Chapter 2 Matrices Miscellaneous Exercise 2B

I) Choose the correct answer from the given alterna-tives in each of the following questions :
Question 1.

word image 18876 1

(a) a = – 2, b = 1
(b) a = 2, b = 4
(c) a = 2, b = –1
(d) a = 1, b = –2
Solution:
(a) a = – 2, b = 1

Question 2.


Solution:

word image 18876 4

Question 3.

(a) 1
(b) -1
(c) 0
(d) -3
word image 18876 6
(d) -3 [Hint : A(adj A) = |A| ∙ I]

Question 4.


Solution:

word image 18876 9

Question 5.

(a) 2
(b) -2
(c) 10
(d) -10
Solution:
(b) -2

Question 6.

(a) 0
(b) ± 1
(c) 2
(d) 3
Solution:
(b) ± 1

Question 7.


Solution:

word image 18876 14

Question 8.
word image 18876 15
(a) F(-∝)
(b) F(∝-1)
(c) F(2∝)
(d) None of these
Solution:
(a) F(-∝)

Question 9.
word image 18876 16

(a) I
(b) A
(c) A’
(d) -I
Solution:
(b) A

Question 10.
The inverse of a symmetric matrix is
(a) Symmetric
(b) Non-symmetric
(c) Null matrix
(d) Diagonal matrix
Solution:
(a) Symmetric

Question 11.
word image 18876 17
(a) 20
(b) 10
(c) 30
(d) 40
Solution:
(b) 10

Question 12.


Solution:

II) Solve the following equations by the methods of inversion.
(i) 2x – y = -2 , 3x + 4y = 5
Solution:

The given equations can be written in the matrix form as :
word image 18876 21
Maharashtra Board 12th Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2B 6
Maharashtra Board 12th Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2B 7
By equality of matrices,
word image 18876 24

(ii) x + y + z = 1, 2x + 3y + 2z = 2 and ax + ay + 2az = 4, a ≠ 0.
Solution:

The given equations can be written in the matrix form as :
word image 18876 25
= 1(6a – 2a) – 1(4a – 2a) + 1(2a – 3a)
= 4a – 2a – a = a ≠ 0 ∴ A-1 exists.
Consider AA-1 = I
Maharashtra Board 12th Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2B 9
Maharashtra Board 12th Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2B 10
word image 18876 28
Maharashtra Board 12th Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2B 12

(iii) 5x – y +4z = 5, 2x + 3y + 5z = 2 and 5x – 2y + 6z = -1
Solution:

The given equations can be written in the matrix form as :
Maharashtra Board 12th Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2B 13
= 5(18 + 10) + 1 (12 – 25) + 4( -4 – 15)
= 140 – 13 – 76 = 51 #0
∴ A-1 exists.
Now, we have to find the cofactor matrix
word image 18876 31
Maharashtra Board 12th Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2B 15
Now, premultiply AX = B by A-1, we get,
A-1(AX) = A-1B
∴ (A-1A)X = A-1B
∴ IX = A-1B
∴ X = 151⎡⎣⎢2813−19−2105−17−1717⎤⎦⎥⎡⎣⎢52−1⎤⎦⎥
Maharashtra Board 12th Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2B 16
By equality of matrices,
x = 3, y = 2, z = -2 is the required solution.

(iv) 2x + 3y = -5, 3x + y = 3
Solution:

(v) x + y + z = -1, y + z = 2 and x + y – z = 3
Solution:

The given equations can be written in the matrix form as :
word image 18876 34
= 1(-1 – 1) – 1 (0 – 1) + 1(0 – 1)
= -2 + 1 – 1 = -2 ≠ 0 ∴ A-1 exists.
Consider AA-1 = I
Maharashtra Board 12th Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2B 18
Maharashtra Board 12th Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2B 19
Now, premultiply AX = B by A-1, we get,
A-1(AX) = A-1B
∴ (A-1A)X = A-1B
∴ IX = A-1B
Maharashtra Board 12th Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2B 20
∴ by equality of the matrices, x= -3, y = 4, z = -2 is the required solution.

Question 2.
Express the following equation in matrix from and solve them by the method of reduction.
(i) x – y + z = 1, 2x – y = 1, 3x + 3y – 4z = 2
Solution:

The given equations can be written in the matrix form as :
word image 18876 38
Maharashtra Board 12th Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2B 22
By equality of matrices,
x – y + z = 1 ……(1)
y – 2z = -1 …..(2)
5z = 5 ….(3)
From (3), z = 1
Substituting z = 1 in (2), we get,
y – 2 = -1 ∴ y = 1
Substituting y = 1, z = 1 in (1), we get,
x – 1 + 1 = 1
∴ x = 1
Hence, x = 1, y = 1, z = 1 is the required solution.

(ii) x + y = 1, y + z = 5/3, z + x = 4/3.
Solution:

The given equations can be written in the matrix form as :
word image 18876 40
By equality of matrices,
x + y = 1 ……(1)
y + z = 53 …(2)
2z = 2 ……..(3)
From (3), z = 1
Substituting z = 1 in (2), we get,
y + 1 = 5/3 ∴ y = 2/3
Substituting y = 2/3 in (1), we get,
x + 2/3 = 1 ∴ x = 1/3
Hence, x = 1/3, y = 2/3, z = 1 is the required solution.

(iii) 2x – y + z = 1, x + 2y + 3z = 8 and 3x + y – 4z = 1
Solution:

The given equations can be written in the matrix form as :
word image 18876 41
word image 18876 42
By equality of matrices,
x + 2y + 3z = 8 …..(1)
-5y – 5z = -15 ….(2)
-8z = -8 …..(3)
From (3), z = 1
Substituting z = 1 in (2), we get,
-5y – 5 = -15
-5y = -10
∴ y = 2
Substituting y = 2, z = 1 in (1), we get,
x + 4 + 3 = 8 ∴ x = 1
Hence, x = 1, y = 2, z = 1 is the required solution.

(iv) x + y + z = 6, 3x – y + 3z =10 and 5x + 5y – 4z = 3.
Solution:

(v) x + 2y + z = 8, 2x + 3y – z =11 and 3x – y – 2z = 5
Solution:

The given equations can be written in the matrix form as :
word image 18876 43
By equality of matrices,
x + 2y + z = 8 … (1)
-y – 3z = -5 … (2)
16z = 16 … (3)
From (3), z = 1
Substituting z = 1 in (2), we get,
-y – 3 = -5, ∴ y = 2
Substituting y = 2, z = 1 in (1), we get,
x + 4 + 1 = 8 ∴ x = 3
Hence, x = 3, y = 2, z = 1 is the required solution.

(vi) x + 3y + 2z = 6, 3x – 2y + 5z =5 and 2x – 3y + 6z = 7.
Solution:

The given equations can be written in the matrix form as :
Maharashtra Board 12th Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2B 26
Maharashtra Board 12th Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2B 27
By equality of matrices,
word image 18876 46
Substituting y = 1, z = 2 in (2), we get,
x + 3y + 2z = 6
x + 3(1) + 2(2) = 6
x + 3 + 4 = 6
x = -1
Hence, x = -1, y = 1, z = 2 is the required solution.

Question 3.
The sum of three numbers is 6. If we multiply third number by 3 and add it to the second number we get 11. By adding first and the third numbers we get a number which is double the second number. Use this information and find a system of linear equations. Find the three numbers using matrices.
Solution:

Let the three numbers be x, y and z. According to the given conditions,
x + y + z = 6.
3z + y = 11, i.e., y + 3z = 11 and x + z = 2y,
i.e., x – 2y + z = 0
Hence, the system of the linear equations is
x + y + z = 6
y + 3z = 11
x – 2y + z = 0
These equations can be written in the matrix form as :
word image 18876 47
By equality of matrices,
x + y + z = 6 …(1)
y + 3z = 11 …(2)
-3y = -6 …(3)
From (3), y = 2
Substituting y = 2 in (2), we get,
2 + 3z = 11
∴ 3z = 9 ∴ z = 3
Put y = 2, z — 3 in (1), we get,
x + 2 + 3 = 6 ∴ x = 1
∴ x = 1, y = 2, z = 3
Hence, the required numbers are 1, 2 and 3.

Question 4.
The cost of 4 pencils, 3 pens and 2 books is ₹ 150. The cost of 1 pencil, 2 pens and 3 books is ₹ 125. The cos of 6 pencils, 2 pens and 3 books is ₹ 175. Fild the cost of each item by using Matrices.
Solution:

Let the cost of 1 pencil, 1 pen and 1 book be ₹x, ₹ y, ₹ z respectively.
According to the given conditions,
4x + 3y + 2z = 150
x + 2y + 3z = 125
6x + 2y + 3z = 175
The equations can be written in matrix form as :
word image 18876 48
By equality of matrices,
x + 2y + 3z = 125 …(1)
-5y – 10z = -350 …(2)
5z = 125 …(3)
From (3), z = 25
Substituting z = 25 in (2), we get
-5y – 10(25) = -350
∴ -5y = -350 + 250 = -100
∴ y = 20
Substituting y = 20, z = 25 in (1), we get
x + 2(20) + 3(25) = 125
∴ x = 125 – 40 – 75 = 10
∴ x = 10, y = 20, z = 25
Hence, the cost of 1 pencil is ₹ 10, 1 pen is ₹ 20 and 1 book is ₹ 25.

Question 5.
The sum of three numbers is 6. Thrice the third number when added to the first number, gives 7. On adding three times first number to the sum of second and third number, we get 12. Find the three numbers by using Matrices.
Solution:

Let the numbers be x, y and z.
According to the given conditions,
x + y + z = 6
3z + x = 7, i.e., x + 3z = 7
and 3x + y + z = 12
Hence, the system of linear equations is
x + y + z = 6
x + 3z = 7
3x + y + z = 12
These equations can be written in matrix form as :
word image 18876 49
By equality of matrices,
word image 18876 50

Question 6.
The sum of three numbers is 2. If twice the second number is added to the sum of first and third number, we get 1 adding five times the first number to the sum of second and third we get 6. Find the three numbers by using matrices.
Solution:

Let the three numbers be x, y and z.
According to the question,
x + y + 2
x + 2y + z = 1
5x + y + z = 6
The given system of equations can be written in matrix form as follows:
Maharashtra Board 12th Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2B 33
Maharashtra Board 12th Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2B 34
Maharashtra Board 12th Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2B 35

Question 7.
An amount of ₹ 5000 is invested in three types of investments, at interest rates 6%, 7%, 8% per annum respectively. The total annual income from these investimest is ₹ 350. If the total annual income from first two investment is ₹ 70 more than the income from the third, find the amount of each investment using matrix method.
Solution:

word image 18876 54
word image 18876 55
Maharashtra Board 12th Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2B 32
By equality of matrices,
x + y + z = 5000 …(1)
y + 2z = 5000 …(2)
-16z = -28000 ….(3)
From (3), z = 1750
Substituting z = 1750 in (2), we get,
y + 2(1750) = 5000
∴ y = 5000 – 3500 = 1500
Substituting y = 1500, z = 1750 in (1), we get,
x + 1500 + 1750 = 5000
∴ x = 5000 – 3250 = 1750
∴ x = 1750, y = 1500, z = 1750
Hence, the amounts of the three investments are ₹ 1750, ₹ 1500 and ₹ 1750 respectively.

Question 8.
The sum of the costs of one ook each of Mathematics, Physics and Chemistry is ₹ 210. Total cost of a mathematics book, 2 physics books, and a chemistry book is ₹ 240 Also the total cost of a Mathematics book, 3 physics book and chemistry books is Rs. 300/-. Find the cost of each book, using Matrices.
Solution: