Class 12 RD Sharma Solutions - Chapter 6 Determinants - Exercise 6.4 | Set 2

Question 17. Solve the following system of the linear equations by Cramer's rule.

2x − 3y − 4z = 29

−2x + 5y − z = −15

3x − y + 5z = −11

Solution:

Using Cramer's Rule, we get,

D=\left|\begin{array}{cc} 2 & -3 & -4 \\ -2 & 5 & -1 \\ 3 & -1 & 5 \end{array} \right|

Expanding along R₁, we get,

= 2 (24) + 3 (−13) + 4 (-13)

= 48 − 21 - 52

= -25

Also, we get,

D_1=\left|\begin{array}{cc} 29 & -3 & -4 \\ -15 & 5 & -1 \\ 11 & -1 & 5 \end{array} \right|

Expanding along R₁, we get,

= 29 (24) + 3 (−64) + 4 (−40)

= 692 − 192 − 160

= 344

D_2=\left|\begin{array}{cc} 2 & 29 & -4 \\ -2 & -15 & -1 \\ 3 & 11 & 5 \end{array} \right|

Expanding along R₁, we get,

= 2 (−64) − 29 (−7) + 4 (23)

= −128 + 203 + 92

= 167

D_3=\left|\begin{array}{cc} 2 & -3 & 29 \\ -2 & 5 & -15 \\ 3 & -1 & 11 \end{array} \right|

Expanding along R₁, we get,

= 2 (40) + 3 (23) + 29 (−13)

= 80 + 69 − 377

= −228

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 Cramer's rule.

x + y = 1

x + z = −6

x − y − 2z = 3

Solution:

Using Cramer's Rule, we get,

D=\left|\begin{array}{cc} 1 & 1 & 0 \\ 1 & 0 & 1 \\ 1 & -1 & -2 \end{array} \right|

= 1(1) - 1(-3)

= 1 + 3

= 4

Also, we get,

D_1=\left|\begin{array}{cc} 1 & 1 & 0 \\ -6 & 0 & 1 \\ 3 & -1 & -2 \end{array} \right|

Expanding along R₁, we get,

= 1 (1) − 1 (9) + 0

= 1 − 9 

= −8

D_2=\left|\begin{array}{cc} 1 & 1 & 0 \\ 1 & -6 & 1 \\ 1 & 3 & -2 \end{array} \right|

Expanding along R₁, we get,

= 1 (9) − 1 (−3)

= 9 + 3

= 12 

D_3=\left|\begin{array}{cc} 1 & 1 & 1 \\ 1 & 0 & -6 \\ 1 & -1 & 3 \end{array} \right|

Expanding along R₁, we get,

= 1 (−6) − 1 (9) + 1 (−1)

= −6 − 9 − 1

= −16

So, x = D₁/D = -8/4 = -2 

y = D₂/D = 12/4 = 3 

z = D₃/D = -16/4 = -4

Therefore, x = −2, y = 3 and z = −4.

Question 19. Solve the following system of linear equations by Cramer's rule.

x + y + z + 1 = 0

ax + by + cz + d = 0

a²x + b²y + c²z + d² = 0

Solution:

Using Cramer's Rule, we get,

D=\left|\begin{array}{cc} 1 & 1 & 1 \\ a & b & c \\ a^2 & b^2 & c^2 \end{array} \right|

c₂ -> c₂ - c₁, c₃ -> c₃ - c₁

D=\left|\begin{array}{cc} 1 & 1 & 1 \\ a & b-a & c-a \\ a^2 & b^2-a^2 & c^2-a^2 \end{array} \right|

Taking common (b-a) from c₂ and (c-a)c₃

D=(b-a)(c-a)\left|\begin{array}{cc} 1 & 1 & 1 \\ a & 1 & 1 \\ a^2 & b+a & c+a \end{array} \right|

Expanding along R₁, we get,

= (b - a)(c - a)(c + a - b - a)

= (b - a)(c - a)(c - b)

= (a - b)(b - c)(c - a)

D_1=-\left|\begin{array}{cc} 1 & 1 & 1 \\ d & b & c \\ d^2 & b^2 & c^2 \end{array} \right|

= -(d - b)(b - c)(c - d)

D_2=-\left|\begin{array}{cc} 1 & 1 & 1 \\ a & b & c \\ a^2 & b^2 & c^2 \end{array} \right|

= -(a - d)(d - c)(c - a)

D_3=-\left|\begin{array}{cc} 1 & 1 & 1 \\ a & b & d \\ a^2 & b^2 & d^2 \end{array} \right|

= -(a - d)(b - d)(d - a)

So, x = D₁/D = -(d - b)(b - c)(c - d)/(a - b)(b - c)(c - a) 

y = D₂/D = -(a - d)(d - c)(c - a)/(a - b)(b - c)(c - a) 

z = D₃/D = -(a - d)(b - d)(d - a)/(a - b)(b - c)(c - a) 

Question 20. Solve the following system of linear equations by Cramer's rule.

x + y + z + w = 2

x − 2y + 2z + 2w = −6

2x + y − 2z + 2w = −5

3x − y + 3z − 3w = −3

Solution:

Using Cramer's Rule, we get,

D=\left|\begin{array}{cc} 1 & 1 & 1 & 1 \\ 1 & -2 & 2 & 2 \\ 2 & 1 & -2 & 2 \\ 3 & -1 & 3 & -3 \end{array} \right|

= \left|\begin{array}{cc} 1 & 0 & 0 & 0 \\ 1 & -3 & 1 & 1 \\ 2 & -1 & -4 & 0 \\ 3 & -4 & 0 & -6 \end{array} \right|

= 1\left|\begin{array}{cc} -3 & 1 & 1 \\ -1 & -4 & 0 \\ -4 & 0 & -6 \\ \end{array} \right|

= −94

Also, we get,

D_1=\left|\begin{array}{cc} 2 & 1 & 1 & 1 \\ -6 & -2 & 2 & 2 \\ -5 & 1 & -2 & 2 \\ -3 & -1 & 3 & -3 \end{array} \right| = 188

D_2=\left|\begin{array}{cc} 1 & 2 & 1 & 1 \\ 1 & -6 & 2 & 2 \\ 2 & -5 & -2 & 2 \\ 3 & -3 & 3 & -3 \end{array} \right| = −282

D_3=\left|\begin{array}{cc} 1 & 1 & 2 & 1 \\ 1 & -2 & -6 & 2 \\ 2 & 1 & -5 & 2 \\ 3 & -1 & -3 & -3 \end{array} \right| = −141

D_4=\left|\begin{array}{cc} 1 & 1 & 1 & 2 \\ 1 & -2 & 2 & -6 \\ 2 & 1 & -2 & -5 \\ 3 & -1 & 3 & -3 \end{array} \right| = 47

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 = −2, y = 3 and z = 3/2 and w = -1/2.

Question 21. Solve the following system of linear equations by Cramer's rule.

2x − 3z + w = 1

x − y + 2w = 1

−3y + z + w = 1

x + y + z = −1

Solution:

Using Cramer's Rule, we get,

D=\left|\begin{array}{cc} 2 & 0 & -3 & 1 \\ 1 & -1 & 0 & 2 \\ 0 & -3 & 1 & 1 \\ 1 & 1 & 1 & 0 \end{array} \right|

= \left|\begin{array}{cc} 2 & -2 & -5 & 1 \\ 1 & -2 & -1 & 2 \\ 0 & -3 & 1 & 1 \\ 1 & 0 & 0 & 0 \end{array} \right|

= -1\left|\begin{array}{cc} -2 & -5 & 1 \\ -2 & -1 & 2 \\ -3 & 1 & 1 \\ \end{array} \right|

= −21

Also, we get,

D_1=\left|\begin{array}{cc} 1 & 0 & -3 & 1 \\ 1 & -1 & 0 & 2 \\ 1 & -3 & 1 & 1 \\ 1 & 1 & 1 & 0 \end{array} \right| = -21

D_2=\left|\begin{array}{cc} 2 & 1 & -3 & 1 \\ 1 & 1 & 0 & 2 \\ 0 & 1 & 1 & 1 \\ 1 & 1 & 1 & 0 \end{array} \right| = −6

D_3=\left|\begin{array}{cc} 2 & 0 & 1 & 1 \\ 1 & -1 & 1 & 2 \\ 0 & -3 & 1 & 1 \\ 1 & 1 & 1 & 0 \end{array} \right| = −6

D_4=\left|\begin{array}{cc} 2 & 0 & -3 & 1 \\ 1 & -1 & 0 & 1 \\ 0 & -3 & 1 & 1 \\ 1 & 1 & 1 & 1 \end{array} \right| = 3

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 − y = 5

4x − 2y = 7

Solution:

Using Cramer's Rule, we get,

D=\left|\begin{array}{cc} 2 & -1 \\ 4 & -2 \end{array} \right|

= −4 + 4

= 0

Also, we get,

D_1=\left|\begin{array}{cc} 5 & -1 \\ 7 & -2 \end{array} \right|

= − 10 + 7

= −3

D_2=\left|\begin{array}{cc} 2 & 5 \\ 4 & 7 \end{array} \right|

= 14 − 20

= −6

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

−6x − 2y = 9

Solution:

Using Cramer's Rule, we get,

D=\left|\begin{array}{cc} 3 & 1 \\ -6 & -2 \end{array} \right|

= −6 + 6

= 0

Also, we get,

D_1=\left|\begin{array}{cc} 5 & 1 \\ 9 & -2 \end{array} \right|

= −10 − 9

= −19

D_2=\left|\begin{array}{cc} 3 & 5 \\ -6 & 9 \end{array} \right|

= 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 − y + 2z = 3

2x + y + 3z = 5

x − 2y − z = 1

Solution:

Using Cramer's Rule, we get,

D=\left|\begin{array}{cc} 3 & -1 & 2 \\ 2 & 1 & 3 \\ 1 & -2 & 1 \end{array} \right|

Expanding along R₁, we get

= 3 (5) + 1 (−5) + 2 (−5)

= 15 − 5 − 10

= 0

Also, we get,

D_1=\left|\begin{array}{cc} 3 & -1 & 2 \\ 5 & 1 & 3 \\ 1 & -2 & 1 \end{array} \right|

Expanding along R₁, we get

= 3 (5) + 1 (−8) + 2 (−11)

= 15 − 8 − 22

= −15

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 − y + 2z = 6

2x − y + z = 2

3x + 6y + 5z = 20

Solution:

Using Cramer's Rule, we get,

D=\left|\begin{array}{cc} 3 & -1 & 2 \\ 2 & -1 & 1 \\ 3 & 6 & 5 \end{array} \right|

Expanding along R₁, we get

= 3 (−11) + 1 (7) + 2 (15)

= −33 + 7 + 30

= 4

Also, we get,

D_1=\left|\begin{array}{cc} 6 & -1 & 2 \\ 2 & -1 & 1 \\ 20 & 6 & 5 \end{array} \right|

Expanding along R₁, we get

= 6 (−11) + 1 (−10) + 2 (32)

= −66 − 10 + 64

= −12

D_2=\left|\begin{array}{cc} 3 & 6 & 2 \\ 2 & 2 & 1 \\ 3 & 20 & 5 \end{array} \right|

Expanding along R₁, we get

= 3 (−10) − 6 (7) + 2 (34)

= −30 − 42 + 68

= −4

D_3=\left|\begin{array}{cc} 3 & -1 & 6 \\ 2 & -1 & 2 \\ 3 & 6 & 20 \end{array} \right|

Expanding along R₁, we get

= 3 (−32) + 1 (34) + 6 (15)

= −96 + 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 = −3, y = −1 and z = 7.

Question 26. Show that the following system of linear equations has infinite number of solutions.

x − y + z = 3

2x + y − z = 2

−x − 2y + 2z = 1

Solution:

Using Cramer's Rule, we get,

D=\left|\begin{array}{cc} 1 & -1 & 1 \\ 2 & 1 & -1 \\ -1 & -2 & 2 \end{array} \right|

= \left|\begin{array}{cc} 0 & -1 & 0 \\ 3 & 1 & 0 \\ -3 & -2 & 0 \end{array} \right|

= 0

Also, we get,

D_1=\left|\begin{array}{cc} 3 & -1 & 1 \\ 2 & 1 & -1 \\ 1 & -2 & 2 \end{array} \right|

= \left|\begin{array}{cc} 3 & -1 & 0 \\ 2 & 1 & 0 \\ 1 & -2 & 0 \end{array} \right|

= 0

D_2=\left|\begin{array}{cc} 1 & 3 & 1 \\ 2 & 2 & -1 \\ -1 & 1 & 2 \end{array} \right|

= \left|\begin{array}{cc} 0 & 0 & 0 \\ 2 & -4 & -3 \\ -1 & 4 & 3 \end{array} \right|

= 0

D_3=\left|\begin{array}{cc} 1 & -1 & 3 \\ 2 & 1 & 2 \\ -1 & -2 & 1 \end{array} \right|

= \left|\begin{array}{cc} 0 & 0 & 0 \\ 2 & 3 & -4 \\ -1 & -3 & 4 \end{array} \right|

= 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 Cramer's Rule, we get,

D=\left|\begin{array}{cc} 1 & 2 \\ 3 & 6 \end{array} \right|

= 6 − 6

= 0

Also, we get,

D_1=\left|\begin{array}{cc} 5 & 2 \\ 15 & 6 \end{array} \right|

= 30 − 30

= 0

D_2=\left|\begin{array}{cc} 1 & 5 \\ 3 & 15 \end{array} \right|

= 15 − 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 − z = 0

x − 2y + z = 0

3x + 6y − 5z = 0

Solution:

Using Cramer's Rule, we get,

D=\left|\begin{array}{cc} 1 & 1 & -1 \\ 1 & -2 & 1 \\ 3 & 6 & -5 \end{array} \right|

= \left|\begin{array}{cc} 1 & 0 & 0 \\ 1 & -3 & 2 \\ 3 & 3 & -2 \end{array} \right|

= 1 (6 − 6)

= 0

Also, we get,

D_1=\left|\begin{array}{cc} 0 & 1 & -1 \\ 0 & -2 & 1 \\ 0 & 6 & -5 \end{array} \right|

= 0

D_2=\left|\begin{array}{cc} 1 & 0 & -1 \\ 1 & 0 & 1 \\ 3 & 0 & -5 \end{array} \right|

= 0

D_3=\left|\begin{array}{cc} 1 & 1 & 0 \\ 1 & -2 & 0 \\ 3 & 6 & 0 \end{array} \right|

= 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 − 2z = 4

x − 2y + z = −2

5x − 5y + z = −2

Solution:

Using Cramer's Rule, we get,

D=\left|\begin{array}{cc} 2 & 1 & -2 \\ 1 & -2 & 1 \\ 5 & -5 & 1 \end{array} \right|

= \left|\begin{array}{cc} 12 & 9 & -2 \\ -4 & -3 & 1 \\ 0 & 0 & 1 \end{array} \right|

= 1 (−36 + 36)

= 0

Also we get,

D_1=\left|\begin{array}{cc} 4 & 1 & -2 \\ -2 & -2 & 1 \\ -2 & -5 & 1 \end{array} \right|

= \left|\begin{array}{cc} 0 & 1 & -2 \\ 0 & -2 & 1 \\ 0 & -5 & 1 \end{array} \right|

= 0

D_2=\left|\begin{array}{cc} 2 & 4 & -2 \\ 1 & -2 & 1 \\ 5 & -2 & 1 \end{array} \right|

= \left|\begin{array}{cc} 2 & 0 & -2 \\ 1 & 0 & 1 \\ 5 & 0 & 1 \end{array} \right|

= 0

D_3=\left|\begin{array}{cc} 2 & 1 & 4 \\ 1 & -2 & -2 \\ 5 & -5 & -2 \end{array} \right|

= \left|\begin{array}{cc} 4 & -3 & 0 \\ 1 & -2 & -2 \\ 4 & -3 & 0 \end{array} \right|

= 2 (−12 + 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 − y + 3z = 6

x + 3y − 3z = −4

5x + 3y + 3z = 10

Solution:

Using Cramer's Rule, we get,

D=\left|\begin{array}{cc} 1 & -1 & 3 \\ 1 & 3 & -3 \\ 5 & 3 & 3 \end{array} \right|

= \left|\begin{array}{cc} 2 & 2 & 0 \\ 1 & 3 & -3 \\ 6 & 6 & 0 \end{array} \right|

= 3 (12 − 12)

= 0

Also we get,

D_1=\left|\begin{array}{cc} 6 & -1 & 3 \\ -4 & 3 & -3 \\ 10 & 3 & 3 \end{array} \right|

= \left|\begin{array}{cc} 2 & 2 & 0 \\ -4 & 3 & -3 \\ 6 & 6 & 0 \end{array} \right|

= 3 (12 − 12)

= 0

D_2=\left|\begin{array}{cc} 1 & 6 & 3 \\ 1 & -4 & -3 \\ 5 & 10 & 3 \end{array} \right|

= \left|\begin{array}{cc} 2 & 2 & 0 \\ 1 & -4 & -3 \\ 6 & 6 & 0 \end{array} \right|

= 3 (12 − 12)

= 0

D_3=\left|\begin{array}{cc} 1 & -1 & 6 \\ 1 & 3 & -4 \\ 5 & 3 & 10 \end{array} \right|

= \left|\begin{array}{cc} 1 & -1 & 6 \\ 0 & 4 & -10 \\ 0 & 8 & -20 \end{array} \right|

= 1 (−80 + 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.

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 Cramer's Rule, we get,

D=\left|\begin{array}{cc} 90 & 100 & 20 \\ 130 & 50 & 40 \\ 60 & 100 & 30 \end{array} \right|

= \left|\begin{array}{cc} -170 & 0 & -60 \\ 130 & 50 & 40 \\ -200 & 0 & -50 \end{array} \right|

= 50 (8500 − 12000)

= −175000

Also we get,

D_1=\left|\begin{array}{cc} 800 & 100 & 20 \\ 900 & 50 & 40 \\ 850 & 100 & 30 \end{array} \right|

= \left|\begin{array}{cc} -1000 & 0 & -60 \\ 900 & 50 & 40 \\ -950 & 0 & -50 \end{array} \right|

= 50 (50000 − 57000)

= −350000

D_2=\left|\begin{array}{cc} 90 & 800 & 20 \\ 130 & 900 & 40 \\ 60 & 850 & 30 \end{array} \right|

= \left|\begin{array}{cc} 900 & 800 & 20 \\ -50 & -700 & 0 \\ -75 & -350 & 0 \end{array} \right|

= 20 (17500 − 52500)

= −700000

D_3=\left|\begin{array}{cc} 90 & 100 & 800 \\ 130 & 50 & 900 \\ 60 & 100 & 850 \end{array} \right|

= \left|\begin{array}{cc} -170 & 0 & -1000 \\ 130 & 50 & 900 \\ -200 & 0 & -950 \end{array} \right|

= 50 (161500 − 200000)

= −1925000

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.

Using Cramer's 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 Cramer's Rule, we get,

D=\left|\begin{array}{cc} 2 & 3 & 4 \\ 1 & 1 & 2 \\ 3 & 2 & 1 \end{array} \right|

= \left|\begin{array}{cc} -10 & -5 & 0 \\ -5 & -3 & 0 \\ 3 & 2 & 1 \end{array} \right|

= 1 (30 − 25)

= 5

Also we get,

D_1=\left|\begin{array}{cc} 29 & 3 & 4 \\ 13 & 1 & 2 \\ 16 & 2 & 1 \end{array} \right|

= \left|\begin{array}{cc} -35 & -5 & 0 \\ -19 & -3 & 0 \\ 16 & 2 & 1 \end{array} \right|

= 1 (105 − 95)

= 10

D_2=\left|\begin{array}{cc} 2 & 29 & 4 \\ 1 & 13 & 2 \\ 3 & 16 & 1 \end{array} \right|

= \left|\begin{array}{cc} -10 & -35 & 0 \\ -5 & -19 & 0 \\ 3 & 16 & 1 \end{array} \right|

= 1 (190 − 175)

= 15

D_3=\left|\begin{array}{cc} 2 & 3 & 29 \\ 1 & 1 & 13 \\ 3 & 2 & 16 \end{array} \right|

= \left|\begin{array}{cc} -2 & 0 & 0 \\ 1 & 1 & 13 \\ 3 & 2 & 16 \end{array} \right|

= −2 (16 − 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.

Summary

Chapter 6, Exercise 6.3 typically focuses on the application of determinants in solving systems of linear equations. The main topics covered usually include:

• Cramer's Rule for solving systems of linear equations
• Consistency and inconsistency of systems of linear equations
• Homogeneous systems of linear equations
• Non-homogeneous systems of linear equations
• Conditions for unique solutions, infinite solutions, or no solutions
• Application of determinants in geometry (e.g., area of triangles, collinearity of points)

This exercise emphasizes the practical application of determinants in solving real-world problems and understanding the nature of solutions to systems of equations.