Matrices Class 12th Mathematics Part I CBSE Solution

Class 12^th Mathematics Part I CBSE Solution

Exercise 3.1

1. In a matrix A = a^n = [cccc 2&5&19&-7 35&-2& 5/2 &12 root 3 &1&-5&17] , (i) The…
2. If a matrix has 24 elements, what are the possible orders it can have? What, if…
3. If a matrix has 18 elements, what are the possible orders it can have? What, if…
4. a_ij = (i+j)^2/2 Construct a 2 × 2 matrix, A = [aij], whose elements are given…
5. a_ij = i/j Construct a 2 × 2 matrix, A = [aij], whose elements are given by:…
6. a_ij = (i+2j)^2/2 Construct a 2 × 2 matrix, A = [aij], whose elements are given…
7. a_ij = 1/2 |-3i+j| Construct a 3 × 4 matrix, whose elements are given by:…
8. a_ij = 2i-j Construct a 3 × 4 matrix, whose elements are given by:…
9. [ll 4&3 x&5] = [ll y 1&5] Find the values of x, y and z from the following…
10. [cc x+y&2 5+z] = [ll 6&2 5&8] Find the values of x, y and z from the following…
11. [c x+y+z x+z y+z] = [9 5 7] Find the values of x, y and z from the following…
12. Find the value of a, b, c and d from the equation: [cc a-b&2a+c 2a-b&3c+d] = [cc…
13. A = [aij]m × n is a square matrix, ifA. m n B. m n C. m = n D. None of these…
14. Which of the given values of x and y make the following pair of matrices equal…
15. The number of all possible matrices of order 3 × 3 with each entry 0 or 1 is:A.…

Exercise 3.2

1. Le t a = [ll 2&4 3&2] , b = [cc 1&3 -2&5] , c = [cc -2&5 3&4] Find each of the…
2. Let a = [ll 2&4 3&2] , b = [rr 1&3 -2&5] , c = [cc -2&5 3&4] Find each of the…
3. Let a = [ll 2&4 3&2] , b = [rr 1&3 -2&5] , c = [cc -2&5 3&4] Find each of the…
4. Let a = [ll 2&4 3&2] , b = [rr 1&3 -2&5] , c = [cc -2&5 3&4] Find each of the…
5. Let a = [ll 2&4 3&2] , b = [rr 1&3 -2&5] , c = [cc -2&5 3&4] Find each of the…
6. [cc a -b] + [ll a b] Compute the following:
7. [cc a^2 + b^2 & b^2 + c^2 a^2 + c^2 & a^2 + b^2] + [cc 2ab&2bc -2ac&-2ab]…
8. [ccc -1&4&-6 8&5&16 2&8&5] + [ccc 12&7&6 8&0&5 3&2&4] Compute the following:…
9. [cc cos^2x^2x sin^2x^2x] + [cc sin^2x^2x cos^2x^2x] Compute the following:…
10. Compute the indicated products.
11. [1 2 3] [lll 2&3&4] Compute the indicated products.
12. [cc 1&-2 2&3] [ccc 1&2&3 2&3&1] Compute the indicated products.
13. [ccc 2&3&4 3&4&5 4&5&6] [ccc 1&-3&5 0&2&4 3&0&5] Compute the indicated…
14. Compute the indicated products.
15. [ccc 3&-1&3 -1&0&2] [cc 2&-3 1&0 3&1] Compute the indicated products.…
16. If a = [ccc 1&2&-3 5&0&2 1&-1&1] , b = [ccc 3&-1&2 4&2&5 2&0&3] c = [ccc 4&1&2…
17. If a = [ccc 2/3 &1& 5/3 1/3 & 2/3 & 4/3 7/3 &2& 2/3] b = [ccc 2/5 & 3/5 &1 1/5 &…
18. Simplify costheta [cc costheta heta -sintegrate heta]+sintegrate heta [cc…
19. x+y = [ll 7&0 2&5] x-y = [ll 3&0 0&3] Find X and Y, if
20. 2x+3y = [ll 2&3 4&0] 3x+2y = [cc 2&-2 -1&5] Find X and Y, if
21. Find X, if y = [ll 3&2 1&4] 2x+y = [cc 1&0 -3&2]
22. Find x and y, if 2 [ll 1&3 0] + [ll y&0 1&2] = [ll 5&6 1&8]
23. Solve the equation for x, y, z and t, if 2 [ll x y]+3 [cc 1&-1 0&2] = 3 [cc 3&5…
24. If x [2 3]+y [c -1 1] = [c 10 5] , find the values of x and y.
25. Given 3 [cc x z] = [cc x&6 -1&2w] + [cc 4+y z+w&3] , find the values of x, y, z…
26. If f (x) = [ll cosx&-sinx&0 sinx&0 0&0&1 show that F(x) F(y) = F(x + y).…
27. [cc 5&-1 6&7] [cc 2&1 3&4] not equal [ll 2&1 3&4] [cc 5&-1 6&7] Show that…
28. [ccc 1&2&3 0&1&0 1&1&0] [ccc -1&1&0 0&-1&1 2&3&4] = [ccc -1&1&0 0&-1&1 2&3&4]…
29. Find A^2 - 5A + 6I, if a = [lll 2&0&1 2&1&3 1&-1&0]
30. If a = [lll 1&0&2 0&2&1 2&0&3] , prove that A3 - 6A^2 + 7A + 2I = 0…
31. If a = [ll 3&-2 4&-2] i = [ll 1&0 0&1] , find k so that A^2 = kA - 2I…
32. If a = [ccc 0& - tan alpha /2 tan alpha /2 &0] and I is the identity matrix of…
33. Rs 1800 A trust fund has Rs. 30,000 that must be invested in two different…
34. Rs. 2000 A trust fund has Rs. 30,000 that must be invested in two different…
35. The bookshop of a particular school has 10 dozen chemistry books, 8 dozen…
36. The restriction on n, k and p so that PY + WY will be defined are:A. k = 3, p =…
37. If n = p, then the order of the matrix 7X - 5Z is:A. p × 2 B. 2 × n C. n × 3 D.…

Exercise 3.3

1. [c 5 1/2 -1] Find the transpose of each of the following matrices:…
2. [cc 1&-1 2&3] Find the transpose of each of the following matrices:…
3. [ccc -1&5&6 root 3 &5&6 2&3&-1] Find the transpose of each of the following…
4. If a = [ccc -1&2&3 5&7&9 -2&1&1] and, then verify that (A + B)’ = A’ + B’,…
5. If a = [ccc -1&2&3 5&7&9 -2&1&1] and, then verify that (A - B)’ = A’- B’…
6. If a^there there eξ sts = [cc 3&4 -1&2 0&1] and, then verify that (A + B). = A’…
7. If a^there there eξ sts = [cc 3&4 -1&2 0&1] and, then verify that (A - B)’ =…
8. If a = [rr -2&3 1&2] and b = [rr -1&0 1&2] , then find (A + 2B)’
9. (i) a = [c 1 -4 3] b = [lll -1&2&1] For the matrices A and B, verify that (AB)’…
10. b = [lll 1&5&7]a = [0 1 2] For the matrices A and B, verify that (AB)’ = B’A’,…
11. If a = [ll cosalpha -sinalpha] , then verify that A’ A = I
12. If a = [cc sinalpha -cosalpha] , then verify that A’ A = I
13. Show that the matrix a = [ccc 1&-1&5 -1&2&1 5&1&3] is a symmetric matrix.…
14. Show that the matrix a = [ccc 0&1&-1 -1&0&1 1&-1&0] is a skew symmetric matrix.…
15. For the matrix a = [ll 1&5 6&7] , verify that (A + A’) is a symmetric matrix…
16. For the matrix a = [ll 1&5 6&7] , verify that (A - A’) is a skew symmetric…
17. Find and 1/2 (a-a^there there eξ sts) , when a = [ccc 0 -a&0 -b&-c&0]…
18. [cc 3&5 1&-1] Express the following matrices as the sum of a symmetric and a…
19. [ccc 6&-2&2 -2&3&-1 2&-1&3] Express the following matrices as the sum of a…
20. [ccc 3&3&-1 -2&-2&1 -4&-5&2] Express the following matrices as the sum of a…
21. [cc 1&5 -1&2] [cc 3&5 1&-1] Express the following matrices as the sum of a…
22. If A, B are symmetric matrices of same order, then AB - BA is aA. Skew…
23. If , and A + A’ = I, if the value of a isA. 3 pi /2 B. pi /3 C. pi D. 3 pi /2…

Exercise 3.4

1. Using elementary transformations, find the inverse of each of the matrices.…
2. [ll 2&1 1&1] Using elementary transformations, find the inverse of each of the…
3. [ll 1&3 2&7] Using elementary transformations, find the inverse of each of the…
4. [ll 2&3 5&7] Using elementary transformations, find the inverse of each of the…
5. [ll 2&1 7&4] Using elementary transformations, find the inverse of each of the…
6. [ll 2&5 1&3] Using elementary transformations, find the inverse of each of the…
7. [ll 3&1 5&2] Using elementary transformations, find the inverse of each of the…
8. [ll 4&5 3&4] Using elementary transformations, find the inverse of each of the…
9. [cc 3&10 2&7] Using elementary transformations, find the inverse of each of the…
10. [cc 3&-1 -4&2] Using elementary transformations, find the inverse of each of…
11. [ll 2&-6 1&-2] Using elementary transformations, find the inverse of each of…
12. [cc 6&-3 -2&1] Using elementary transformations, find the inverse of each of…
13. [cc 2&-1 -3&2] Using elementary transformations, find the inverse of each of…
14. [ll 2&1 4&2] Using elementary transformations, find the inverse of each of the…
15. [ccc 2&-3&3 2&2&3 3&-2&2] Using elementary transformations, find the inverse of…
16. [ccc 1&3&-2 -3&0&-5 2&5&0] Using elementary transformations, find the inverse…
17. [ccc 2&0&-1 5&1&0 0&1&3] Using elementary transformations, find the inverse of…
18. Matrices A and B will be inverse of each other only ifA. AB = BA B. AB = BA = 0…

Miscellaneous Exercise

1. Let, show that (aI + bA)n = anI + nan-1bA, where I is the identity matrix of…
2. If a = [lll 1&1&1 1&1&1 1&1&1] , prove that .
3. If a = [ll 3&-4 1&-1] then prove that a^n = [cc 1+2n&-4n n&1-2n] where n is any…
4. If A and B are symmetric matrices, prove that AB - BA is a skew symmetric…
5. Show that the matrix B’AB is symmetric or skew symmetric according as A is…
6. Find the values of x, y, z if the matrix satisfy a = [ccc 0&2y x&-z x&-y] the…
7. For what values of x: [lll 1&2&1] [lll 1&2&0 2&0&1 1&0&2] [0 2 x] = 0?…
8. If a = [cc 3&1 -1&2] show that A^2 - 5A + 7I =0.
9. Find x, if [rrr x&-5&-1] [lll 1&0&2 0&2&1 2&0&3] [x 4 1] = 0
10. A manufacturer produces three products x, y, z which he sells in two markets.…
11. Find the matrix X so that x [ccc 1&2&3 4&5&6] = [ccc -7&-8&-9 2&4&6]…
12. If A and B are square matrices of the same order such that AB = BA, then prove…
13. If a = [ll alpha & beta gamma & - alpha] is such that A � = I, thenA. 1 +…
14. If the matrix A is both symmetric and skew symmetric, thenA. A is a diagonal…
15. If A is square matrix such that A^2 = A, then (I + A)^3 7 A is equal toA. A B.…

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Exercise 3.1

Question 1.

In a matrix A =



(i) The order of the matrix,

(ii) The number of elements,

(iii) Write the elements a₁₃, a₂₁, a₃₃, a₂₄, a₂₃.

Answer:

(i) In the given matrix, the number of rows is 3 and the number of columns is 4.

Order of a matrix =No of rows × No of columns

Therefore, the order of the matrix is 3 × 4.

(ii) Since, the order of the matrix is 3 × 4, there are 3 × 4 = 12 elements in it.

(iii) a₁₃ = 19, a₂₁ = 35, a₃₃ = -5, a₂₄ = 12, a₂₃ = .

Question 2.

If a matrix has 24 elements, what are the possible orders it can have? What, if it has 13 elements?

Answer:

It is known that if a matrix is of the order m × n, then it has mn elements.

Therefore, to find all the possible orders of a matrix having 24 elements, we had to find all the ordered pairs of natural numbers whose product is 24.

The ordered pairs are: (1, 24), (24, 1), (2, 12), (12, 2), (3, 8), (8, 3), (4, 6), and (6, 4).

Therefore, the possible orders of a matrix having 24 elements are;

1 × 24, 24 × 1, 2 × 12, 12 × 2, 3 × 8, 8 × 3, 4 × 6, 6 × 4

(1, 13) and (13, 1) are the ordered pairs of natural numbers whose product is 13.

Therefore, the possible orders of a matrix having 13 elements are 1 × 13 and 13 × 1.

Question 3.

If a matrix has 18 elements, what are the possible orders it can have? What, if it has 5 elements?

Answer:

It is known that if a matrix s of the order m × n, then it has mn elements.

Therefore, to find all the possible orders of a matrix having 18 elements, we had to find all the ordered pairs of natural numbers whose product is 18.

The ordered pairs are: (1, 18), (18, 1), (2, 9), (9, 2), (3, 6) and (6, 3).

Therefore, the possible orders of a matrix having 24 elements are;

1 × 18, 18 × 1, 2 × 9, 9 × 2, 3 × 6, 6 × 3

(1, 5) and (5, 1) are the ordered pairs of natural numbers whose product is 5.

Therefore, the possible orders of a matrix having 5 elements are 1 × 5 and 5 × 1.

Question 4.

Construct a 2 × 2 matrix, A = [a_ij], whose elements are given by:

Answer:

In general, a 2 × 2 matrix is given by A = 

a_ij = 

Therefore,

a₁₁ = 

a₁₂ = 

a₂₁ = 

a₂₂ = 

Therefore, the required matrix is A = 

Question 5.

Construct a 2 × 2 matrix, A = [a_ij], whose elements are given by:

Answer:

In general, a 2 × 2 matrix is given by A = 

a_ij = 

Therefore,

a₁₁ = 

a₁₂ = 

a₂₁ = 

a₂₂ = 

Therefore, the required matrix is A = .

Question 6.

Construct a 2 × 2 matrix, A = [a_ij], whose elements are given by:

Answer:

In general, a 2 × 2 matrix is given by A = 

a_ij = 

Therefore,

a₁₁ = 

a₁₂ = 

a₂₁ = 

a₂₂ = 

Therefore, the required matrix is A = 

Question 7.

Construct a 3 × 4 matrix, whose elements are given by:

Answer:

In general 3 × 4 matrix is given by A = 

Therefore,

a₃₂ = 

a₁₃ = 

a₂₃ = 

a₃₃ = 

a₁₄ = 

a₂₄ = 

a₃₄ = 

Therefore, required matrix is A = 

Question 8.

Construct a 3 × 4 matrix, whose elements are given by:

Answer:

In general 3 × 4 matrix is given by A = 

a_ij = 2i-j, i = 1,2,3 and j = 1,2,3,4

Therefore,

a₁₁ = 2 × 1 - 1 = 2 - 1 = 1

a₂₁ = 2 × 2 - 1 = 4 - 1 = 3

a₃₁ = 2 × 3 - 1 = 6 - 1 = 5

a₁₂ = 2 × 1 - 2 = 2 - 2 = 0

a₂₂ = 2 × 2 - 2 = 4 - 2 = 2

a₃₂ = 2 × 3 - 2 = 6 - 2 = 4

a₁₃ = 2 × 1 - 3 = 2 - 3 = -1

a₂₃ = 2 × 2 - 3 = 4 - 3 = 1

a₃₃ = 2 × 3 - 3 = 6 - 3 = 3

a₁₄ = 2 × 1 - 4 = 2 - 4 = -2

a₂₄ = 2 × 2 - 4 = 4 - 4 = 0

a₃₄ = 2 × 3 - 4 = 6 - 4 = 2

Therefore, required matrix is A = 

Question 9.

Find the values of x, y and z from the following equations:

Answer:

Since, the given matrices are equal, their corresponding elements are also equal.

Comparing the corresponding element, we have:

x = 1, y = 4 and z = 3

Question 10.

Find the values of x, y and z from the following equations:

Answer:

Since, the given matrices are equal, their corresponding elements are also equal.

Comparing the corresponding element, we have:

x + y = 6, xy = 8, 5 + z = 5

Now, 5 + z = 5
⇒ z = 0

x+ y = 6
⇒ x = 6 - y 

Also, 
xy = 8
⇒ (6 - y)y = 8
⇒ 6y - y²= 8
⇒ y² - 6y + 8 = 0
⇒ y² - 4y - 2y + 8 = 0
⇒ y(y - 4) - 2(y - 4) = 0
⇒ (y - 2)(y - 4) = 0

Hence, y = 2 or y = 4
when y = 2 
x = 6 - 2 = 4

when x = 4
x = 6 - 4 = 2


Question 11.

Find the values of x, y and z from the following equations:

Answer:

Since, the given matrices are equal, their corresponding elements are also equal.

Comparing the corresponding element, we have:

x + y + z = 9...(1)

x + z = 5.....(2)

y + z = 7.......(3)

Putting the value of equation 2 in equation 1,

y + 5 = 9

⇒ y = 4

Then, putting the value of y in equation 3, we get,

4 + z = 7

⇒ z = 3

Therefore, x + z = 5

⇒ x = 2

Therefore, x =2, y = 4 and z = 3.

Question 12.

Find the value of a, b, c and d from the equation:

Answer:

Since, the given matrices are equal, their corresponding elements are also equal.

Comparing the corresponding element, we have:

a –b = -1 …(1)

2a – b = 0 …(2)

2a + c = 5 …(3)

3c + d = 13 …(4)

From equation (2), we get:

b = 2a

Then, from eq. (1), we get,

a -2a = -1

⇒ a = 1

⇒ b = 2

Now, from eq. (3), we get:

2 × 1 + c = 5

⇒ c = 3

From (4), we get,

3 × 3 + d = 13

⇒ 9 + d = 13

⇒ d = 4

Therefore, a = 1, b = 2, c = 3 and d = 4.

Question 13.

A = [a_ij]_{m × n} is a square matrix, if
A. m < n

B. m > n

C. m = n

D. None of these

Answer:

We know that if a given matrix is said to be square matrix if the number of rows is equal to the number of columns.

Therefore, A = [a_ij]_{m × n} is a square matrix, if m = n.

Question 14.

Which of the given values of x and y make the following pair of matrices equal


A. 

B. Not possible to find

C. y = 7, 

D. 

Answer:

Now, 

Since, the given matrices are equal, their corresponding elements are also equal.

Comparing the corresponding element, we have:

3x + 7 = 0

5 = y – 2

⇒ y = 7

y + 1 = 8

⇒ y = 7

And 2 – 3x = 4

Thus, on comparing the corresponding elements of the two matrices, we get different values of x, which is not possible.

Therefore, it is not possible to find the values of x and y for which the given matrices are equal.

Question 15.

The number of all possible matrices of order 3 × 3 with each entry 0 or 1 is:
A. 27

B. 18

C. 81

D. 512

Answer:

The given matrix of the order 3 × 3 has 9 elements with each entry 0 or 1.

Now, each of the 9 elements can be filled in two possible ways.

Hence, the required number of possible matrices is 2⁹ = 512.

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Exercise 3.2

Question 1.

Le t 
Find each of the following:

i) A + B
ii) A - B
iii) 3A - C
iv) AB
v) BA

Answer:

i) A + B = 

ii) A - B








iii) 3A - C







iv) AB





v) BA

Question 2.

Let 

Find each of the following:

A – B

Answer:

A – B = 

Question 3.

Let 

Find each of the following:

3A – C

Answer:

3A – C = 

Question 4.

Let 

Find each of the following:

AB

Answer:

AB 

Question 5.

Let 

Find each of the following:

BA

Answer:

BA = 

Question 6.

Compute the following:

Answer:

Question 7.

Compute the following:

Answer:

.

Question 8.

Compute the following:

Answer:

Question 9.

Compute the following:

Answer:

Question 10.

Compute the indicated products.

Answer:

Question 11.

Compute the indicated products.

Answer:

Question 12.

Compute the indicated products.

Answer:

Question 13.

Compute the indicated products.

Answer:

.

.

Question 14.

Compute the indicated products.

Answer:

Question 15.

Compute the indicated products.

Answer:

Question 16.

If  then compute (A + B) and (B – C). Also, verify that A + (B – C) = (A + B) – C.

Answer:

Now, A+ B 

B – C 

A + (B –C) 

.

+ B) – C = 

.

Therefore, A + (B –C) = (A + B) – C

Question 17.

If  then compute 3A – 5B.

Answer:

3A – 5B 

Question 18.

Simplify 

Answer:

.

Question 19.

Find X and Y, if

Answer:

Now, X + Y =  …(1)

Adding (1) and (2), we get,

2X = 

.

Now, X + Y = 

Question 20.

Find X and Y, if

Answer:

2X + 3Y  …(1)

3X + 2Y  …(2)

Now, multiply equation (1) by 2 and equation (2) by 3, we get,

4X + 6Y  …(3)

9X + 6Y  …(4)

Subtracting equation (4) from (3), we get,

(4X + 6Y) – (9X + 6Y)

Now, 2X + 3Y 

Question 21.

Find X, if 

Answer:

2X + Y 

Question 22.

Find x and y, if 

Answer:

Now, on comparing elements of these two matrices, we get,

2 + y = 5

=> y = 3

And 2x + 2 = 8

=> x = 3

Therefore, x = 3 and y = 3.

Question 23.

Solve the equation for x, y, z and t, if 

Answer:

On comparing the elements of these two matrices, we get,

2x + 3 = 9

⇒ 2x = 6

⇒ x = 3

2y = 12

⇒ y = 6

2z -3 = 15

⇒ 2z = 18

⇒ z = 9

2t +6 = 18

⇒ 2t = 12

⇒ t = 6

Therefore, x = 3, y =6, z = 9 and t = 6.

Question 24.

If , find the values of x and y.

Answer:

On comparing the corresponding elements of these two matrices, we get,

2x –y = 10 and 3x + y =5

Now, adding above two equations, we get

5x = 15

⇒ x = 3

Now, 3x + y = 5

⇒ y = 5 - 3x

⇒ y = 5 - 9 = -4

Therefore, x = 3 and y = -4.

Question 25.

Given , find the values of x, y, z and w.

Answer:

On comparing the corresponding elements of these two matrices, we get,

3x = x + 4

⇒ 2x + 4

⇒ x =2

3y = 6 + x + y

⇒ 2y = 6 + x = 6 + 2 = 8

⇒ y = 4

3w = 2w + 3

⇒ w = 3

3z = -1 + z + w

⇒ 2z = -1 + w = -1 +3 = 2

⇒ z = 1

Therefore, x = 2, y = 4, z = 1 and w = 3.

Question 26.

If  show that F(x) F(y) = F(x + y).

Answer:

Therefore, F(x)F(y) = F(x+y)

Question 27.

Show that

Answer:

Question 28.

Show that

Answer:

Question 29.

Find A² – 5A + 6I, if 

Answer:

A² = A.A 

Now, A² – 5A + 6I 

Question 30.

If, prove that A₃ – 6A² + 7A + 2I = 0

Answer:

A² = A.A 

Now, A³ = A². A

Now, A³ – 6A² + 7A + 2I 

Therefore, A³ – 6A² + 7A + 2I = 0

Question 31.

If , find k so that A² = kA – 2I

Answer:

A² = A.A 

Now, A² = kA – 2I

Comparing the corresponding elements, we get,

3k -2 = 1

⇒ 3k = 3

⇒ k = 1

Therefore, the value of k is 1.

Question 32.

If  and I is the identity matrix of order 2, show that 

Answer:

We know that I 

Now, LHS = I + A 

=

And on RHS = (I – A)

As we know,

cos 2θ = 1 - 2 sin² θ

cos 2θ = 2 cos² θ - 1

and sin2θ = 2 sinθ cosθ

As tanθ = sinθ/cosθ

=

= LHS

Hence Proved.

Question 33.

A trust fund has Rs. 30,000 that must be invested in two different types of bonds. The first bond pays 5% interest per year, and the second bond pays 7% interest per year. Using matrix multiplication, determine how to divide Rs. 30,000 among the two types of bonds. If the trust fund must obtain an annual total interest of:

Rs 1800

Answer:

(a) Let Rs x be invested in the first bond.

Then, the sum of money invested in the second bond will be Rs. (30000 – x).

It is given that the first bond pays 5% interest per year, and the second bond pays 7% interest per year.

Thus, in order to obtain an annual total interest of Rs. 1800, we get:

⇒ 5x + 210000 -7x = 180000

⇒ 210000 -2x = 180000

⇒ 2x = 210000 – 180000

⇒ 2x = 30000

⇒ x = 15000

Therefore, in order to obtain an annual total interest of Rs. 1800, the trust fund should invest Rs. 15000 in the first bond and the remaining Rs. 15000 in the second bond.

Question 34.

A trust fund has Rs. 30,000 that must be invested in two different types of bonds. The first bond pays 5% interest per year, and the second bond pays 7% interest per year. Using matrix multiplication, determine how to divide Rs. 30,000 among the two types of bonds. If the trust fund must obtain an annual total interest of:

Rs. 2000

Answer:

(a) Let Rs x be invested in the first bond.

Then, the sum of money invested in the second bond will be Rs (30000 – x).

It is given that the first bond pays 5% interest per year, and the second bond pays 7% interest per year.

Thus, in order to obtain an annual total interest of Rs 1800, we get:

⇒ 5x + 210000 -7x = 200000

⇒ 210000 -2x = 200000

⇒ 2x = 210000 – 200000

⇒ 2x = 10000

⇒ x = 5000

Therefore, in order to obtain an annual total interest of Rs 2000, the trust fund should invest Rs 5000 in the first bond and the remaining Rs 25000 in the second bond.

Question 35.

The bookshop of a particular school has 10 dozen chemistry books, 8 dozen physics books, 10 dozen economics books. Their selling prices are Rs 80, Rs 60 and Rs 40 each respectively. Find the total amount the bookshop will receive from selling all the books using matrix algebra.

Assume X, Y, Z, W and P are matrices of order 2 × n, 3 × k, 2 × p, n × 3 and p × k, respectively.

Answer:

It is given that the bookshop of a particular school has 10 dozen chemistry books, 8 dozen physics books, 10 dozen economics books.

Number of chemistry book = 10 × 12
= 120

Number of physics book = 8 × 12

= 96

Number of economics book = 10 × 12

= 120

Their selling prices are Rs 80, Rs 60 and Rs 40 each respectively.

Let A be the matrix of books per subject.4

A=

Let B denotes selling price of the books.

The total amount of money = number of books × selling price

= [120 × 80 + 96 × 60 + 120 × 40]

= (9600 + 5760 + 4800)

= 20160

Therefore, the bookshop will receive Rs. 20160 from the sale of all these books.

Question 36.

The restriction on n, k and p so that PY + WY will be defined are:
A. k = 3, p = n

B. k is arbitrary, p = 2

C. p is arbitrary, k = 3

D. k = 2, p = 3

Answer:

Matrices P and Y are of the orders p × k and 3 × k respectively.

Therefore, matrix PY will be defined if k = 3.

Then, PY will be of the order p × k.

Matrices W and Y are of the orders n × 3 and 3 × k respectively.

As, the number of columns in W is equal to the number of rows in Y, Matrix WY is well defined and is of the order n × k.

Matrices PY and WY can be added only when their orders are the same.

Therefore, PY is of the order p × k and WY is of the order n × k.

Thus, we must have p = n.

Therefore, k = 3 and p = n are the restrictions on n, k and p so that

PY + WY will be defined.

Question 37.

If n = p, then the order of the matrix 7X – 5Z is:
A. p × 2

B. 2 × n

C. n × 3

D. p × n

Answer:

Matrix X is of the order 2 × n.

Therefore, matrix 7X is also of the same order.

Matrix Z is of order 2 × p

⇒ 2 × n

⇒ matrix 5Z is also of the same order.

Now, both the matrices 7X and 5Z are of the order 2 × n.

Thus, matrix 7X – 5Z is well- defined and is of the order 2 × n.

---

Exercise 3.3

Question 1.

Find the transpose of each of the following matrices:

Answer:

We know that transpose of a matrix is obtained by interchanging the elements of the rows and columns. In other words, we can say, if 

So, let

Therefore, transpose of the given matrix A is denoted by A’

Hence, 

The transpose of the given matrix is .

Question 2.

Find the transpose of each of the following matrices:

Answer:

We know that transpose of a matrix is obtained by interchanging the elements of the rows and columns. In other words, we can say, if 

So, let

Therefore, transpose of the given matrix B is denoted by B’.

The transpose of the given matrix is .

Question 3.

Find the transpose of each of the following matrices:

Answer:

We know that transpose of a matrix is obtained by interchanging the elements of the rows and columns. In other words, we can say, if 

Therefore, the transpose of the given matrix C is denoted by C’.

The transpose of the given matrix is 

Question 4.

If  and, then verify that

(A + B)’ = A’ + B’,

Answer:

(A+B)’ = A’+B’

Explanation: We will first calculate L.H.S i.e. (A+B)’ and then consecutively we will calculate R.H.S and verify that both are equal.

Therefore, 

Now,

So,

From equation 1 & 2 we verify that

(A+B)’ = A’+B’. Hence verified.

Question 5.

If  and, then verify that

(A – B)’ = A’– B’

Answer:

We will first calculate L.H.S i.e. (A+B)’ and then consecutively we will calculate R.H.S and verify that both are equal.

From equation 1 & 2 we verify that

(A-B)’ = A’-B’. Hence verified.

Question 6.

If  and, then verify that

(A + B). = A’ + B’

Answer:

(A+B)’ = A’+B’

Explanation: Calculate matrix A by taking transpose of A’ and also find the transpose of matrix B and the solve L.H.S and R.H.S separately and then compare the result.

So, A = transpose of A’

B’ = transpose of B

From equation 1 & 2 we verify that

(A+B)’ = A’+B’. Hence verified.

Question 7.

If  and, then verify that

(A – B)’ = A’– B’ 

Answer:

(A-B)’ = A’-B’

Explanation: Calculate matrix A by taking transpose of A’ and also find the transpose of matrix B and the solve L.H.S and R.H.S separately and then compare the result.

From equation 1 & 2 we verify that

(A - B)’ = A’ - B’. Hence verified.

Question 8.

Ifand , then find (A + 2B)’

Answer:

Explanation: Whenever a constant term is multiplied with a matrix then it implies that every elements in each rows and columns are to be multiplied with that constant term. So in order to solve this question we have to multiplied every element of the matrix B with constant term that is 2.

Now, (A+2B)’ = transpose of A+2B

Question 9.

For the matrices A and B, verify that (AB)’ = B’A’, where

(i) 

Answer:

Explanation: The product of two matrices is defined or possible only if the number of columns of the former matrix is equal to number of rows of the latter matrix.

From equation 1 & 2 we verify that

(AB)’ = B’A’. Hence verified.

Question 10.

For the matrices A and B, verify that (AB)’ = B’A’, where

Answer:

Explanation: The product of two matrices is defined or possible only if the number of columns of the former matrix is equal to number of rows of the latter matrix.

From equation 1 & 2 we verify that

(AB)’ = B’A’. Hence verified.

Question 11.

If , then verify that A’ A = I

Answer:

We know A’ can be calculated by taking the transpose of the given matrix A.

Now multiply A and A’. So,

And we know ‘I’ represents an identity matrix

From equation 1 & 2 we can say that

AA’ = I

AA’ = I. Hence verified.

Question 12.

If, then verify that A’ A = I

Answer:

.

And we know ‘I’ represents an identity matrix

From equation 1 & 2 we can say that

AA’ = I

AA’ = I. Hence verified.

Question 13.

Show that the matrix is a symmetric matrix.

Answer:

A matrix is aid to be symmetric only if the transpose of matrix and the matrix itself are equal or same. This means that A = A’.

Now, we know that the transpose of matrix A is

So, from equation 1 & 2 we get

A = A’, hence we can say that Matrix A is a symmetric matrix.

Hence proved.

Question 14.

Show that the matrix is a skew symmetric matrix.

Answer:

A matrix is said to be skew symmetric if the transpose of the matrix is equal to the negative of the matrix. This means that A’ = -A.

Now, we know that the transpose of matrix A is

Now if we carefully look at the equation 2 we can rewrite it as

So, A’ = (-1) × A (from equation 1)

⇒ A’ = -A, hence we can say that Matrix A is a skew symmetric matrix.

Hence proved

Question 15.

For the matrix , verify that

(A + A’) is a symmetric matrix

Answer:

(A + A’) is a symmetric matrix.

On adding them we get,

Explanation: Now to show that the matrix obtained i.e. (A + A’) is symmetric we need to calculate its transpose and prove that the matrix (A + A’) and its transpose are equal. This means that (A + A’) = (A + A’)’.

. → eqⁿ 2

So, from equation 1 & 2 we get,

(A + A’) = (A + A’)’, hence we can say that (A + A’) is a symmetric matrix.

Ans. Hence proved.

Question 16.

For the matrix , verify that

(A – A’) is a skew symmetric matrix

Answer:

(A – A’) is a skew symmetric matrix.

.

subtracting A’ from A, we get,

Explanation: Now to show that the matrix obtained i.e. (A + A’) is skew symmetric we need to calculate its transpose and prove that the matrix (A + A’) is equal to the negative of its transpose are equal. This means that (A + A’) = -(A + A’)’.

We can rewrite above equation as

Also, (A – A’)’ = (-1) × (A – A’) (from equation 1)

(A – A’)’ = -(A – A’), hence we can say that Matrix A is a skew symmetric matrix.

Hence proved.

Question 17.

Find  and , when 

Answer:

(i) 

On adding A and A’, we get,

Therefore, (A + A’) = 0

(ii)

On adding A and A’, we get,

We can rewrite the above equation as’

Therefore, (A - A’) = 2(A) (from equation 1)

Question 18.

Express the following matrices as the sum of a symmetric and a skew symmetric matrix:

Answer:

As per Theorem 2 “Any square matrix can be expressed as the sum of a symmetric and skew symmetric matrix.” So in order to prove this we will be using Theorem 1 which states that “For any square matrix A with real number entries, A + A’ is a symmetric matrix and A – A’ is a skew symmetric matrix.”

Now, on adding A and A’ we will get,

.

.

.

.

 on subtracting A’ from A we will get,

Now, Add M and N, we get,

.

us, A is represented as the sum of a symmetric matrix M and a skew symmetric matrix N.

Ans. Hence proved

Question 19.

Express the following matrices as the sum of a symmetric and a skew symmetric matrix:

Answer:

As per Theorem 2 “Any square matrix can be expressed as the sum of a symmetric and skew symmetric matrix.” So in order to prove this we will be using Theorem 1 which states that “For any square matrix A with real number entries, A + A’ is a symmetric matrix and A – A’ is a skew symmetric matrix.”

Now, on adding A and A’ we will get,

.

Now, on subtracting A’ from A we will get,

⇒ N’ = -N

Now, Add M and N, we get,

 .

Thus, A is represented as the sum of a symmetric matrix M and a skew symmetric matrix N.

Question 20.

Express the following matrices as the sum of a symmetric and a skew symmetric matrix:

Answer:

As per Theorem 2 “Any square matrix can be expressed as the sum of a symmetric and skew symmetric matrix.” So in order to prove this we will be using Theorem 1 which states that “For any square matrix A with real number entries, A + A’ is a symmetric matrix and A – A’ is a skew symmetric matrix.”

Now, on adding A and A’ we will get,

Now, on subtracting A’ from A we will get,

.

w, Add M and N, we get,

Thus, A is represented as the sum of a symmetric matrix M and a skew symmetric matrix N.

Question 21.

Express the following matrices as the sum of a symmetric and a skew symmetric matrix:

Answer:

Explanation: As per Theorem 2 “Any square matrix can be expressed as the sum of a symmetric and skew symmetric matrix.” So in order to prove this we will be using Theorem 1 which states that “For any square matrix A with real number entries, A + A’ is a symmetric matrix and A – A’ is a skew symmetric matrix.”

Now, on adding A and A’ we will get,

Now, on subtracting A’ from A we will get,

Now, Add M and N, we get,

So we see here, 

Thus, A is represented as the sum of a symmetric matrix M and a skew symmetric matrix N.

Question 22.

If A, B are symmetric matrices of same order, then AB – BA is a
A. Skew symmetric matrix

B. Symmetric matrix

C. Zero matrix

D. Identity matrix

Answer:

Given A and B are symmetric matrix of same order

⇒ A = A’ → eqn1

⇒ B = B’ → eqn2

So, AB – BA = A’B’ – B’A’ (from 1 & 2)

⇒ AB –BA = (BA)’ – (AB)’ ()

⇒ AB – BA = (-1) ((AB)’ – (BA)’) (taking -1 common)

⇒ AB – BA = -(AB – BA)’ ()

Here we see that the relation between (AB – BA) and its transpose i.e. (AB – BA)’ is (AB –BA) = -(AB – BA)’, this implies that (AB – BA is a skew symmetric matrix.

Hence proved.

Question 23.

If , and A + A’ = I, if the value of a is
A. 

B. 

C. 

D. 

Answer:

Given A = 

Therefore, A’ = 

Also given that A + A’ = I

(Putting the values in the above equation)

We know when two matrices are equal only when all their corresponding elements or entries are equal i.e. if A = B, then a_ij = b_ij for all i and j.

This implies,

---

Exercise 3.4

Question 1.

Using elementary transformations, find the inverse of each of the matrices.

Answer:

First of all we need to check whether the matrix is invertible or not. For that-

For the inverse of a matrix A to exist,

Determinant of A ≠ 0

Here ∣A∣ = (1)(3) – (-1)(2) = 5

So the matrix is invertible.

Now to find the inverse of the matrix,

We know AA^-1 = I

Let’s make augmented matrix-

→ [ A : I ]

→

Apply row operation- R₂→ R₂ – 2R₁

→ 

Apply row operation- R₂→ R₂/5

→

Apply row operation- R₁→ R₁ + R₂

→

The matrix so obtained is of the form –

→ [ I : A^-1 ]

Hence inverse of the given matrix-

→ 

Question 2.

Using elementary transformations, find the inverse of each of the matrices.

Answer:

First of all we need to check whether the matrix is invertible or not. For that-

For the inverse of a matrix A to exist,

Determinant of A ≠ 0

Here ∣A∣ = (2)(1) – (1)(1) = 1

So the matrix is invertible.

Now to find the inverse of the matrix,

We know AA^-1 = I

Let’s make augmented matrix-

→ [ A : I ]

→

Apply row operation- R₁→ R₁ – R₂

→ 

Apply row operation- R₂→ R₂ – R₁

→

The matrix so obtained is of the form –

→ [ I : A^-1 ]

Hence inverse of the given matrix-

→ 

Question 3.

Using elementary transformations, find the inverse of each of the matrices.

Answer:

First of all we need to check whether the matrix is invertible or not. For that-

For the inverse of a matrix A to exist,

Determinant of A ≠ 0

Here ∣A∣ = (1)(7) – (2)(3) = 1

So the matrix is invertible.

Now to find the inverse of the matrix,

We know AA^-1 = I

Let’s make augmented matrix-

→ [ A : I ]

→

Apply row operation- R₂→ R₂ – 2R₁

→ 

Apply row operation- R₁→ R₁ - 3R₂

→

The matrix so obtained is of the form –

→ [ I : A^-1 ]

Hence inverse of the given matrix-

→ 

Question 4.

Using elementary transformations, find the inverse of each of the matrices.

Answer:

First of all we need to check whether the matrix is invertible or not. For that-

For the inverse of a matrix A to exist,

Determinant of A ≠ 0

Here ∣A∣ = (2)(7) – (5)(3) = -1

So the matrix is invertible.

Now to find the inverse of the matrix,

We know AA^-1 = I

Let’s make augmented matrix-

→ [ A : I ]

→

Apply row operation- R₂→ R₂ – R₁

→ 

Apply row operation- R₁→ R₁/2

→

Apply row operation- R₁→ R₁ + 3R₂

→

Apply row operation- R₂→ -2R₂

→

The matrix so obtained is of the form –

→ [I : A^-1]

Hence inverse of the given matrix-

→ 

Question 5.

Using elementary transformations, find the inverse of each of the matrices.

Answer:

First of all we need to check whether the matrix is invertible or not. For that-

For the inverse of a matrix A to exist,

Determinant of A ≠ 0

Here ∣A∣ = (2)(4) – (1)(7) = 1

So the matrix is invertible.

Now to find the inverse of the matrix,

We know AA^-1 = I

Let’s make augmented matrix-

→ [ A : I ]

→

Apply row operation- R₂→ R₂ – R₁

→ 

Apply row operation- R₂→ 2R₂

→

Apply row operation- R₁→ R₁ –R₂

→

Apply row operation- R₁→ R₁

→

The matrix so obtained is of the form –

→ [I : A^-1]

Hence inverse of the given matrix-

→ 

Question 6.

Using elementary transformations, find the inverse of each of the matrices.

Answer:

First of all we need to check whether the matrix is invertible or not. For that-

For the inverse of a matrix A to exist,

Determinant of A ≠ 0

Here ∣A∣ = (2)(3) – (1)(5) = 1

So the matrix is invertible.

Now to find the inverse of the matrix,

We know AA^-1 = I

Let’s make augmented matrix-

→ [ A : I ]

→

Apply row operation- R₂→ R₂ – R₁

→ 

Apply row operation- R₁→ R₁/2

→

Apply row operation- R₁→ R₁ - 5R₂

→

Apply row operation- R₂→ 2R₂

→

The matrix so obtained is of the form –

→ [ I : A^-1 ]

Hence inverse of the given matrix-

→ 

Question 7.

Using elementary transformations, find the inverse of each of the matrices.

Answer:

First of all we need to check whether the matrix is invertible or not. For that-

For the inverse of a matrix A to exist,

Determinant of A ≠ 0

Here ∣A∣ = (3)(2) – (5)(1) = 1

So the matrix is invertible.

Now to find the inverse of the matrix,

We know AA^-1 = I

Let’s make augmented matrix-

→ [ A : I ]

→

Apply row operation- R₂→ R₂ –R₁

→ 

Apply row operation- R₂→ 3R₂

→

Apply row operation- R₁→ R₁ - R₂

→

Apply row operation- R₁→ R₂

→

The matrix so obtained is of the form –

→ [I : A^-1]

Hence inverse of the given matrix-

→ 

Question 8.

Using elementary transformations, find the inverse of each of the matrices.

Answer:

First of all we need to check whether the matrix is invertible or not. For that-

For the inverse of a matrix A to exist,

Determinant of A ≠ 0

Here ∣A∣ = (4)(4) – (5)(3) = 1

So the matrix is invertible.

Now to find the inverse of the matrix,

We know AA^-1 = I

Let’s make augmented matrix-

→ [ A : I ]

→

Apply row operation- 

→ 

Apply row operation- R₁→ R₁/4

→

Apply row operation- R₁→ R₁ - 5R₂

→

Apply row operation- R₂→ 4R₂

→

The matrix so obtained is of the form –

→ [ I : A^-1 ]

Hence inverse of the given matrix -

→ 

Question 9.

Using elementary transformations, find the inverse of each of the matrices.

Answer:

First of all we need to check whether the matrix is invertible or not. For that-

For the inverse of a matrix A to exist,

Determinant of A ≠ 0

Here ∣A∣ = (3)(7) – (2)(10) = 1

So the matrix is invertible.

Now to find the inverse of the matrix,

We know AA^-1 = I

Let’s make augmented matrix-

→ [A : I]

→

Apply row operation- R₂→ R₂ – R₁

→ 

Apply row operation- R₁→ R₁/3

→

Apply row operation- R₁→ R₁ - 10R₂

→

Apply row operation- R₂→ 3R₂

→

The matrix so obtained is of the form –

→ [I : A^-1]

Hence inverse of the given matrix-

→ 

Question 10.

Using elementary transformations, find the inverse of each of the matrices.

Answer:

First of all we need to check whether the matrix is invertible or not. For that-

For the inverse of a matrix A to exist,

Determinant of A ≠ 0

Here ∣A∣ = (3)(2) – (-1)(-4) = 2

So the matrix is invertible.

Now to find the inverse of the matrix,

We know AA^-1 = I

Let’s make augmented matrix-

→ [ A : I ]

→

Apply row operation- R₂→ R₂ + R₁

→ 

Apply row operation- R₁→ R₁/3

→

Apply row operation- R₁→ R₁ + R₂

→

Apply row operation- R₂→ R₂

→

The matrix so obtained is of the form –

→ [ I : A^-1 ]

Hence inverse of the given matrix-

→ 

Question 11.

Using elementary transformations, find the inverse of each of the matrices.

Answer:

First of all we need to check whether the matrix is invertible or not. For that-

For the inverse of a matrix A to exist,

Determinant of A ≠ 0

Here ∣A∣ = (2)(-2) – (-6)(1) = 2

So the matrix is invertible.

Now to find the inverse of the matrix,

We know AA^-1 = I

Let’s make augmented matrix-

→ [ A : I ]

→

Apply row operation- R₂→ R₂ – R₁

→ 

Apply row operation- R₁→ R₁/2

→

Apply row operation- R₁→ R₁ + 3R₂

→

The matrix so obtained is of the form –

→ [I : A^-1]

Hence inverse of the given matrix-

→ 

Question 12.

Using elementary transformations, find the inverse of each of the matrices.

Answer:

First of all we need to check whether the matrix is invertible or not. For that-

For the inverse of a matrix A to exist,

Determinant of A ≠ 0

Here ∣A∣ = (6)(1) – (-2)(-3) = 0

So the matrix is not invertible.

∴ the inverse of the given matrix does not exist.

Question 13.

Using elementary transformations, find the inverse of each of the matrices.

Answer:

First of all we need to check whether the matrix is invertible or not. For that-

For the inverse of a matrix A to exist,

Determinant of A ≠ 0

Here ∣A∣ = (2)(2) – (-1)(-3) = 1

So the matrix is invertible.

Now to find the inverse of the matrix,

We know AA^-1 = I

Let’s make augmented matrix-

→ [ A : I ]

→

Apply row operation- R₂→ R₂ + R₁

→ 

Apply row operation- R₁→ R₁/2

→

Apply row operation- R₁→ R₁ + R₂

→

Apply row operation- R₂→ 2R₂

→

The matrix so obtained is of the form –

→ [ I : A^-1 ]

Hence inverse of the given matrix-

→ 

Question 14.

Using elementary transformations, find the inverse of each of the matrices.

Answer:

First of all we need to check whether the matrix is invertible or not. For that-

For the inverse of a matrix A to exist,

Determinant of A ≠ 0

Here ∣A∣ = (2)(2) – (1)(4) = 0

So the matrix is not invertible.

∴ the inverse of the given matrix does not exist.

Question 15.

Using elementary transformations, find the inverse of each of the matrices.

Answer:

First of all we need to check whether the matrix is invertible or not. For that-

For the inverse of a matrix A to exist,

Determinant of A ≠ 0

Here ∣A∣ = [ (2) {2×2- 3×(-2)} – (-3) {2 × 2 - 3×3} + (3) {2× (-2) – 2×3}]

= [2 {4-(-6)} + 3 {4-9} + 3 { -4-6}]

= [2(10) + 3(-5) + 3(-10)]

= [20-15-30]

= -25

So the matrix is invertible.

Now to find the inverse of the matrix,

We know AA^-1 = I

Let’s make augmented matrix-

→ [ A : I ]

→

Apply row operation- R₁→ R₁

→

Apply row operation- R₂→ R₂ -2R₁

→

Apply row operation- R₃→ R₃ - 3R₁

→.

Apply row operation- R₂→  R₂

Apply row operation- R₁→ R₁ +  R₂

→

Apply row operation- R₃→ R₃ -  R₂

→

Apply row operation- R₃→  R₃

→

Apply row operation- R₁→ R₁ R₃

→

The matrix so obtained is of the form –

→ [I : A^-1]

Hence inverse of the given matrix-

→ 

Question 16.

Using elementary transformations, find the inverse of each of the matrices.

Answer:

First of all we need to check whether the matrix is invertible or not. For that-

For the inverse of a matrix A to exist,

Determinant of A ≠ 0

Here ∣A∣ = [ (1) {0- 5 × (-5)} – (3) {(-3) × 0 – (-5) × 2} + (-2) {5 × (-3) – 2 × 0}]

= [1 {25} - 3 {0 + 10} - 2 {-15}]

= [1(25) - 3(10) – 2 (-15)]

= [25-30+30]

= 25

So the matrix is invertible.

Now to find the inverse of the matrix,

We know AA^-1 = I

Let’s make augmented matrix-

→ [A : I]

→

Apply row operation- R₂→ R₂ + 3R₁

→

Apply row operation- R₃→ R₃ -2R₁

→

Apply row operation- R₂→ R₂

→

Apply row operation- R₁→ R₁ – 3R₂

→

Apply row operation- R₃→ R₃ + R₂

→

Apply row operation- R₁→ R₁ - R₃

→

Apply row operation- R₂→ R₂ + R₃

→

Apply row operation- R₃→  R₃

→

The matrix so obtained is of the form –

→ [I : A^-1]

Hence inverse of the given matrix-

→ 

Question 17.

Using elementary transformations, find the inverse of each of the matrices.

Answer:

First of all we need to check whether the matrix is invertible or not. For that-

For the inverse of a matrix A to exist,

Determinant of A ≠ 0

Here |A| = [(2) {1×3 – 1 × (0)} – (0) {3 × 5 – 0 × 0} + (-1) {5 × 1 – 1 × 0}]

= [2{3} - 0 {15} - 1 {5}]

= [6-0-5]

= 1

So the matrix is invertible.

Now to find the inverse of the matrix,

We know AA^-1 = I

Let’s make augmented matrix-

→ [A : I]

→

Apply row operation- R₁→ R₁

→

Apply row operation- R₂→ R₂ -5R₁

→

Apply row operation- R₃→ R₃ – R₂

→

Apply row operation- R₁→ R₁ + R₃

→

Apply row operation- R₂→ R₂ –5R₃

→

Apply row operation- R₃ → 2R₃

→

The matrix so obtained is of the form –

→ [I : A^-1]

Hence inverse of the given matrix-

→ 

Question 18.

Matrices A and B will be inverse of each other only if
A. AB = BA

B. AB = BA = 0

C. AB = 0, BA = I

D. AB = BA = I

Answer:

Here it is given that A & B are inverse of each other.

∴ A^-1 = B -(i)

Also B^-1 = A -(ii)

From definition of inverse matrix, we know that-

→ AA^-1 = I

∵ A^-1 = B {from eq(i)}

→ AB = I -(iii)

Similarly, BB^-1 = I

∵ B^-1 = A {from eq(ii)}

→ BA = I -(iv)

So AB = BA = I {from eq(iii) and eq(iv)}

Hence option D is the correct answer.

---

Miscellaneous Exercise

Question 1.

Let, show that (aI + bA)ⁿ = aⁿI + na^n-1bA, where I is the identity matrix of order 2 and nϵN.

Answer:

To prove: (aI + bA)ⁿ = aⁿI + na^n-1bA

Proof: Given A 

We will be proving the above equation using mathematical induction.

Steps involved in mathematical inductionare-

1. Prove the equation for n=1

2. Assume the equation to be true for n=k, where kϵN

3. Finally prove the equation for n=k+1

I is the identity matrix of order 2,

i.e. I 

Let P(n): (aI + bA)ⁿ = aⁿI + na^n-1bA, n ϵ N

For n=1,

L.H.S: (aI + bA)¹ = aI + bA

R.H.S: a¹I + 1a^1-1bA= aI + a⁰bA= aI + bA

So, L.H.S = R.H.S

∴ P(n) is true for n=1

Now assuming P(n) to be true for n=k, where k ϵ N

P(k) : (aI + bA)^k = a^kI + ka^k-1bA …… (1)

Now proving for n=k+1, i.e. P(k+1) is also true

L.H.S = (aI + bA)^k+1

= (aI + bA)^k . (aI + bA)¹

= (a^kI + ka^k-1bA). (aI + bA) ……from (1)

= aI(a^kI + ka^k-1bA) + bA(a^kI + ka^k-1bA)

= aI(a^kI) + aI(ka^k-1bA) + bA(a^kI) + bA(ka^k-1bA)

= (a.a^k) (I×I) + kb(a.a^k-1)(IA) + (ba^k)(AI) + (bb) ka^k-1(AA)

= a^k+1 I² + ka^1+k-1 bA + ba^kA + b²ka^k-1A² (IA = AI= A & I² = I)

= a^k+1 I + ka^k bA + ba^kA + b²ka^k-1A²

Calculating A²

A² = A.A

∴ A² = O (O is the null matrix)

Putting value of A² in L.H.S

L.H.S = a^k+1 I + ka^k bA + ba^kA + b²ka^k-1(O)

= a^k+1 I + ka^k bA + ba^kA + 0

= a^k+1 I + ka^k bA + ba^kA

= a^k+1 I + (k+1)a^k bA

Putting n=k+1 in R.H.S

R.H.S = a^k+1 I + (k+1)a^k bA

∴ L.H.S = R.H.S

All conditions are proved. Hence P(k+1) is true.

∴ By mathematical induction we have proved that P(n) is true for all n ϵ N.

Thus,(aI + bA)ⁿ = aⁿI + na^n-1bA, where I is the identity matrix of order 2 and nϵN.

Question 2.

If  , prove that .

Answer:

We will be proving the above equation by putting different values of n (i.e. n = 1, 2, 3 ….n)

For n=1,

For n=2,

A² = A.A

For n=3,

A³ = A².A

r n=4,

A⁴ = A³.A

d so on for other values of n.

If we notice each result, then we will see that it is of same type that we are trying to prove.

So we can generalise the above results for all n ϵ N

Hence Proved

Question 3.

If  then prove that  where n is any positive integer.

Answer:

We will be proving the above equation by putting different values of n. Because n is a positive integer so it will take values which are greater than 0 i.e. n = 1, 2, 3 …n

For n=1,

L.H.S = Aⁿ = A¹ = A = 

R.H.S = 

∴ L.H.S = R.H.S

For n=2,

L.H.S = A²

= A.A

∴ L.H.S = R.H.S

For n = 3,

L.H.S = A³

= A².A

∴ L.H.S = R.H.S

For n = 4,

L.H.S = A⁴

= A³.A

=

==

R.H.S = 

=

=

∴ L.H.S = R.H.S

And so on for other values of n.

If we notice each result then we will see that it is of same type that we are trying to prove.

So we can generalise the above results for all positive integer values of n.

Hence Proved

Question 4.

If A and B are symmetric matrices, prove that AB – BA is a skew symmetric matrix.

Answer:

To prove: AB – BA is a skew symmetric matrix.

Symmetric matrix: A symmetric matrix is a square matrix that is equal to its transpose. In simple words, matrix A is symmetric if

A = A’

where A’ is the transpose of matrix A.

Skew Symmetric matrix: A skewsymmetric matrix is a square matrix that is equal to minus of its transpose. In simple words, matrix A is skew symmetric if

A = -A’

Given: A and B are symmetric matrices i.e.

A = A’ …(1)

B = B’ …(2)

Now calculating the transpose of AB – BA,

(AB – BA)’ = (AB)’ – (BA)’

(By property of transpose i.e. (A – B)’ = A’ – B’)

= B’A’ – A’B’

(By property of transpose i.e. (AB)’ = B’A’)

= BA – AB

= -(AB – BA)

Or we can say that: (AB – BA) = - (AB – BA)’

Clearly it satisfies the condition of skew symmetric matrix.

Hence AB–BA is a skew symmetric matrix.

Question 5.

Show that the matrix B’AB is symmetric or skew symmetric according as A is symmetric or skew symmetric.

Answer:

Case 1: When A is a symmetric matrix i.e.

A = A’……. (1)

where A’ is the transpose of A

To prove: B’AB is also a symmetric matrix.

Calculating the transpose of B’AB

(B’AB)’= B’A’(B’)’ (By property of transpose i.e. (AB)’ = B’A’)

= B’A’B (By property of transpose i.e. (A’)’ = A)

= B’AB (from (1))

It satisfies the condition of symmetric matrix as matrix B’AB is equal to its transpose.

HenceB’ABis a symmetric matrix when A is symmetric.

Case 2: When A is a skew symmetric matrix i.e.

A = -A’……. (2)

where A’ is the transpose of A.

To prove: B’AB is also a skew symmetric matrix.

Calculating the transpose of B’AB

(B’AB)’= B’A’(B’)’ (By property of transpose i.e. (AB)’ = B’A’)

= B’A’B (By property of transpose i.e. (A’)’ = A)

= B’(-A)B (from (2))

= - (B’AB)

It satisfies the condition of skew symmetric matrix as matrix (B’AB) is equal to its transpose.

Hence(B’AB)is a skew symmetric matrix when A is skew symmetric.

∴ Both results are proved.

Question 6.

Find the values of x, y, z if the matrix satisfy  the equation A’A =I.

Answer:

Given A 

Transpose of a matrix: If A be an m×n matrix, then the matrix obtained by interchanging the rows and columns of A is called the transpose of A. It is denoted by A’ or A^T.

∴ Transpose of A = A’ =

Given equation

A’A =I

 =

=

. =

As these matrices are equal to each other that means each element of matrix on L.H.S is equal each element of matrix on R.H.S.

∴ On comparing elements on both sides we get

4y² + z² = 1 …… (1)

2y² – z² = 0 …… (2)

x² + y² + z² = 1 …… (3)

– y² – z² = 0 …… (4)

From equation (4) we get,

x² = y² + z² …… (5)

Substituting this value in equation (3) we get,

2y² + 2z² = 1 …… (6)

Subtracting equation (2) and (6) we get,

3z² = 1

z² = 1/3

z =1/3

Substituting value of z in equation (2) we get,

2y² = 1/3

y² = 1/6

y = �1/6

Substituting values of y and z in equation (5) we get,

x² = 1/2

x = 1/2

Hence values of x, y, z are 1/2, 1/6, 1/3 respectively.

Question 7.

For what values of 

Answer:

Multiplying matrices on the left hand side

L.H.S = 

= [6.0 + 2.2 + 4.x]

= [4 + 4x]

R.H.S = O (Ois the null matrix)

= 0

∴ L.H.S = R.H.S, so we get,

[4+4x] = O

4 + 4x = 0

4x = -4

x = -1

Hence value of x is equal to -1.

Question 8.

If  show that A² – 5A + 7I =0.

Answer:

To prove: A² – 5A + 7I = 0

Given: A = 

L.H.S: A² – 5A + 7I

R.H.S = 0

I = 

Calculating value of A²:

A² = A.A

=

=

=

=

Substituting value in L.H.S we get,

= A² – 5A + 7I

.

= 0 = R.H.S

L.H.S = R.H.S

Hence A² – 5A + 7I = 0 is proved.

Question 9.

Find x, if 

Answer:

Multiplying matrices on the left hand side

L.H.S = 

= 

= 

= 

= 

=

=

R.H.S = O

=

∴ L.H.S = R.H.S we get,

x² – 48 =0

x² = 48

Hence the required value of

Question 10.

A manufacturer produces three products x, y, z which he sells in two markets. Annual sales are indicated below:



(a) If unit sale prices of x, y and z are Rs. 2.50, Rs. 1.50 and Rs. 1.00, respectively, find the total revenue in each market with the help of matrix algebra.

(b) If the unit costs of the above three commodities are Rs. 2.00, Rs. 1.00 and 50 paise respectively. Find the gross profit.

Answer:

(a) Given the unit sale prices of x, y and z as Rs. 2.50, Rs. 1.50 and Rs. 1.00 respectively.

Unit sale prices can be represented in form of matrix as: 

Calculating total revenue in market I:

Number of products in the form of matrix:

So the total revenue is given by:

=

=

=

=

∴ Total revenue in market is Rs 46000.

Calculating total revenue in market II:

Number of products in the form of matrix: 

So the total revenue is given by:

=

=

=

=.

∴ Total revenue in market is Rs. 53000.

(b) Given the unit cost prices of x, y and z as Rs. 2.00, Rs. 1.00 and 50 paisa respectively.

Calculating gross profit in market I:

Unit cost prices can be represented in form of matrix as: 

So the total cost of products in market I is given by:

=

=

=

Since the total revenue in market I is Rs. 46000, the gross profit in this market is given by:

(Rs. 46000 – Rs. 31000)

= Rs. 15000.

Calculating gross profit in market II:

The total cost of products in market II is given by:

=

=

=

=

Since the total revenue in market II is Rs. 53000, the gross profit in this market is given by:

(Rs. 53000 – Rs. 36000)

= Rs. 17000.

Question 11.

Find the matrix X so that 

Answer:

Given X

From above equation it can be observed that matrix on R.H.S is a 2×3 matrix and that on the L.H.S is also a 2×3 matrix. Therefore, X must be a 2×2 matrix.

Let X 

So the equation is given by:

Now equating the corresponding elements of both the matrices we get,

a + 4b = -7, 2a + 5b = -8, 3a + 6b = -9

c + 4d = 2, 2c + 5d = 4, 3c + 6d = 6

Now, a + 4b = -7 ⇒ a = -4b -7

∴ 2a + 5b = -8 ⇒ 2.(-4b -7) + 5b = -8

⇒ -8b -14 + 5b = -8

⇒ -3b = 6

⇒ b = -2

∴ a = -4b -7 ⇒ a = -4.(-2) -7

⇒ a = 1

Now, c + 4d = 2 ⇒ c = -4d + 2

∴ 2c + 5d = 4 ⇒ 2.(-4d + 2) + 5d = 4

⇒ -8d + 4 + 5d = 4

⇒ -3d = 0

⇒ d = 0

∴ c = -4d + 2 ⇒ c = -4.0 + 2

⇒ c = 2

Thus, a = 1, b = -2, c = 2, d = 0.

Hence X becomes

Question 12.

If A and B are square matrices of the same order such that AB = BA, then prove by induction that ABⁿ = BⁿA. Further, prove that (AB)ⁿ = AⁿBⁿ for all nϵN.

Answer:

To prove: ABⁿ = BⁿA

Given A and B are square matrices of same order such that AB = BA.

We have to prove it using mathematical induction.

Steps involved in mathematical induction are-

1. Prove the equation for n=1

2. Assume the equation to be true for n=k, where kϵN

3. Finally prove the equation for n=k+1

Let P(n): ABⁿ = BⁿA

For n=1,

L.H.S: ABⁿ = AB¹ = AB

R.H.S: BⁿA = B¹A = BA = AB

So, L.H.S = R.H.S

∴ P(n) is true for n=1.

Now assuming P(n) to be true for n=k, where k ϵ N

P(k): AB^k = B^kA …… (1)

Now proving for n=k+1, i.e. P(k+1) is also true

L.H.S = ABⁿ

= AB^k+1

= (AB^k).B

= (B^kA).B …… from (1)

= B^k(A.B)

= B^k(BA) (∵AB = BA)

= B^k+1A

R.H.S = BⁿA

= B^k+1A

∴ L.H.S = R.H.S

All conditions are proved. Hence P(k+1) is true.

∴ By mathematical induction we have proved that ABⁿ = BⁿA.

Now, to prove: (AB)ⁿ = AⁿBⁿ for all n ϵ N

For n=1,

L.H.S = (AB)ⁿ = (AB)¹ = AB

R.H.S = AⁿBⁿ = A¹B¹ = AB

∴ L.H.S = R.H.S

∴ It is true for n=1

Assuming it to be true for n=k then,

(AB)^k = A^kB^k ……(2)

Now proving for n=k+1,

L.H.S = (AB)ⁿ

= (AB)^k+1

= (AB)^k(AB)¹

= (A^kB^k)AB

= A^k(B^k.A)B

= A^k(A.B^k)B (ABⁿ = BⁿA)

= (A^kA)(B^kB)

= A^k+1B^k+1

R.H.S = AⁿBⁿ

= A^k+1B^k+1

∴ L.H.S = R.H.S

All conditions are proved. Hence P(k+1) is true.

∴ By mathematical induction we have proved that (AB)ⁿ = AⁿBⁿ for all nϵN

Hence proved.

Question 13.

If  is such that A² = I, then
A. 

B. 

C. 

D. 

Answer:

Given A = 

Calculating A²:

A² = A.A

=

=

=

=

And given that A² = I

Then 

Comparing corresponding elements we get,

Question 14.

If the matrix A is both symmetric and skew symmetric, then
A. A is a diagonal matrix

B. A is a zero matrix

C. A is a square matrix

D. None of these

Answer:

Given A is both symmetric and skew symmetric matrix then, A = A’ and also A = -A’

⇒ A’ = -A’

⇒ 2A’ = 0

⇒ A’ = O

Clearly it is observed that transpose of A is a null matrix or zero matrix then matrix A must also be a zero matrix.

Hence A is a zero matrix.

Question 15.

If A is square matrix such that A² = A, then (I + A)³ – 7 A is equal to
A. A

B. I – A

C. I

D. 3A

Answer:

Given that A² = A

Calculating value of (I + A)³ – 7 A:

⇒ I³ + A³ + 3I²A + 3IA² – 7A

⇒ I + A².A + 3A + 3A² – 7A (Iⁿ = I and I.A = A)

⇒ I + A.A + 3A + 3A – 7A (A² = A)

⇒ I + A² + 3A + 3A – 7A

⇒ I + A - A

⇒ I

Hence(I + A)³ – 7 A = I.