第 12 类 RD Sharma 解决方案 - 第 6 章行列式 - 练习 6.4 |设置 2

问题 17. 用克莱默法则求解下列线性方程组。

2x - 3y - 4z = 29

-2x + 5y - z = -15

3x - y + 5z = -11

解决方案：

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.

问题 18. 用克莱默法则求解下列线性方程组。

x + y = 1

x + z = -6

x - y - 2z = 3

解决方案：

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.

问题 19. 用克莱默法则求解下列线性方程组。

x + y + z + 1 = 0

ax + by + cz + d = 0

a ² x + b ² y + c ² z + d ² = 0

解决方案：

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) 

问题 20. 用克莱默法则求解下列线性方程组。

x + y + z + w = 2

x - 2y + 2z + 2w = -6

2x + y - 2z + 2w = -5

3x - y + 3z - 3w = -3

解决方案：

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.

问题 21. 用克莱默法则求解下列线性方程组。

2x - 3z + w = 1

x - y + 2w = 1

−3y + z + w = 1

x + y + z = -1

解决方案：

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.

问题 22. 证明下列线性方程组是不一致的。

2x - y = 5

4x - 2y = 7

解决方案：

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.

问题 23. 证明下列线性方程组是不一致的。

3x + y = 5

−6x − 2y = 9

解决方案：

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.

问题 24. 证明下列线性方程组是不一致的。

3x - y + 2z = 3

2x + y + 3z = 5

x - 2y - z = 1

解决方案：

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.

问题 25. 证明下列线性方程组是一致的并求解。

3x - y + 2z = 6

2x - y + z = 2

3x + 6y + 5z = 20

解决方案：

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.

问题 26. 证明下列线性方程组有无限个解。

x - y + z = 3

2x + y - z = 2

-x - 2y + 2z = 1

解决方案：

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.

问题 27. 证明下列线性方程组有无限个解。

x + 2y = 5

3x + 2y = 15

解决方案：

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.

问题 28. 证明下列线性方程组有无限个解。

x + y - z = 0

x − 2y + z = 0

3x + 6y - 5z = 0

解决方案：

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.

问题 29. 证明下列线性方程组有无限个解。

2x + y - 2z = 4

x - 2y + z = -2

5x - 5y + z = -2

解决方案：

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.

问题 30. 证明下列线性方程组有无限个解。

x - y + 3z = 6

x + 3y - 3z = -4

5x + 3y + 3z = 10

解决方案：

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.

问题 31. 一位推销员在三个月内对三个不同佣金率的项目 A、B 和 C 的销售记录如下。

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

使用行列式方法找出项目 A、B 和 C 的佣金率。

解决方案：

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.

问题32、某汽车公司使用三种钢材S ₁ 、S ₂和S ₃生产三种汽车C ₁ 、C ₂和C ₃ 。每种汽车的钢材要求（吨）如下所示。

Steel	Cars
	C1	C2	C3
S1	2	3	4
S2	1	1	2
S3	3	2	1

使用克莱默法则，找出每种类型的汽车的数量，分别使用 29、13 和 16 吨三种钢材可以生产。

解决方案：

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.