Class 12 RD Sharma Solutions – Chapter 5 Algebra of Matrices – Exercise 5.3 | Set 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.

Question 33. If A =find A² – 5A – 14I.

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,

=> A² = 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.