Here we provide RD Sharma Class 12 Ex 5.3 Solutions Chapter 5 Algebra of Matrices for English medium students, Which will very helpful for every student in their exams. Students can download the RD Sharma Class 12 Ex 5.3 Solutions Chapter 5 Algebra of Matrices book pdf download. Now you will get step-by-step solutions to each question.

Textbook | NCERT |

Class | Class 12th |

Subject | Maths |

Chapter | 5 |

Exercise | 5.3 |

Category | RD Sharma Solutions |

**RD Sharma Class 12 Ex 5.3 Solutions Chapter 5 Algebra of Matrices **

**Question 1. Compute the indicated products:**

**(i) **

**Solution:**

We have,

=

=

**(ii) **

**Solution:**

We have,

=

=

**(iii) **

**Solution:**

We have,

=

=

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

**(i) ****and **

**Solution:**

We have,

A =and B =

AB =

=

=

And we have,

BA =

=

=

Therefore, AB ≠ BA.

Hence, proved.

**(ii) ****and **

**Solution:**

We have,

A =and B =

AB =

=

=

And we have,

BA =

=

=

Therefore, AB ≠ BA.

Hence proved.

**(iii) ****and **

**Solution:**

We have,

A =and B =

AB =

=

=

And we have,

BA =

=

=

Therefore, AB ≠ BA.

Hence proved.

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

**(i) ****and **

**Solution:**

We have,

A =and B =

As A is of order 2 × 2 and B is of order 2 × 3, AB is possible but BA is not possible.

So, we get

AB =

=

=

**(ii) ****and **

**Solution:**

We have,

A =and B =

As A is of order 3 × 2 and B is of order 2 × 3, AB and BA both are possible.

So, we get,

AB =

=

=

Also we have,

BA =

=

=

**(iii) ****and **

**Solution:**

We have,

A =and B =

As A is of order 1 × 4 and B is of order 4 × 1, AB and BA both are possible.

So, we get,

AB =

=

=

=

Also, we have,

BA =

=

=

**(iv) **

**Solution:**

We have,

=

=

=

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

**(i) ****and **

**Solution:**

We have,

A =and B =

AB =

=

=

And we have,

BA =

=

=

Therefore, AB ≠ BA.

Hence proved.

**(ii) ****and **

**Solution:**

We have,

A =and B =

AB =

=

=

And we have,

BA =

=

=

Therefore, AB ≠ BA.

Hence proved.

**Question 5. Evaluate the following:**

**(i) **

**Solution:**

We have,

=

=

=

=

=

**(ii) **

**Solution:**

We have,

=

=

=

=

=

**(iii) **

**Solution:**

We have,

=

=

=

=

=

**Question 6. If A =****, B =****and C =****, then show that A**^{2} = B^{2} = C^{2} = I_{2}.

^{2}= B

^{2}= C

^{2}= I

_{2}.

**Solution:**

We have,

A =, B =and C =

A^{2} =

=

=

Therefore, A^{2} = I_{2}

B^{2} =

=

=

Therefore, B^{2} = I_{2}

C^{2} =

=

=

Therefore, C^{2} = I_{2}

So, we get A^{2} = B^{2} = C^{2} = I_{2}

**Hence proved.**

**Question 7. If A =****and B =****, find 3A**^{2} – 2B + I.

^{2}– 2B + I.

**Solution:**

We are given,

A =and B =

So, we get,

3A

^{2}– 2B + I ==

=

=

=

=

**Question 8. If A =****, prove that (A – 2I) (A – 3I) = 0.**

**Solution:**

We are given,

A =

L.H.S. = (A – 2I) (A – 3I)

=

=

=

=

=

=

= 0

= R.H.S.

Hence proved.

**Question 9. If A =****, show that A**^{2} =**and A**^{3} =**.**

^{2}=

^{3}=

**Solution:**

We have,

A =

So, A

^{2}==

=

Hence, A

^{3}= A^{2}. A=

=

=

Hence proved.

**Question 10. If A =****, show that A**^{2} = 0.

^{2}= 0.

**Solution:**

We have,

A =

So, we get

L.H.S. = A

^{2 }==

=

= 0

= R.H.S.

Hence proved.

**Question 11. If A =****, find A**^{2}.

^{2}.

**Solution:**

We have,

A =

So, we get

A

^{2}==

=

=

**Question 12. If A =****and B =****, show that AB = BA = O**_{3×3}.

_{3×3}.

**Solution:**

We have,

A =and B =

So, we get

AB =

=

=

= O

_{3×3}And we have,

BA =

=

=

= O

_{3×3}Therefore, AB = BA = O

_{3×3}.

Hence proved.

**Question 13. If A =****and B =****, show that AB = BA = O**_{3×3}.

_{3×3}.

**Solution:**

We have,

A =and B =

So, we have,

AB =

=

=

And we have,

BA =

=

=

= O

_{3×3}Therefore, AB = BA = O

_{3×3}.

Hence proved.

**Question 14. If A =****and B =****, show that AB = A and BA = B.**

**Solution:**

We have,

A =and B =

AB =

=

=

= A

And we have,

BA =

=

=

= B

Hence proved.

**Question 15. If A =****and B =****, compute A**^{2} – B^{2}.

^{2}– B

^{2}.

**Solution:**

We have,

A =and B =

A

^{2}==

=

And we have,

B

^{2}==

=

So, we get

A

^{2}– B^{2}==

=

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

**(i) A =****, B =****, C =**

**Solution:**

We are given,

A =, B =, C =

L.H.S. = (AB) C

=

=

=

=

And R.H.S. = A (BC)

=

=

=

=

=

= L.H.S.

Hence proved.

**(ii) A =****, B =****, C =**

**Solution:**

We are given,

A =, B =, C =

L.H.S. = (AB) C

=

=

=

=

=

And R.H.S. = A (BC)

=

=

=

=

=

= L.H.S.

Hence proved.

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

**(i) A =****, B =****, C =**

**Solution:**

We have,

A =, B =, C =

L.H.S. = A (B + C)

=

=

=

=

=

R.H.S. = AB + AC

=

=

=

=

=

= L.H.S.

Hence proved.

**(ii) A =****, B =****, C =**

**Solution:**

We have,

A =, B =, C =

L.H.S. = A (B + C)

=

=

=

=

=

R.H.S. = AB + AC

=

=

=

=

=

= L.H.S.

Hence proved.

**Question 18. If A =****, B =****and C =****, show that A (B – C) = AB –** **AC.**

**Solution:**

We have,

A =, B =and C =

L.H.S. = A (B – C)

=

=

=

=

=

R.H.S. = AB – AC

=

=

=

=

=

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

_{43}and a

_{22}of the matrix:

**A =**

**Solution:**

We are given,

A =

=

=

=

=

Therefore, a_{43}= 8 and a_{22}= 0.

**Question 20. If A =****and I is the identity matrix of order 3, show that A**^{3} = pI + qA + rA^{2}.

^{3}= pI + qA + rA

^{2}.

**Solution:**

We have,

A =

L.H.S. = A

^{3}=

=

=

=

=

And R.H.S. = pI + qA + rA

^{2}=

=

=

=

=

=

= L.H.S.

Hence proved.

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

**Solution:**

We have,

L.H.S. =

=

=

=

=

=

=

= R.H.S.

Hence proved.

**Question 22. If A =****, prove that A**^{2} = A.

^{2}= A.

**Solution:**

We have,

A =

So, A

^{2}==

=

= A

Hence proved.

**Question 23. If A =****, show that A**^{2} = I_{3}.

^{2}= I

_{3}.

**Solution:**

We have,

A =

So, A

^{2}==

=

= I

_{3}

Hence proved.

**Question 24.**

**(i) If****= 0, find x.**

**Solution:**

We have,

=>= 0

=>= 0

=>= 0

=>= 0

=> [3x + 6] = 0

=> 3x = –6

=> x = –6/3

=> x = –2

Therefore, the value of x is –2.

**(ii) If****= 0, find x.**

**Solution:**

We have,

=>

=>

=>

On comparing the above matrix we get,

x = 13

Therefore, the value of x is –13.

**Question 25. If****, find x.**

**Solution:**

We have,

=>

=>

=>

=>

=> 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

Therefore, the value of x is –1 or –2.

**Question 26. If****= 0, find x.**

**Solution:**

We have,

=>= 0

=>

=>

=>

=> 2x – 4 = 0

=> 2x = 4

=> x = 2

Therefore, the value of x is 2.

**Question 27. If A =****and I =****, then prove that A**^{2} – A + 2I = 0.

^{2}– A + 2I = 0.

**Solution:**

We have,

A =

A

^{2}==

=

L.H.S. = A

^{2}– A + 2I=

=

=

=

=

=

= 0

= R.H.S.

Hence proved.

**Question 28. If A =****and I =****, then find **λ** so that A**^{2} = 5A + λI.

^{2}= 5A + λI.

**Solution:**

We have,

A =

A

^{2}==

=

We are given,

=> A

^{2}= 5A + λI=>

=>

=>

=>

On comparing both sides, we get

=> 8 = 15 + λ

=> λ = –7

Therefore, the value of λ is –7.

**Question 29. If A =****, show that A**^{2} – 5A + 7I_{2} = 0.

^{2}– 5A + 7I

_{2}= 0.

**Solution:**

We have,

A =

A2 =

=

=

L.H.S. = A

^{2}– 5A + 7I_{2}=

=

=

=

= 0

= R.H.S.

Hence proved.

**Question 30. If A =****, show that A**^{2} – 2A + 3I_{2} = 0.

^{2}– 2A + 3I

_{2}= 0.

**Solution:**

We have,

A =

A

^{2}==

=

L.H.S. = A

^{2}– 2A + 3I_{2}=

=

=

=

= 0

= R.H.S.

Hence proved.

**Question 31. Show that the matrix A =****satisfies the equation A**^{3} – 4A^{2} + A = 0.

^{3}– 4A

^{2}+ A = 0.

**Solution:**

We have,

A =

A

^{2}==

=

A

^{3}= A^{2}. A=

=

=

L.H.S. = A

^{3}– 4A^{2}+ A=

=

=

=

= 0

= R.H.S.

Hence proved.

**Question 32. Show that the matrix A =****is root of the equation A**^{2 }– 12A – I = 0

^{2 }– 12A – I = 0

**Solution:**

We have,

A =

A

^{2}==

=

L.H.S. = A

^{2}– 12A – I=

=

=

=

= 0

= R.H.S.

Hence proved.

**Question 33. If A =****find A**^{2} – 5A – 14I.

^{2}– 5A – 14I.

**Solution:**

We have,

A =

A

^{2}==

=

A

^{2}– 5A – 14I ==

=

=

**Question 34. If A =****, find A**^{2} – 5A + 7I = 0. Use this to find A^{4}.

^{2}– 5A + 7I = 0. Use this to find A

^{4}.

**Solution:**

We have,

A =

A

^{2}==

=

L.H.S. = A

^{2}– 5A + 7I = 0=

=

=

=

= 0

= R.H.S.

Hence proved.Now we have A

^{2}– 5A + 7I = 0=> A

^{2}= 5A – 7I=> A

^{4}= (5A – 7I) (5A – 7I)=> A

^{4}= 25A^{2}– 35AI – 35AI + 49I=> A

^{4}= 25A^{2}– 70AI + 49I=> A

^{4}= 25 (5A – 7I) – 70AI + 49I=> A

^{4}= 125A – 175I – 70A + 49I=> A

^{4}= 55A – 126I=> A

^{4}==> A

^{4}==> A

^{4}==> A

^{4}=

**Question 35. If A =****, find k such that A**^{2 }= kA – 2I_{2}.

^{2 }= kA – 2I

_{2}.

**Solution:**

We have,

A =

A

^{2}==

=

We are given,

=> A

^{2 }= kA – 2I_{2}=>

=>

=>

On comparing both sides, we get

=> 3k – 2 = 1

=> 3k = 3

=> k = 1

Therefore, the value of k is 1.

**Question 36. If A =****, find k such that A**^{2} – 8A + kI = 0.

^{2}– 8A + kI = 0.

**Solution:**

We have,

A =

A

^{2}==

=

We are given,

=> A

^{2}– 8A + kI = 0=>

=>

=>

=>

On comparing both sides, we get

=> –k + 7 = 0

=> k = 7

Therefore, the value of k is 7.

**Question 37. If A =****and f(x) = x**^{2} – 2x – 3, show that f(A) = 0.

^{2}– 2x – 3, show that f(A) = 0.

**Solution:**

We have,

A =and f(x) = x

^{2}– 2x – 3A

^{2}==

=

L.H.S. = f(A) = A

^{2}– 2A – 3I_{2}=

=

=

=

= 0

= R.H.S.

Hence proved.

**Question 38. If A =****and I =****, find λ, μ so that A**^{2} = λA + μI.

^{2}= λA + μI.

**Solution:**

We have,

A =

A

^{2}==

=

We are given,

=> A

^{2}= λA + μI=>

=>

=>

=>

On comparing both sides, we get,

=> 2λ + μ = 7 and λ = 4

=> 2(4) + μ = 7

=> μ = 7 – 8

=> μ = –1

Therefore, the value of λ is 4 and μ is –1.

**Question 39. Find the value of x for which the matrix product****equals an identity matrix.**

**Solution:**

We have,

=>

=>

=>

On comparing both sides, we get,

=> 5x = 1

=> x = 1/5

Therefore, the value of x is 1/5.

**Question 40. Solve the following matrix equations:**

**(i)**

**Solution:**

We have,

=>

=>

=>

=>

=> x

^{2}– 2x – 15 = 0=> x

^{2}– 5x + 3x – 15 = 0=> x (x – 5) + 3 (x – 5) = 0

=> (x – 5) (x + 3) = 0

=> x = 5 or –3

Therefore, the value of x is 5 or –3.

**(ii)**

**Solution:**

We have,

=>

=>

=>

=>

=> 4 + 4x = 0

=> 4x = –4

=> x = –1

Therefore, the value of x is –1.

**(iii)**

**Solution:**

We have,

=>

=>

=>

=>

=> x

^{2}– 48 = 0=> x

^{2}= 48=> x = ±4√3

Therefore, the value of x is ±4√3.

**(iv)**

**Solution:**

We have,

=>

=>

=>

=>

=> 2x

^{2}+ 23x = 0=> x (2x + 23) = 0

=> x = 0 or x = –23/2

Therefore, the value of x is 0 or –23/2.

**Question 41. If A =****, compute A**^{2} – 4A + 3I_{3}.

^{2}– 4A + 3I

_{3}.

**Solution:**

We have,

A =

A

^{2}==

=

So, A

^{2}– 4A + 3I_{3}==

=

=

**Question 42. If f(x) = x**^{2} – 2x, find f(A), where A =**.**

^{2}– 2x, find f(A), where A =

**Solution:**

We have,

A =and f(x) = x

^{2}– 2xA

^{2}==

=

So, f(A) = A

^{2}– 2A=

=

=

=

**Question 43. If f(x) = x**^{3} + 4x^{2} – x, find f(A) where A =**.**

^{3}+ 4x

^{2}– x, find f(A) where A =

**Solution:**

We have,

A =and f(x) = x

^{3}+ 4x^{2}– xA

^{2}==

=

A

^{3}= A^{2}. A=

=

=

Now, f(A) = A

^{3}+ 4A^{2}– A=

=

=

=

**Question 44. If A =****, then show that A is a root of the polynomial f(x) = x**^{3} – 6x^{2} + 7x +2.

^{3}– 6x

^{2}+ 7x +2.

**Solution:**

We have,

A =and f(x) = x

^{3}– 6x^{2}+ 7x +2.A

^{2}==

=

A

^{3}= A^{2}. A=

=

=

In order to show that A is a root of above polynomial, we need to prove that f(A) = 0.

Now, f(A) = A

^{3}– 6A^{2}+ 7A + 2I=

=

=

=

= 0

Hence proved.

**Question 45. If A =****, prove that A**^{2} – 4A – 5I = 0.

^{2}– 4A – 5I = 0.

**Solution:**

We have,

A =

A

^{2}==

=

Now, L.H.S. = A

^{2}– 4A – 5I=

=

=

=

= 0

= R.H.S.

Hence proved.

**Question 46. If A =****, show that A**^{2} – 7A + 10I_{3} = 0.

^{2}– 7A + 10I

_{3}= 0.

**Solution:**

We have,

A =

A

^{2}==

=

Now, L.H.S. = A

^{2}– 7A + 10I_{3}=

=

=

=

= 0

= R.H.S.

Hence proved.

**Question 47. Without using the concept of inverse of a matrix, find the matrix****such that,**

**Solution:**

We have,

=>

=>

On comparing both sides, we get,

5x – 7z = –16

5y – 7u = –6

–2x + 3z = 7

–2y + 3u = 2

On solving the above equations, we get

=> x = 1, y = –4, z = 3 and u = –2.

So, we get.

**Question 48. Find the matrix A such that**

**(i)**

**Solution:**

Let A =

Given equation is,

=>

=>

=>

=>

On comparing both sides, we get, a = 1, b = 0 and c = 1.

And x + 1 = 3 => x = 2

Also, y = 3 and

z + 1 = 5 => z = 4

So, we have A =

**(ii)**

**Solution:**

Let A =

Given equation is,

=>

=>

=>

On comparing both sides, we get,

w + 4x = 7

2w + 5x = –6

y + 4z = 2

2y + 5z = 4

On solving the above equations, we get

=> x = –2, y = 2, w = 1 and z = 0.

So, we get A =

**(iii)**

**Solution:**

Let A =

Given equation is,

=>

=>

=>

On comparing both sides, we get,

=> 4x = – 4, 4y = 8 and 4z = 4.

=> x = –1, y = 2 and z = 1.

So, we get A =

**(iv)**

**Solution:**

We have,

A =

A =

A =

A =

**(v)**

**Solution:**

Let A =

Given equation is,

=>

=>

=>

On comparing both sides, we get,

=> x = 1, y = –2 and z = –5

And also we have,

2x – a = –1

2y – b = –8

2z – c = –10

On solving these, we get,

=> a = 3, b = 4 and c = 0.

So, we get A =

**(vi)**

**Solution:**

Let A =

Given equation is,

=>

=>

=>

On comparing both sides, we get

x + 4a = –7 and 2x + 5a = –8

=> x = 1 and a = –2

y + 4b = 2 and 2y + 5b = 4

=> b = 0 and y = 2

z + 4c = 11 and 2z + 5c = 10

=> c = 4 and z = –5

So, we get A =

**Question 49. Find a 2 × 2 matrix A such that****= 6I**_{2}.

_{2}.

**Solution:**

Let A =

Given equation is,

=>= 6I

=>

=>

=>

On comparing both sides, we get

w + x = 6 and –2w + 4x = 0

=> w = 4 and x = 2

y + z = 0 and –2y + 4z = 6

=> y = –1 and z = 1

So, we get A =

**Question 50. If A =****, find A**^{16}.

^{16}.

**Solution:**

We have,

A =

A

^{2}==

=

A

^{16}= A^{2}A^{2}A^{2}A^{2}=

=

**Question 51. If A =****, B =****and x**^{2} = –1, then show that (A + B)^{2} = A^{2} + B^{2}.

^{2}= –1, then show that (A + B)

^{2}= A

^{2}+ B

^{2}.

**Solution:**

We have,

A =, B =and x

^{2}= –1L.H.S. = (A + B)

^{2}=

=

=

=

=

=

=

=

R.H.S. = A

^{2}+ B^{2}=

=

=

=

=

=

= L.H.S.

Hence proved.

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