Class 12 RD Sharma Solutions β Chapter 6 Determinants β Exercise 6.4 | Set 2
Question 17. Solve the following system of the linear equations by Cramerβs rule.
2x β 3y β 4z = 29
β2x + 5y β z = β15
3x β y + 5z = β11
Solution:
Using Cramerβs Rule, we get,
Expanding along R1, we get,
= 2 (24) + 3 (β13) + 4 (-13)
= 48 β 21 β 52
= -25
Also, we get,
Expanding along R1, we get,
= 29 (24) + 3 (β64) + 4 (β40)
= 692 β 192 β 160
= 344
Expanding along R1, we get,
= 2 (β64) β 29 (β7) + 4 (23)
= β128 + 203 + 92
= 167
Expanding along R1, we get,
= 2 (40) + 3 (23) + 29 (β13)
= 80 + 69 β 377
= β228
So, x = D1/D = -344/25
y = D2/D = -167/25
z = D3/D = 228/25
Therefore, x = -344/25, y = -167/25, and z = 228/25.
Question 18. Solve the following system of the linear equations by Cramerβs rule.
x + y = 1
x + z = β6
x β y β 2z = 3
Solution:
Using Cramerβs Rule, we get,
= 1(1) β 1(-3)
= 1 + 3
= 4
Also, we get,
Expanding along R1, we get,
= 1 (1) β 1 (9) + 0
= 1 β 9
= β8
Expanding along R1, we get,
= 1 (9) β 1 (β3)
= 9 + 3
= 12
Expanding along R1, we get,
= 1 (β6) β 1 (9) + 1 (β1)
= β6 β 9 β 1
= β16
So, x = D1/D = -8/4 = -2
y = D2/D = 12/4 = 3
z = D3/D = -16/4 = -4
Therefore, x = β2, y = 3 and z = β4.
Question 19. Solve the following system of linear equations by Cramerβs rule.
x + y + z + 1 = 0
ax + by + cz + d = 0
a2x + b2y + c2z + d2 = 0
Solution:
Using Cramerβs Rule, we get,
c2 -> c2 β c1, c3 -> c3 β c1
Taking common (b-a) from c2 and (c-a)c3
Expanding along R1, we get,
= (b β a)(c β a)(c + a β b β a)
= (b β a)(c β a)(c β b)
= (a β b)(b β c)(c β a)
= -(d β b)(b β c)(c β d)
= -(a β d)(d β c)(c β a)
= -(a β d)(b β d)(d β a)
So, x = D1/D = -(d β b)(b β c)(c β d)/(a β b)(b β c)(c β a)
y = D2/D = -(a β d)(d β c)(c β a)/(a β b)(b β c)(c β a)
z = D3/D = -(a β d)(b β d)(d β a)/(a β b)(b β c)(c β a)
Question 20. Solve the following system of linear equations by Cramerβs rule.
x + y + z + w = 2
x β 2y + 2z + 2w = β6
2x + y β 2z + 2w = β5
3x β y + 3z β 3w = β3
Solution:
Using Cramerβs Rule, we get,
=
=
= β94
Also, we get,
So, x = D1/D = -188/94 = -2
y = D2/D = -282/-94 = 3
z = D3/D = -141/-94 = 3/2
w = D4/D = -47/94 = -1/2
Therefore, x = β2, y = 3 and z = 3/2 and w = -1/2.
Question 21. Solve the following system of linear equations by Cramerβs rule.
2x β 3z + w = 1
x β y + 2w = 1
β3y + z + w = 1
x + y + z = β1
Solution:
Using Cramerβs Rule, we get,
=
=
= β21
Also, we get,
So, x = D1/D = -21/-21 = 1
y = D2/D = -6/-21 = 2/7
z = D3/D = -6/-21 = 2/7
w = D4/D = -3/21 = -1/7
Therefore, x = 1, y = 2/7 and z = 2/7 and w = -1/7.
Question 22. Show that the following system of linear equations is inconsistent.
2x β y = 5
4x β 2y = 7
Solution:
Using Cramerβs Rule, we get,
= β4 + 4
= 0
Also, we get,
= β 10 + 7
= β3
= 14 β 20
= β6
Since D = 0 and D1 and D2 both are non-zero, the given system of equations is inconsistent.
Hence proved.
Question 23. Show that the following system of linear equations is inconsistent.
3x + y = 5
β6x β 2y = 9
Solution:
Using Cramerβs Rule, we get,
= β6 + 6
= 0
Also, we get,
= β10 β 9
= β19
= 27 + 30
= 57
Since D = 0 and D1 and D2 both are non-zero, the given system of equations is inconsistent.
Hence proved.
Question 24. Show that the following system of linear equations is inconsistent.
3x β y + 2z = 3
2x + y + 3z = 5
x β 2y β z = 1
Solution:
Using Cramerβs Rule, we get,
Expanding along R1, we get
= 3 (5) + 1 (β5) + 2 (β5)
= 15 β 5 β 10
= 0
Also, we get,
Expanding along R1, we get
= 3 (5) + 1 (β8) + 2 (β11)
= 15 β 8 β 22
= β15
Since D = 0 and D1 are non-zero, the given system of equations is inconsistent.
Hence proved.
Question 25. Show that the following system of linear equations is consistent and solve.
3x β y + 2z = 6
2x β y + z = 2
3x + 6y + 5z = 20
Solution:
Using Cramerβs Rule, we get,
Expanding along R1, we get
= 3 (β11) + 1 (7) + 2 (15)
= β33 + 7 + 30
= 4
Also, we get,
Expanding along R1, we get
= 6 (β11) + 1 (β10) + 2 (32)
= β66 β 10 + 64
= β12
Expanding along R1, we get
= 3 (β10) β 6 (7) + 2 (34)
= β30 β 42 + 68
= β4
Expanding along R1, we get
= 3 (β32) + 1 (34) + 6 (15)
= β96 + 34 + 90
= 28
As D, D1, D2 and D3 all are non-zero, the given system of equations is consistent.
So, x = D1/D = -12/4 = -3
y = D2/D = -4/4 = -1
z = D3/D = 28/4 = 7
Therefore, x = β3, y = β1 and z = 7.
Question 26. Show that the following system of linear equations has infinite number of solutions.
x β y + z = 3
2x + y β z = 2
βx β 2y + 2z = 1
Solution:
Using Cramerβs Rule, we get,
=
= 0
Also, we get,
=
= 0
=
= 0
=
= 0
As D, D1, D2 and D3 all are zero, the given system of equations has infinite number of solutions.
Hence proved.
Question 27. Show that the following system of linear equations has infinite number of solutions.
x + 2y = 5
3x + 2y = 15
Solution:
Using Cramerβs Rule, we get,
= 6 β 6
= 0
Also, we get,
= 30 β 30
= 0
= 15 β 15
= 0
As D, D1 and D2 all are zero, the given system of equations has infinite number of solutions.
Hence proved.
Question 28. Show that the following system of linear equations has infinite number of solutions.
x + y β z = 0
x β 2y + z = 0
3x + 6y β 5z = 0
Solution:
Using Cramerβs Rule, we get,
=
= 1 (6 β 6)
= 0
Also, we get,
= 0
= 0
= 0
As D, D1, D2 and D3 all are zero, the given system of equations has infinite number of solutions.
Hence proved.
Question 29. Show that the following system of linear equations has infinite number of solutions.
2x + y β 2z = 4
x β 2y + z = β2
5x β 5y + z = β2
Solution:
Using Cramerβs Rule, we get,
=
= 1 (β36 + 36)
= 0
Also we get,
=
= 0
=
= 0
=
= 2 (β12 + 12)
= 0
As D, D1, D2 and D3 all are zero, the given system of equations has infinite number of solutions.
Hence proved.
Question 30. Show that the following system of linear equations has infinite number of solutions.
x β y + 3z = 6
x + 3y β 3z = β4
5x + 3y + 3z = 10
Solution:
Using Cramerβs Rule, we get,
=
= 3 (12 β 12)
= 0
Also we get,
=
= 3 (12 β 12)
= 0
=
= 3 (12 β 12)
= 0
=
= 1 (β80 + 80)
= 0
As D, D1, D2 and D3 all are zero, the given system of equations has infinite number of solutions.
Hence proved.
Question 31. A salesman has the following record of sales during three months for three items A, B and C which have different rates of commission.
Month | Sale of units | Total Commission (in Rs) | ||
---|---|---|---|---|
A | B | C | ||
Jan | 90 | 100 | 20 | 800 |
Feb | 130 | 50 | 40 | 900 |
Mar | 60 | 100 | 30 | 850 |
Find out the rates of commission on items A, B and C by using the determinant method.
Solution:
Let the rates of commission on items A, B and C be x, y and z respectively.
According to the question, we have,
90x + 100y + 20z = 800
130x + 50y + 40z = 900
60x + 100y + 30z = 850
Using Cramerβs Rule, we get,
=
= 50 (8500 β 12000)
= β175000
Also we get,
=
= 50 (50000 β 57000)
= β350000
=
= 20 (17500 β 52500)
= β700000
=
= 50 (161500 β 200000)
= β1925000
So, x = D1/D = -350000/-175000 = 2
y = D2/D = -700000/-175000 = 4
z = D3/D = -1925000/-175000 = 11
Therefore, the rates of commission on items A, B and C are 2%, 4% and 11% respectively.
Question 32. An automobile company uses three types of steel S1, S2, and S3 for producing three types of cars C1, C2 and C3. Steel requirements (in tons) for each type of cars are given below.
Steel | Cars | ||
---|---|---|---|
C1 | C2 | C3 | |
S1 | 2 | 3 | 4 |
S2 | 1 | 1 | 2 |
S3 | 3 | 2 | 1 |
Using Cramerβs Rule, find the number of cars of each type which can be produced using 29, 13 and 16 tons of steel of three types respectively.
Solution:
Let x, y and z be the number of cars C1, C2 and C3 produced respectively.
According to the question, we have,
2x + 3y + 4z = 29
x + y + 2z = 13
3x + 2y + z = 16
Using Cramerβs Rule, we get,
=
= 1 (30 β 25)
= 5
Also we get,
=
= 1 (105 β 95)
= 10
=
= 1 (190 β 175)
= 15
=
= β2 (16 β 26)
= 20
So, x = D1/D = 10/5 = 2
y = D2/D = 15/5 = 3
z = D3/D = 20/5 = 4
Therefore, the number of cars produced of type C1, C2 and C3 are 2, 3 and 4.