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• RD Sharma Class 12 Solutions for Maths

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

### 3x − y + 5z = −11

Solution:

Using Cramer’s Rule, we get, Expanding along R1, we get,

= 2 (24) + 3 (−13) + 4 (-13)

= 48 − 21 – 52

= -25

Also, we get, Expanding along R1, we get,

= 29 (24) + 3 (−64) + 4 (−40)

= 692 − 192 − 160

= 344 Expanding along R1, we get,

= 2 (−64) − 29 (−7) + 4 (23)

= −128 + 203 + 92

= 167 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.

### x − y − 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) − 1 (9) + 0

= 1 − 9

= −8 Expanding along R1, we get,

= 1 (9) − 1 (−3)

= 9 + 3

= 12 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.

### 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 − y + 3z − 3w = −3

Solution:

Using Cramer’s Rule, we get,   = −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 = −2, y = 3 and z = 3/2 and w = -1/2.

### x + y + z = −1

Solution:

Using Cramer’s Rule, we get, =  = −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 − 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 D1 and D2 both are non-zero, the given system of equations is inconsistent.

Hence proved.

### −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 D1 and D2 both are non-zero, the given system of equations is inconsistent.

Hence proved.

### x − 2y − z = 1

Solution:

Using Cramer’s Rule, we get, Expanding along R1, we get

= 3 (5) + 1 (−5) + 2 (−5)

= 15 − 5 − 10

= 0

Also, we get, 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.

### 3x + 6y + 5z = 20

Solution:

Using Cramer’s Rule, we get, Expanding along R1, we get

= 3 (−11) + 1 (7) + 2 (15)

= −33 + 7 + 30

= 4

Also, we get, Expanding along R1, we get

= 6 (−11) + 1 (−10) + 2 (32)

= −66 − 10 + 64

= −12 Expanding along R1, we get

= 3 (−10) − 6 (7) + 2 (34)

= −30 − 42 + 68

= −4 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.

### −x − 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 − 6

= 0

Also, we get, = 30 − 30

= 0 = 15 − 15

= 0

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

Hence proved.

### 3x + 6y − 5z = 0

Solution:

Using Cramer’s Rule, we get,  = 1 (6 − 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 − 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, 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 − 12)

= 0

Also we get,  = 3 (12 − 12)

= 0  = 3 (12 − 12)

= 0  = 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.

### 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 = 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 − 25)

= 5

Also we get,  = 1 (105 − 95)

= 10  = 1 (190 − 175)

= 15  = −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.

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