Class 12 RD Sharma Solutions â Chapter 5 Algebra of Matrices â Exercise 5.3 | Set 2
Question 26. If= 0, find x.
Solution:
We have,
=>= 0
=>
=>
=>
=> 2x â 4 = 0
=> 2x = 4
=> x = 2
Therefore, the value of x is 2.
Question 27. If A =and I =, then prove that A2 â A + 2I = 0.
Solution:
We have,
A =
A2 =
=
=
L.H.S. = A2 â A + 2I
=
=
=
=
=
=
= 0
= R.H.S.
Hence proved.
Question 28. If A =and I =, then find Îŧ so that A2 = 5A + ÎŧI.
Solution:
We have,
A =
A2 =
=
=
We are given,
=> A2 = 5A + ÎŧI
=>
=>
=>
=>
On comparing both sides, we get
=> 8 = 15 + Îŧ
=> Îŧ = â7
Therefore, the value of Îŧ is â7.
Question 29. If A =, show that A2 â 5A + 7I2 = 0.
Solution:
We have,
A =
A2 =
=
=
L.H.S. = A2 â 5A + 7I2
=
=
=
=
= 0
= R.H.S.
Hence proved.
Question 30. If A =, show that A2 â 2A + 3I2 = 0.
Solution:
We have,
A =
A2 =
=
=
L.H.S. = A2 â 2A + 3I2
=
=
=
=
= 0
= R.H.S.
Hence proved.
Question 31. Show that the matrix A =satisfies the equation A3 â 4A2 + A = 0.
Solution:
We have,
A =
A2 =
=
=
A3 = A2. A
=
=
=
L.H.S. = A3 â 4A2 + A
=
=
=
=
= 0
= R.H.S.
Hence proved.
Question 32. Show that the matrix A =is root of the equation A2 â 12A â I = 0
Solution:
We have,
A =
A2 =
=
=
L.H.S. = A2 â 12A â I
=
=
=
=
= 0
= R.H.S.
Hence proved.
Question 33. If A =find A2 â 5A â 14I.
Solution:
We have,
A =
A2 =
=
=
A2 â 5A â 14I =
=
=
=
Question 34. If A =, find A2 â 5A + 7I = 0. Use this to find A4.
Solution:
We have,
A =
A2 =
=
=
L.H.S. = A2 â 5A + 7I = 0
=
=
=
=
= 0
= R.H.S.
Hence proved.
Now we have A2 â 5A + 7I = 0
=> A2 = 5A â 7I
=> A4 = (5A â 7I) (5A â 7I)
=> A4 = 25A2 â 35AI â 35AI + 49I
=> A4 = 25A2 â 70AI + 49I
=> A4 = 25 (5A â 7I) â 70AI + 49I
=> A4 = 125A â 175I â 70A + 49I
=> A4 = 55A â 126I
=> A4 =
=> A4 =
=> A4 =
=> A4 =
Question 35. If A =, find k such that A2 = kA â 2I2.
Solution:
We have,
A =
A2 =
=
=
We are given,
=> A2 = kA â 2I2
=>
=>
=>
On comparing both sides, we get
=> 3k â 2 = 1
=> 3k = 3
=> k = 1
Therefore, the value of k is 1.
Question 36. If A =, find k such that A2 â 8A + kI = 0.
Solution:
We have,
A =
A2 =
=
=
We are given,
=> A2 â 8A + kI = 0
=>
=>
=>
=>
On comparing both sides, we get
=> âk + 7 = 0
=> k = 7
Therefore, the value of k is 7.
Question 37. If A =and f(x) = x2 â 2x â 3, show that f(A) = 0.
Solution:
We have,
A =and f(x) = x2 â 2x â 3
A2 =
=
=
L.H.S. = f(A) = A2 â 2A â 3I2
=
=
=
=
= 0
= R.H.S.
Hence proved.
Question 38. If A =and I =, find Îŧ, Ξ so that A2 = ÎŧA + ΞI.
Solution:
We have,
A =
A2 =
=
=
We are given,
=> A2 = ÎŧA + ΞI
=>
=>
=>
=>
On comparing both sides, we get,
=> 2Îŧ + Ξ = 7 and Îŧ = 4
=> 2(4) + Ξ = 7
=> Ξ = 7 â 8
=> Ξ = â1
Therefore, the value of Îŧ is 4 and Ξ is â1.
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 â 2x â 15 = 0
=> x2 â 5x + 3x â 15 = 0
=> x (x â 5) + 3 (x â 5) = 0
=> (x â 5) (x + 3) = 0
=> x = 5 or â3
Therefore, the value of x is 5 or â3.
(ii)
Solution:
We have,
=>
=>
=>
=>
=> 4 + 4x = 0
=> 4x = â4
=> x = â1
Therefore, the value of x is â1.
(iii)
Solution:
We have,
=>
=>
=>
=>
=> x2 â 48 = 0
=> x2 = 48
=> x = Âą4â3
Therefore, the value of x is Âą4â3.
(iv)
Solution:
We have,
=>
=>
=>
=>
=> 2x2 + 23x = 0
=> x (2x + 23) = 0
=> x = 0 or x = â23/2
Therefore, the value of x is 0 or â23/2.
Question 41. If A =, compute A2 â 4A + 3I3.
Solution:
We have,
A =
A2 =
=
=
So, A2 â 4A + 3I3 =
=
=
=
Question 42. If f(x) = x2 â 2x, find f(A), where A =.
Solution:
We have,
A =and f(x) = x2 â 2x
A2 =
=
=
So, f(A) = A2 â 2A
=
=
=
=
Question 43. If f(x) = x3 + 4x2 â x, find f(A) where A =.
Solution:
We have,
A =and f(x) = x3 + 4x2 â x
A2 =
=
=
A3 = A2. A
=
=
=
Now, f(A) = A3 + 4A2 â A
=
=
=
=
Question 44. If A =, then show that A is a root of the polynomial f(x) = x3 â 6x2 + 7x +2.
Solution:
We have,
A =and f(x) = x3 â 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 â 6A2 + 7A + 2I
=
=
=
=
= 0
Hence proved.
Question 45. If A =, prove that A2 â 4A â 5I = 0.
Solution:
We have,
A =
A2 =
=
=
Now, L.H.S. = A2 â 4A â 5I
=
=
=
=
= 0
= R.H.S.
Hence proved.
Question 46. If A =, show that A2 â 7A + 10I3 = 0.
Solution:
We have,
A =
A2 =
=
=
Now, L.H.S. = A2 â 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 â 7z = â16
5y â 7u = â6
â2x + 3z = 7
â2y + 3u = 2
On solving the above equations, we get
=> x = 1, y = â4, z = 3 and u = â2.
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 = â6
y + 4z = 2
2y + 5z = 4
On solving the above equations, we get
=> x = â2, 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 = â 4, 4y = 8 and 4z = 4.
=> x = â1, 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 = â2 and z = â5
And also we have,
2x â a = â1
2y â b = â8
2z â c = â10
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 = â7 and 2x + 5a = â8
=> x = 1 and a = â2
y + 4b = 2 and 2y + 5b = 4
=> b = 0 and y = 2
z + 4c = 11 and 2z + 5c = 10
=> c = 4 and z = â5
So, we get A =
Question 49. Find a 2 Ã 2 matrix A such that= 6I2.
Solution:
Let A =
Given equation is,
=>= 6I
=>
=>
=>
On comparing both sides, we get
w + x = 6 and â2w + 4x = 0
=> w = 4 and x = 2
y + z = 0 and â2y + 4z = 6
=> y = â1 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 = â1, then show that (A + B)2 = A2 + B2.
Solution:
We have,
A =, B =and x2 = â1
L.H.S. = (A + B)2
=
=
=
=
=
=
=
=
R.H.S. = A2 + B2
=
=
=
=
=
=
= L.H.S.
Hence proved.