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

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.

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

(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.

(i)

Solution:

We have,

=

=

=

=

=

(ii)

Solution:

We have,

=

=

=

=

=

(iii)

Solution:

We have,

=

=

=

=

=

### Question 6. If A =, B =and C =, then show that A2 = B2 = C2 = I2.

Solution:

We have,

A =, B =and C =

A2 =

=

=

Therefore, A2 = I2

B2 =

=

=

Therefore, B2 = I2

C2 =

=

=

Therefore, C2 = I2

So, we get A2 = B2 = C2 = I2

Hence proved.

Solution:

We are given,

A =and B =

So, we get,

3A2 – 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 A2 =and A3 =.

Solution:

We have,

A =

So, A2 =

=

=

Hence, A3 = A2 . A

=

=

=

Hence proved.

Solution:

We have,

A =

So, we get

L.H.S. = A=

=

=

= 0

= R.H.S.

Hence proved.

Solution:

We have,

A =

So, we get

A2 =

=

=

=

### Question 12. If A =and B =, show that AB = BA = O3×3.

Solution:

We have,

A =and B =

So, we get

AB =

=

=

= O3×3

And we have,

BA =

=

=

= O3×3

Therefore, AB = BA = O3×3.

Hence proved.

### Question 13. If A =and B =, show that AB = BA = O3×3.

Solution:

We have,

A =and B =

So, we have,

AB =

=

=

And we have,

BA =

=

=

= O3×3

Therefore, AB = BA = O3×3.

Hence proved.

Solution:

We have,

A =and B =

AB =

=

=

= A

And we have,

BA =

=

=

= B

Hence proved.

Solution:

We have,

A =and B =

A2 =

=

=

And we have,

B2 =

=

=

So, we get

A2 – B2 =

=

=

### 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 a43 and a22 of the matrix:

A =

Solution:

We are given,

A =

=

=

=

=

Therefore, a43 = 8 and a22 = 0.

### Question 20. If A =and I is the identity matrix of order 3, show that A3 = pI + qA + rA2.

Solution:

We have,

A =

L.H.S. = A3

=

=

=

=

=

And R.H.S. = pI + qA + rA2

=

=

=

=

=

=

= L.H.S.

Hence proved.

Solution:

We have,

L.H.S. =

=

=

=

=

=

=

= R.H.S.

Hence proved.

Solution:

We have,

A =

So, A2 =

=

=

= A

Hence proved.

Solution:

We have,

A =

So, A2 =

=

=

= I3

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,

=>

=>

=>

=>

=> 2x2 + 4x + 4x + 8 – 2x – 4 = 0

=> 2x2 + 6x + 4 = 0

=> 2x2 + 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 A2 – A + 2I = 0.

Solution:

We have,

A =

A2 =

=

=

L.H.S. = A2 – A + 2I

=

=

=

=

=

=

= 0

= R.H.S.

Hence proved.

### Question 28. If A =and I =, then find λ so that A2 = 5A + λI.

Solution:

We have,

A =

A2 =

=

=

We are given,

=> A2 = 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 A2 – 5A + 7I2 = 0.

Solution:

We have,

A =

A2 =

=

=

L.H.S. = A2 – 5A + 7I2

=

=

=

=

= 0

= R.H.S.

Hence proved.

### Question 30. If A =, show that A2 – 2A + 3I2 = 0.

Solution:

We have,

A =

A2 =

=

=

L.H.S. = A2 – 2A + 3I2

=

=

=

=

= 0

= R.H.S.

Hence proved.

### Question 31. Show that the matrix A =satisfies the equation A3 – 4A2 + A = 0.

Solution:

We have,

A =

A2 =

=

=

A3 = A2. A

=

=

=

L.H.S. = A3 – 4A2 + A

=

=

=

=

= 0

= R.H.S.

Hence proved.

### Question 32. Show that the matrix A =is root of the equation A2 – 12A – I = 0

Solution:

We have,

A =

A2 =

=

=

L.H.S. = A2 – 12A – I

=

=

=

=

= 0

= R.H.S.

Hence proved.

Solution:

We have,

A =

A2 =

=

=

A2 – 5A – 14I =

=

=

=

### Question 34. If A =, find A2 – 5A + 7I = 0. Use this to find A4.

Solution:

We have,

A =

A2 =

=

=

L.H.S. = A2 – 5A + 7I = 0

=

=

=

=

= 0

= R.H.S.

Hence proved.

Now we have A2 – 5A + 7I = 0

=> A2 = 5A – 7I

=> A4 = (5A – 7I) (5A – 7I)

=> A4 = 25A2 – 35AI – 35AI + 49I

=> A4 = 25A2 – 70AI + 49I

=> A4 = 25 (5A – 7I) – 70AI + 49I

=> A4 = 125A – 175I – 70A + 49I

=> A4 = 55A – 126I

=> A4 =

=> A4 =

=> A4 =

=> A4 =

### Question 35. If A =, find k such that A2 = kA – 2I2.

Solution:

We have,

A =

A2 =

=

=

We are given,

=> A= kA – 2I2

=>

=>

=>

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 A2 – 8A + kI = 0.

Solution:

We have,

A =

A2 =

=

=

We are given,

=> A2 – 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) = x2 – 2x – 3, show that f(A) = 0.

Solution:

We have,

A =and f(x) = x2 – 2x – 3

A2 =

=

=

L.H.S. = f(A) = A2 – 2A – 3I2

=

=

=

=

= 0

= R.H.S.

Hence proved.

### Question 38. If A =and I =, find λ, μ so that A2 = λA + μI.

Solution:

We have,

A =

A2 =

=

=

We are given,

=> A2 = 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 productequals 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,

=>

=>

=>

=>

=> x2 – 2x – 15 = 0

=> x2 – 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,

=>

=>

=>

=>

=> x2 – 48 = 0

=> x2 = 48

=> x = ±4√3

Therefore, the value of x is ±4√3.

(iv)

Solution:

We have,

=>

=>

=>

=>

=> 2x2 + 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 A2 – 4A + 3I3.

Solution:

We have,

A =

A2 =

=

=

So, A2 – 4A + 3I3 =

=

=

=

### Question 42. If f(x) = x2 – 2x, find f(A), where A =.

Solution:

We have,

A =and f(x) = x2 – 2x

A2 =

=

=

So, f(A) = A2 – 2A

=

=

=

=

### Question 43. If f(x) = x3 + 4x2 – x, find f(A) where A =.

Solution:

We have,

A =and f(x) = x3 + 4x2 – x

A2 =

=

=

A3 = A2. A

=

=

=

Now, f(A) = A3 + 4A2 – A

=

=

=

=

### Question 44. If A =, then show that A is a root of the polynomial f(x) = x3 – 6x2 + 7x +2.

Solution:

We have,

A =and f(x) = x3 – 6x2 + 7x +2.

A2 =

=

=

A3 = A2. A

=

=

=

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

Now, f(A) = A3 – 6A2 + 7A + 2I

=

=

=

=

= 0

Hence proved.

### Question 45. If A =, prove that A2 – 4A – 5I = 0.

Solution:

We have,

A =

A2 =

=

=

Now, L.H.S. = A2 – 4A – 5I

=

=

=

=

= 0

= R.H.S.

Hence proved.

### Question 46. If A =, show that A2 – 7A + 10I3 = 0.

Solution:

We have,

A =

A2 =

=

=

Now, L.H.S. = A2 – 7A + 10I3

=

=

=

=

= 0

= R.H.S.

Hence proved.

### Question 47. Without using the concept of inverse of a matrix, find the matrixsuch 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= 6I2.

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 A16.

Solution:

We have,

A =

A2 =

=

=

A16 = A2 A2 A2 A2

=

=

### Question 51. If A =, B =and x2 = –1, then show that (A + B)2 = A2 + B2.

Solution:

We have,

A =, B =and x2 = –1

L.H.S. = (A + B)2

=

=

=

=

=

=

=

=

R.H.S. = A2 + B2

=

=

=

=

=

=

= L.H.S.

Hence proved.

I think you got complete solutions for this chapter. If You have any queries regarding this chapter, please comment in the below section our subject teacher will answer you. We tried our best to give complete solutions so you got good marks in your exam.

If these solutions have helped you, you can also share rdsharmasolutions.in to your friends.