

a_{11} =

a_{12} =

a_{13} =

a_{21} =

a_{22} =

a_{23} =

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

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

a_{11} =

a_{12} =

a_{21} =

a_{22} =

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

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

a_{11} =

a_{12} =

a_{21} =

a_{22} =

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

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

a_{11} =

a_{12} =

a_{21} =

a_{22} =

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

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

a_{11} =

a_{12} =

a_{21} =

a_{22} =

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

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

a_{11} =

a_{12} =

a_{21} =

a_{22} =

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

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

a_{11} =

a_{12} =

a_{21} =

a_{22} =

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

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

a_{11} =

a_{12} =

a_{21} =

a_{22} =

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

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

A =

a_{11} = 1 + 1 = 2

a_{12} = 1 + 2 = 3

a_{13} = 1 + 3 = 4

a_{14} = 1 + 4 = 5

a_{21} = 2 + 1 = 3

a_{22} = 2 + 2 = 4

a_{23} = 2 + 3 = 5

a_{24 =} 2 + 4 = 6

a_{31} = 3 + 1 = 4

a_{32} = 3 + 2 = 5

a_{33} = 3 + 3 = 6

a_{34} = 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

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

A =

a_{11} = 1 – 1 = 0

a_{12} = 1 – 2 = – 1

a_{13} = 1 – 3 = – 2

a_{14} = 1 – 4 = – 3

a_{21} = 2 – 1 = 1

a_{22} = 2 – 2 = 0

a_{23} = 2 – 3 = – 1

a_{24 =} 2 – 4 = – 2

a_{31} = 3 – 1 = 2

a_{32} = 3 – 2 = 1

a_{33} = 3 – 3 = 0

a_{34} = 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

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

A =

a_{11} = 2×1 = 2

a_{12} = 2×1 = 2

a_{13} = 2×1 = 2

a_{14} = 2×1 = 2

a_{21} = 2×2 = 4

a_{22} = 2×2 = 4

a_{23} = 2×2 = 4

a_{24 =} 2×2 = 4

a_{31} = 2×3 = 6

a_{32} = 2×3 = 6

a_{33} = 2×3 = 6

a_{34} = 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

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

A =

a_{11} = 1

a_{12} = 2

a_{13} = 3

a_{14} = 4

a_{21} = 1

a_{22} = 2

a_{23} = 3

a_{24 =} 4

a_{31} = 1

a_{32} = 2

a_{33} = 3

a_{34} = 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

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

A =

a_{11} =

a_{12} =

a_{13} =

a_{14} =

a_{21} =

a_{22} =

a_{23} =

a_{24 =}

a_{31} =

a_{32} =

a_{33} =

a_{34} =

Substituting these values in matrix A we get,

A =

Multiplying by 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

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

A =

a_{11} =

a_{12} =

a_{13} =

a_{21} =

a_{22} =

a_{23} =

a_{31} =

a_{32} =

a_{33} =

a_{41} =

a_{42} =

a_{43} =

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

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

A =

a_{11} =

a_{12} =

a_{13} =

a_{21} =

a_{22} =

a_{23} =

a_{31} =

a_{32} =

a_{33} =

a_{41} =

a_{42} =

a_{43} =

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

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

A =

a_{11} = 1

a_{12} = 1

a_{13} = 1

a_{21} = 2

a_{22} = 2

a_{23} = 2

a_{31} = 3

a_{32} = 3

a_{33} = 3

a_{41} = 4

a_{42} = 4

a_{43} = 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 matrices are also equal.

Therefore 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)

3 × 2 + 4y = 2

⇒ 6 + 4y = 2

⇒ 4y = 2 – 6

⇒ 4y = – 4

⇒ y = – 1

Now by adding equation (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 matrices 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 matrices 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

Exercise 5.2 Page No: 5.18

**1. Compute the following sums:**

**Solution:**

(i) Given

Corresponding elements of two matrices should be added

Therefore, we get

Therefore,

(ii) Given

Therefore,

** **

**Find each of the following:**

**(i) 2A – 3B**

**(ii) B – 4C**

**(iii) 3A – C**

**(iv) 3A – 2B + 3C**

**Solution:**

(i) Given

First we have to compute 2A

Now by computing 3B we get,

Now by we have to compute 2A – 3B we get

Therefore

(ii) Given

First we have to compute 4C,

Now,

Therefore we get,

(iii) Given

First we have to compute 3A,

Now,

Therefore,

(iv) Given

First we have to compute 3A

Now we have to compute 2B

By computing 3C we get,

Therefore,

**(i) A + B and B + C**

**(ii) 2B + 3A and 3C – 4B**

**Solution:**

(i) Consider A + B,

A + B is not possible because matrix A is an order of 2 x 2 and Matrix B is an order of 2 x 3, so the Sum of the matrix is only possible when their order is same.

Now consider B + C

(ii) Consider 2B + 3A

2B + 3A also does not exist because the order of matrix B and matrix A is different, so we cannot find the sum of these matrix.

Now consider 3C – 4B,

**Solution:**

Given

Now we have to compute 2A – 3B + 4C

**5. If A = diag (2 -5 9), B = diag (1 1 -4) and C = diag (-6 3 4), find**

**(i) A – 2B**

**(ii) B + C – 2A**

**(iii) 2A + 3B – 5C**

**Solution:**

(i) Given A = diag (2 -5 9), B = diag (1 1 -4) and C = diag (-6 3 4)

(ii) Given A = diag (2 -5 9), B = diag (1 1 -4) and C = diag (-6 3 4)

We have to find B + C – 2A

Here,

Now we have to compute B + C – 2A

(iii) Given A = diag (2 -5 9), B = diag (1 1 -4) and C = diag (-6 3 4)

Now we have to find 2A + 3B – 5C

Here,

Now consider 2A + 3B – 5C

**6. Given the matrices**

**Verify that (A + B) + C = A + (B + C)**

**Solution:**

Given

Now we have to verify (A + B) + C = A + (B + C)

First consider LHS, (A + B) + C,

Now consider RHS, that is A + (B + C)

Therefore LHS = RHS

Hence (A + B) + C = A + (B + C)

**7. Find the matrices X and Y, **

**Solution:**

Consider,

Now by simplifying we get,

Therefore,

Again consider,

Now by simplifying we get,

Therefore,

**Solution:**

Given

Now by transposing, we get

Therefore,

**Solution:**

Given

Now by multiplying equation (1) and (2) we get,

Now by adding equation (2) and (3) we get,

Now by substituting X in equation (2) we get,

**Solution:**

Consider

Now, again consider

Therefore,

And

Exercise 5.3 Page No: 5.41

**1. Compute the indicated products:**

**Solution:**

(i) Consider

On simplification we get,

(ii) Consider

On simplification we get,

(iii) Consider

On simplification we get,

**2. Show that AB ≠ BA in each of the following cases:**

**Solution:**

(i) Consider,

Again consider,

From equation (1) and (2), it is clear that

AB ≠ BA

(ii) Consider,

Now again consider,

From equation (1) and (2), it is clear that

AB ≠ BA

(iii) Consider,

Now again consider,

From equation (1) and (2), it is clear that

AB ≠ BA

**3. Compute the products AB and BA whichever exists in each of the following cases:**

**Solution:**

(i) Consider,

BA does not exist

Because the number of columns in B is greater than the rows in A

(ii) Consider,

Again consider,

(iii) Consider,

AB = [0 + (-1) + 6 + 6]

AB = 11

Again consider,

(iv) Consider,

**4. Show that AB ≠ BA in each of the following cases:**

**Solution:**

(i) Consider,

Again consider,

From equation (1) and (2), it is clear that

AB ≠ BA

(ii) Consider,

Again consider,

From equation (1) and (2) it is clear that,

AB ≠ BA

**5. Evaluate the following:**

**Solution:**

(i) Given

First we have to add first two matrix,

On simplifying, we get

(ii) Given,

First we have to multiply first two given matrix,

= 82

(iii) Given

First we have subtract the matrix which is inside the bracket,

**Solution:**

Given

We know that,

Again we know that,

Now, consider,

We have,

Now, from equation (1), (2), (3) and (4), it is clear that A^{2 }= B^{2}= C^{2}= I_{2}

**Solution:**

Given

Consider,

Now we have to find,

**Solution:**

Given

Consider,

Hence the proof.

**Solution:**

Given,

Consider,

Again consider,

Hence the proof.

**Solution:**

Given,

Consider,

Hence the proof.

**Solution:**

Given,

Consider,

We know that,

Again we have,

**Solution:**

Given,

Consider,

Again consider,

From equation (1) and (2) AB = BA = 0_{3×3}

**Solution:**

Given

Consider,

Again consider,

From equation (1) and (2) AB = BA = 0_{3×3}

**Solution:**

Given

Now consider,

Therefore AB = A

Again consider, BA we get,

Hence BA = B

Hence the proof.

**Solution:**

Given,

Consider,

Now again consider, B^{2}

Now by subtracting equation (2) from equation (1) we get,

**16. For the following matrices verify the associativity of matrix multiplication i.e. (AB) C = A (BC)**

**Solution:**

(i) Given

Consider,

Now consider RHS,

From equation (1) and (2), it is clear that (AB) C = A (BC)

(ii) Given,

Consider the LHS,

Now consider RHS,

From equation (1) and (2), it is clear that (AB) C = A (BC)

**17. For the following matrices verify the distributivity of matrix multiplication over matrix addition i.e. A (B + C) = AB + AC.**

**Solution:**

(i) Given

Consider LHS,

Now consider RHS,

From equation (1) and (2), it is clear that A (B + C) = AB + AC

(ii) Given,

Consider the LHS

Now consider RHS,

**Solution:**

Given,

Consider the LHS,

Now consider RHS

From the above equations LHS = RHS

Therefore, A (B – C) = AB – AC.

**19. Compute the elements a _{43} and a_{22} of the matrix:**

**Solution:**

Given

From the above matrix, a_{43} = 8and a_{22} = 0

**Solution:**

Given

Consider,

Again consider,

Now, consider the RHS

Therefore, A^{3} = p I + q A + rA^{2}

Hence the proof.

**21. If ω is a complex cube root of unity, show that**

**Solution:**

Given

It is also given that ω is a complex cube root of unity,

Consider the LHS,

We know that 1 + ω + ω^{2} = 0 and ω^{3} = 1

Now by simplifying we get,

Again by substituting 1 + ω + ω^{2} = 0 and ω^{3} = 1 in above matrix we get,

Therefore LHS = RHS

Hence the proof.

**Solution:**

Given,

Consider A^{2}

Therefore A^{2} = A

**Solution:**

Given

Consider A^{2},

Hence A^{2} = I_{3}

**Solution:**

(i) Given

= [2x + 1 + 2 + x + 3] = 0

= [3x + 6] = 0

= 3x = -6

x = -6/3

x = -2

(ii) Given,

On comparing the above matrix we get,

x = 13

**Solution:**

Given

⇒ [(2x + 4) x + 4 (x + 2) – 1(2x + 4)] = 0

⇒ 2x^{2} + 4x + 4x + 8 – 2x – 4 = 0

⇒ 2x^{2} + 6x + 4 = 0

⇒ 2x^{2} + 2x + 4x + 4 = 0

⇒ 2x (x + 1) + 4 (x + 1) = 0

⇒ (x + 1) (2x + 4) = 0

⇒ x = -1 or x = -2

Hence, x = -1 or x = -2

**Solution:**

Given

By multiplying we get,

**Solution:**

Given

Now we have to prove A^{2} – A + 2 I = 0

**Solution:**

Given

**Solution:**

Given

Hence the proof.

**Solution:**

Given

Hence the proof.

**Solution:**

Given

**Solution:**

Given

**Solution:**

Given

**Solution:**

Given

**Solution:**

Given

**Solution:**

Given

I is identity matrix, so

Also given,

Now, we have to find A^{2}, we get

Now, we will find the matrix for 8A, we get

So,

Substitute corresponding values from eqn (i) and (ii), we get

And to satisfy the above condition of equality, the corresponding entries of the matrices should be equal

Hence,

Therefore, the value of k is 7

**Solution:**

Given

To show that f (A) = 0

Substitute x = A in f(x), we get

I is identity matrix, so

Now, we will find the matrix for A^{2}, we get

Now, we will find the matrix for 2A, we get

Substitute corresponding values from eqn (ii) and (iii) in eqn (i), we get

So,

Hence Proved

**Solution:**

Given

So

Now, we will find the matrix for A^{2}, we get

Now, we will find the matrix for λ A, we get

But given, A^{2} = λ A + μ I

Substitute corresponding values from equation (i) and (ii), we get

And to satisfy the above condition of equality, the corresponding entries of the matrices should be equal

Hence, λ + 0 = 4 ⇒ λ = 4

And also, 2λ + μ = 7

Substituting the obtained value of λ in the above equation, we get

2(4) + μ = 7 ⇒ 8 + μ = 7 ⇒ μ = – 1

Therefore, the value of λ and μ are 4 and – 1 respectively

**39. Find the value of x for which the matrix product**

**Solution:**

We know,

is identity matrix of size 3.

So according to the given criteria

Now we will multiply the two matrices on LHS using the formula c_{ij} = a_{i1}b_{1j} + a_{i2}b_{2j} + … + a_{in }b_{nj}, we get

And to satisfy the above condition of equality, the corresponding entries of the matrices should be equal

So we get

So the value of x is

Exercise 5.4 Page No: 5.54

**(i) (2A) ^{T} = 2 A^{T}**

**(ii) (A + B) ^{T} = A^{T} + B^{T}**

**(iii) (A – B) ^{T} = A^{T} – B^{T}**

**(iv) (AB) ^{T} = B^{T} A^{T}**

**Solution:**

(i) Given

Consider,

Put the value of A

L.H.S = R.H.S

(ii) Given

Consider,

L.H.S = R.H.S

Hence proved.

(iii) Given

Consider,

L.H.S = R.H.S

(iv) Given

So,

**Solution:**

Given

L.H.S = R.H.S

So,

**(i) A + B) ^{T} = A^{T} + B^{T}**

**(ii) (AB) ^{T} = B^{T} A^{T}**

**(iii) (2A) ^{T} = 2 A^{T}**

**Solution:**

(i) Given

Consider,

L.H.S = R.H.S

So,

(ii) Given

Consider,

L.H.S = R.H.S

So,

(iii) Given

Consider,

L.H.S = R.H.S

So,

**Solution:**

Given

Consider,

L.H.S = R.H.S

So,

**Solution:**

Given

Now we have to find (AB)^{T}

So,

Exercise 5.5 Page No: 5.60

**Solution:**

Given

Consider,

… (i)

… (ii)

From (i) and (ii) we can see that

A skew-symmetric matrix is a square matrix whose transpose equal to its negative, that is,

X = – X^{T}

So, A – A^{T} is a skew-symmetric.

**Solution:**

Given

Consider,

… (i)

… (ii)

From (i) and (ii) we can see that

A skew-symmetric matrix is a square matrix whose transpose equals its negative, that is,

X = – X^{T}

So, A – A^{T} is a skew-symmetric matrix.

**Solution:**

Given,

is a symmetric matrix.

We know that A = [a_{ij}]_{m × n }is a symmetric matrix if a_{ij }= a_{ji}

So,

Hence, x = 4, y = 2, t = -3 and z can have any value.

**4. Let. Find matrices X and Y such that X + Y = A, where X is a symmetric and y is a skew-symmetric matrix.**

**Solution:**

Given,

Then

Now,

Now,

X is a symmetric matrix.

Now,

-Y ^{T} = Y

Y is a skew symmetric matrix.

Hence, X + Y = A

### Also, access exercises of RD Sharma Solutions for Class 12 Maths Chapter 5 Algebra of Matrices

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

Let us have a look at some of the important concepts that are discussed in this chapter.

- Definition and meaning of matrix
- Types of matrices
- Equality of matrices
- Addition of matrices
- Properties of matrix addition
- Multiplication of a matrix by a scalar
- Properties of scalar multiplication
- Subtraction of matrices
- Multiplication of matrices
- Properties of matrix multiplication
- Transpose of a matrix
- Properties of transpose
- Symmetric and skew-symmetric matrices

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