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

• Last Updated : 26 May, 2021

3x âˆ’ y + 5z = âˆ’11

Solution:

Using Cramer’s Rule, we get,

Expanding along R1, we get,

= 2 (24) + 3 (âˆ’13) + 4 (-13)

= 48 âˆ’ 21 – 52

= -25

Also, we get,

Expanding along R1, we get,

= 29 (24) + 3 (âˆ’64) + 4 (âˆ’40)

= 692 âˆ’ 192 âˆ’ 160

= 344

Expanding along R1, we get,

= 2 (âˆ’64) âˆ’ 29 (âˆ’7) + 4 (23)

= âˆ’128 + 203 + 92

= 167

Expanding along R1, we get,

= 2 (40) + 3 (23) + 29 (âˆ’13)

= 80 + 69 âˆ’ 377

= âˆ’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.

x âˆ’ y âˆ’ 2z = 3

Solution:

Using Cramer’s Rule, we get,

= 1(1) – 1(-3)

= 1 + 3

= 4

Also, we get,

Expanding along R1, we get,

= 1 (1) âˆ’ 1 (9) + 0

= 1 âˆ’ 9

= âˆ’8

Expanding along R1, we get,

= 1 (9) âˆ’ 1 (âˆ’3)

= 9 + 3

= 12

Expanding along R1, we get,

= 1 (âˆ’6) âˆ’ 1 (9) + 1 (âˆ’1)

= âˆ’6 âˆ’ 9 âˆ’ 1

= âˆ’16

So, x = D1/D = -8/4 = -2

y = D2/D = 12/4 = 3

z = D3/D = -16/4 = -4

Therefore, x = âˆ’2, y = 3 and z = âˆ’4.

a2x + b2y + c2z + d2 = 0

Solution:

Using Cramer’s Rule, we get,

c2 -> c2 – c1, c3 -> c3 – c1

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

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 – b)(b – c)(c – d)

= -(a – d)(d – c)(c – a)

= -(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)

3x âˆ’ y + 3z âˆ’ 3w = âˆ’3

Solution:

Using Cramer’s Rule, we get,

= âˆ’94

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

x + y + z = âˆ’1

Solution:

Using Cramer’s Rule, we get,

=

= âˆ’21

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.

4x âˆ’ 2y = 7

Solution:

Using Cramer’s Rule, we get,

= âˆ’4 + 4

= 0

Also, we get,

= âˆ’ 10 + 7

= âˆ’3

= 14 âˆ’ 20

= âˆ’6

Since D = 0 and D1 and D2 both are non-zero, the given system of equations is inconsistent.

Hence proved.

âˆ’6x âˆ’ 2y = 9

Solution:

Using Cramer’s Rule, we get,

= âˆ’6 + 6

= 0

Also, we get,

= âˆ’10 âˆ’ 9

= âˆ’19

= 27 + 30

= 57

Since D = 0 and D1 and D2 both are non-zero, the given system of equations is inconsistent.

Hence proved.

x âˆ’ 2y âˆ’ z = 1

Solution:

Using Cramer’s Rule, we get,

Expanding along R1, we get

= 3 (5) + 1 (âˆ’5) + 2 (âˆ’5)

= 15 âˆ’ 5 âˆ’ 10

= 0

Also, we get,

Expanding along R1, we get

= 3 (5) + 1 (âˆ’8) + 2 (âˆ’11)

= 15 âˆ’ 8 âˆ’ 22

= âˆ’15

Since D = 0 and D1 are non-zero, the given system of equations is inconsistent.

Hence proved.

3x + 6y + 5z = 20

Solution:

Using Cramer’s Rule, we get,

Expanding along R1, we get

= 3 (âˆ’11) + 1 (7) + 2 (15)

= âˆ’33 + 7 + 30

= 4

Also, we get,

Expanding along R1, we get

= 6 (âˆ’11) + 1 (âˆ’10) + 2 (32)

= âˆ’66 âˆ’ 10 + 64

= âˆ’12

Expanding along R1, we get

= 3 (âˆ’10) âˆ’ 6 (7) + 2 (34)

= âˆ’30 âˆ’ 42 + 68

= âˆ’4

Expanding along R1, we get

= 3 (âˆ’32) + 1 (34) + 6 (15)

= âˆ’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 = âˆ’3, y = âˆ’1 and z = 7.

âˆ’x âˆ’ 2y + 2z = 1

Solution:

Using Cramer’s 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.

3x + 2y = 15

Solution:

Using Cramer’s Rule, we get,

= 6 âˆ’ 6

= 0

Also, we get,

= 30 âˆ’ 30

= 0

= 15 âˆ’ 15

= 0

As D, D1 and D2 all are zero, the given system of equations has infinite number of solutions.

Hence proved.

3x + 6y âˆ’ 5z = 0

Solution:

Using Cramer’s Rule, we get,

= 1 (6 âˆ’ 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.

5x âˆ’ 5y + z = âˆ’2

Solution:

Using Cramer’s Rule, we get,

= 1 (âˆ’36 + 36)

= 0

Also we get,

= 0

= 0

= 2 (âˆ’12 + 12)

= 0

As D, D1, D2 and D3 all are zero, the given system of equations has infinite number of solutions.

Hence proved.

5x + 3y + 3z = 10

Solution:

Using Cramer’s Rule, we get,

= 3 (12 âˆ’ 12)

= 0

Also we get,

= 3 (12 âˆ’ 12)

= 0

= 3 (12 âˆ’ 12)

= 0

= 1 (âˆ’80 + 80)

= 0

As D, D1, D2 and D3 all are zero, the given system of equations has infinite number of solutions.

Hence proved.

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 âˆ’ 12000)

= âˆ’175000

Also we get,

= 50 (50000 âˆ’ 57000)

= âˆ’350000

= 20 (17500 âˆ’ 52500)

= âˆ’700000

= 50 (161500 âˆ’ 200000)

= âˆ’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.

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,

= 1 (30 âˆ’ 25)

= 5

Also we get,

= 1 (105 âˆ’ 95)

= 10

= 1 (190 âˆ’ 175)

= 15

= âˆ’2 (16 âˆ’ 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.

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