# 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,

Expanding along R

_{1}, we get,= 2 (24) + 3 (−13) + 4 (-13)

= 48 − 21 – 52

= -25

Also, we get,

Expanding along R

_{1}, we get,= 29 (24) + 3 (−64) + 4 (−40)

= 692 − 192 − 160

= 344

Expanding along R

_{1}, we get,= 2 (−64) − 29 (−7) + 4 (23)

= −128 + 203 + 92

= 167

Expanding along R

_{1}, we get,= 2 (40) + 3 (23) + 29 (−13)

= 80 + 69 − 377

= −228

So, x = D

_{1}/D = -344/25y = D

_{2}/D = -167/25z = D

_{3}/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,

= 1(1) – 1(-3)

= 1 + 3

= 4

Also, we get,

Expanding along R

_{1}, we get,= 1 (1) − 1 (9) + 0

= 1 − 9

= −8

Expanding along R

_{1}, we get,= 1 (9) − 1 (−3)

= 9 + 3

= 12

Expanding along R

_{1}, we get,= 1 (−6) − 1 (9) + 1 (−1)

= −6 − 9 − 1

= −16

So, x = D

_{1}/D = -8/4 = -2y = D

_{2}/D = 12/4 = 3z = D

_{3}/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

### a^{2}x + b^{2}y + c^{2}z + d^{2} = 0

**Solution:**

Using Cramer’s Rule, we get,

c

_{2}-> c_{2 }– c_{1}, c_{3}-> c_{3}– c_{1}Taking common (b-a) from c

_{2}and (c-a)c_{3}

Expanding along R_{1}, we get,

= (b – a)(c – a)(c + a – b – a)

= (b – a)(c – a)(c – b)

= (a – b)(b – c)(c – a)

= -(d – b)(b – c)(c – d)

= -(a – d)(d – c)(c – a)

= -(a – d)(b – d)(d – a)So, x = D

_{1}/D =-(d – b)(b – c)(c – d)/(a – b)(b – c)(c – a)y = D

_{2}/D =-(a – d)(d – c)(c – a)/(a – b)(b – c)(c – a)z = D

_{3}/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,

=

=

= −94

Also, we get,

So, x = D

_{1}/D = -188/94 = -2y = D

_{2}/D = -282/-94 = 3z = D

_{3}/D = -141/-94 = 3/2w = D

_{4}/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,

=

=

= −21

Also, we get,

So, x = D

_{1}/D = -21/-21 = 1y = D

_{2}/D = -6/-21 = 2/7z = D

_{3}/D = -6/-21 = 2/7w = D

_{4}/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,

= −4 + 4

= 0

Also, we get,

= − 10 + 7

= −3

= 14 − 20

= −6

Since D = 0 and D

_{1}and D_{2}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,

= −6 + 6

= 0

Also, we get,

= −10 − 9

= −19

= 27 + 30

= 57

Since D = 0 and D

_{1}and D_{2}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,

Expanding along R

_{1}, we get= 3 (5) + 1 (−5) + 2 (−5)

= 15 − 5 − 10

= 0

Also, we get,

Expanding along R

_{1}, we get= 3 (5) + 1 (−8) + 2 (−11)

= 15 − 8 − 22

= −15

Since D = 0 and D

_{1}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,

Expanding along R

_{1}, we get= 3 (−11) + 1 (7) + 2 (15)

= −33 + 7 + 30

= 4

Also, we get,

Expanding along R

_{1}, we get= 6 (−11) + 1 (−10) + 2 (32)

= −66 − 10 + 64

= −12

Expanding along R

_{1}, we get= 3 (−10) − 6 (7) + 2 (34)

= −30 − 42 + 68

= −4

Expanding along R

_{1}, we get= 3 (−32) + 1 (34) + 6 (15)

= −96 + 34 + 90

= 28

As D, D

_{1}, D_{2}and D_{3}all are non-zero, the given system of equations is consistent.So, x = D

_{1}/D = -12/4 = -3y = D

_{2}/D = -4/4 = -1z = D

_{3}/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,

=

= 0

Also, we get,

=

= 0

=

= 0

=

= 0

As D, D

_{1}, D_{2}and D_{3}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,

= 6 − 6

= 0

Also, we get,

= 30 − 30

= 0

= 15 − 15

= 0

As D, D

_{1 }and D_{2 }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,

=

= 1 (6 − 6)

= 0

Also, we get,

= 0

= 0

= 0

As D, D

_{1}, D_{2}and D_{3}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,

=

= 1 (−36 + 36)

= 0

Also we get,

=

= 0

=

= 0

=

= 2 (−12 + 12)

= 0

As D, D

_{1}, D_{2}and D_{3}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,

=

= 3 (12 − 12)

= 0

Also we get,

=

= 3 (12 − 12)

= 0

=

= 3 (12 − 12)

= 0

=

= 1 (−80 + 80)

= 0

_{1}, D_{2}and D_{3}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 Cramer’s Rule, we get,

=

= 50 (8500 − 12000)

= −175000

Also we get,

=

= 50 (50000 − 57000)

= −350000

=

= 20 (17500 − 52500)

= −700000

=

= 50 (161500 − 200000)

= −1925000

So, x = D

_{1}/D = -350000/-175000 = 2y = D

_{2}/D = -700000/-175000 = 4z = D

_{3}/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 S_{1}, S_{2}, and S_{3} for producing three types of cars C_{1}, C_{2} and C_{3}. Steel requirements (in tons) for each type of cars are given below.

_{1}

Steel | Cars | ||
---|---|---|---|

C_{1} | C_{2} | C_{3} | |

S_{1} | 2 | 3 | 4 |

S_{2} | 1 | 1 | 2 |

S_{3} | 3 | 2 | 1 |

### 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 C

_{1}, C_{2}and C_{3}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,

=

= 1 (30 − 25)

= 5

Also we get,

=

= 1 (105 − 95)

= 10

=

= 1 (190 − 175)

= 15

=

= −2 (16 − 26)

= 20

So, x = D

_{1}/D = 10/5 = 2y = D

_{2}/D = 15/5 = 3z = D

_{3}/D = 20/5 = 4

Therefore, the number of cars produced of type C_{1}, C_{2}and C_{3}are 2, 3 and 4.