Class 12 RD Sharma Solutions β Chapter 6 Determinants β Exercise 6.4 | Set 1
Question 1. Solve the following system of linear equations by Cramerβs rule.
x β 2y = 4
β3x + 5y = β7
Solution:
Using Cramerβs Rule, we get,
= 5 β 6
= β1
Also, we get,
= 20 β 14
= 6
= β7 + 12
= 5
So, x = D1/D = 6/-1 = -6
And y = D2/D = 5/-1 = -5
Therefore, x = β6 and y = β5.
Question 2. Solve the following system of linear equations by Cramerβs rule.
2x β y = 1
7x β 2y = β7
Solution:
Using Cramerβs Rule, we get,
= β4 + 7
= 3
Also, we get,
= β2 β 7
= β9
= β14 β 7
= β21
So, x = D1/D = -9/3 = -3
And y = D2/D = -21/3 = -7
Therefore, x = β3 and y = β7.
Question 3. Solve the following system of linear equations by Cramerβs rule.
2x β y = 17
3x + 5y = 6
Solution:
Using Cramerβs Rule, we get,
= 10 + 3
= 13
Also, we get,
= 85 + 6
= 91
= 12 β 51
= β39
So, x = D1/D = 91/3 = 7
And y = D2/D = -39/13 = -3
Therefore, x = 7 and y = β3.
Question 4. Solve the following system of linear equations by Cramerβs rule.
3x + y = 19
3x β y = 23
Solution:
Using Cramerβs Rule, we get,
= β3 β 3
= β6
Also, we get,
= β19 β 23
= β42
= 69 β 57
= 12
So, x = D1/D = -42/-6 = 7
And y = D2/D = 12/-6 = -2
Therefore, x = 7 and y = β2.
Question 5. Solve the following system of linear equations by Cramerβs rule.
2x β y = β2
3x + 4y = 3
Solution:
Using Cramerβs Rule, we get,
= 8 + 3
= 11
Also, we get,
= β8 + 3
= β5
= 6 + 6
= 12
So, x = D1/D = -5/11
And y = D2/D = 12/11
Therefore, x = -5/11 and y = 12/11.
Question 6. Solve the following system of linear equations by Cramerβs rule.
3x + ay = 4
2x + ay = 2, a β 0
Solution:
Using Cramerβs Rule, we get,
= 3a β 2a
= a
Also, we get,
= 4a β 2a
= 2a
= 6 β 8
= β2
So, x = D1/D = 2a/a = 2
And y = D2/D = -2/a
Therefore, x = a and y = -2/a.
Question 7. Solve the following system of linear equation by Cramerβs rule.
2x + 3y = 10
x + 6y = 4
Solution:
Using Cramerβs Rule, we get,
= 12 β 3
= 9
Also, we get,
= 60 β 12
= 48
= 8 β 10
= β2
So, x = D1/D = 48/9 = 16/3
And y = D2/D = -2/9
Therefore, x = 4/3 and y = -2/9.
Question 8. Solve the following system of linear equation by Cramerβs rule.
5x + 7y = β2
4x + 6y = β3
Solution:
Using Cramerβs Rule, we get,
= 30 β 28
= 2
Also, we get,
= β12 + 21
= 9
= β15 + 8
= β7
So, x = D1/D = 9/2
And y = D2/D = -7/2
Therefore, x = 9/2 and y = -7/2.
Question 9. Solve the following system of linear equation by Cramerβs rule.
9x + 5y = 10
3y β 2x = 8
Solution:
Using Cramerβs Rule, we get,
= 27 + 10
= 37
Also, we get,
= 30 β 40
= β10
= 72 + 20
= 92
So, x = D1/D = -10/37
And y = D2/D = 92/37
Therefore, x = -10/37 and y = 92/37.
Question 10. Solve the following system of linear equations by Cramerβs rule.
x + 2y = 1
3x + y = 4
Solution:
Using Cramerβs Rule, we get,
= 1 β 6
= β5
Also, we get,
= 1 β 8
= β7
= 4 β 3
= 1
So, x = D1/D = -7/-5 = 7/5
And y = D2/D = -1/5
Therefore, x = 7/5 and y = -1/5.
Question 11. Solve the following system of linear equations by Cramerβs rule.
3x + y + z = 2
2x β 4y + 3z = β1
4x + y β 3z = β11
Solution:
Using Cramerβs Rule, we get,
Expanding along R1, we get,
= 3 (12 β 3) + (β1) (β6 β 12) + 1 (2 + 16)
= 27 + 18 + 18
= 63
Also, we get,
Expanding along R1, we get,
= 2 (12 β 3) + (β1) (3 + 33) + 1 (β1 β 44)
= 18 β 36 β 45
= β63
Expanding along R1, we get,
= 3 (3 + 33) + (β2) (β6 β 12) + 1 (β22 + 4)
= 108 + 36 β 18
= 126
Expanding along R1, we get,
= 3 (44 + 1) + (β1) (β22 + 4) + 2 (2 + 16)
= 135 + 18 + 36
= 189
So, x = D1/D = -63/63 = -1
y = D2/D = 126/63 = 2
z = D3/D = 189/63 = 3
Therefore, x = β1, y = 2 and z = 3.
Question 12. Solve the following system of linear equations by Cramerβs rule.
x β 4y β z = 11
2x β 5y + 2z = 39
β3x + 2y + z = 1
Solution:
Using Cramerβs Rule, we get,
Expanding along R1, we get,
= 1 (β5 β 4) + 4 (2 + 6) β 1 (4 β 15)
= β9 + 32 + 11
= 34
Also, we get,
Expanding along R1, we get,
= 11 (β5 β 4) + 4 (39 β 2) β 1 (78 + 5)
= β99 + 148 β 83
= β34
Expanding along R1, we get,
= 1 (39 β 2) β 11 (2 + 6) β1 (2 + 117)
= 37 β 88 β 119
= β170
Expanding along R1, we get,
= 1 (β5 β 78) + 4 (2 + 117) + 11 (4 β 15)
= β83 + 476 β 121
= 272
So, x = D1/D = -34/34 = -1
y = D2/D = -170/34 = -5
z = D3/D = 272/34 = 8
Therefore, x = β1, y = β5 and z = 8.
Question 13. Solve the following system of linear equations by Cramerβs rule.
6x + y β 3z = 5
x + 3y β 2z = 5
2x + y + 4z = 8
Solution:
Using Cramerβs Rule, we get,
Expanding along R1, we get,
= 6 (12 + 2) β 1 (4 + 4) β 3 (1 β 6)
= 84 β 8 + 15
= 91
Also, we get,
Expanding along R1, we get,
= 5 (12 + 2) β 1 (20 + 16) β 3 (5 β 24)
= 70 β 36 + 57
= 91
Expanding along R1, we get,
= 6 (20 + 16) β 5 (4 + 4) β 3 (8 β 10)
= 216 β 40 + 6
= 182
Expanding along R1, we get,
= 6 (24 β 5) β 1 (8 β 10) + 5 (1 β 6)
= 114 + 2 β 25
= 91
So, x = D1/D = 91/91 = 1
y = D2/D = 182/91 = 2
z = D3/D = 92/92 =1
Therefore, x = 1, y = 2 and z = 1.
Question 14. Solve the following system of linear equations by Cramerβs rule.
x + y = 5
y + z = 3
x + z = 4
Solution:
Using Cramerβs Rule, we get,
Expanding along R1, we get,
= 1 (1) β 1 (β1) + 0 (β1)
= 1 + 1
= 2
Also, we get,
Expanding along R1, we get,
= 5 (1) β 1 (β1) + 0 (β4)
= 5 + 1 + 0
= 6
Expanding along R1, we get,
= 1 (β1) β 5 (β1) + 0 (β3)
= β1 + 5 + 0
= 4
Expanding along R1, we get,
= 1 (4) β 1 (β3) + 5 (β1)
= 4 + 3 β 5
= 2
So, x = D1/D = 6/2 = 3
y = D2/D = 4/2 = 2
z = D3/D = 2/2 = 1
Therefore, x = 3, y = 2 and z = 1.
Question 15. Solve the following system of linear equations by Cramerβs rule.
2y β 3z = 0
x + 3y = β4
3x + 4y = 3
Solution:
Using Cramerβs Rule, we get,
Expanding along R1, we get,
= 0 (0) β 2 (0) β 3 (β5)
= 0 β 0 + 15
= 15
Also, we get,
Expanding along R1, we get,
= 0 (0) β 2 (0) β 3 (β25)
= 0 β 0 + 75
= 75
Expanding along R1, we get,
= 0 (0) β 0 (0) β 3 (15)
= 0 β 0 β 45
= β45
Expanding along R1, we get,
= 0 (25) β 2 (15) + 0 (1)
= 0 β 30 + 0
= β30
So, x = D1/D = 75/15 = 5
y = D2/D = -45/15 = -3
z = D3/D = -30/15 = -2
Therefore, x = 5, y = β3 and z = β2.
Question 16. Solve the following system of linear equations by Cramerβs rule.
5x β 7y + z = 11
6x β 8y β z = 15
3x + 2y β 6z = 7
Solution:
Using Cramerβs Rule, we get,
Expanding along R1, we get,
= 5 (50) + 7 (β33) + 1 (36)
= 250 β 231 + 36
= 55
Also, we get,
Expanding along R1, we get,
= 11 (50) + 7 (β83) + 1 (86)
= 550 β 581 + 86
= 55
Expanding along R1, we get,
= 5 (β83) β 11 (β33) + 1 (β3)
= β415 + 363 β 3
= β55
Expanding along R1, we get,
= 5 (β86) + 7 (β3) + 11 (36)
= β430 β 21 + 396
= β55
So, x = D1/D = 55/55 = 1
y = D2/D = -55/55 = -1
z = D3/D = -55/55 = -1
Therefore, x = 1, y = β1 and z = β1.