第 12 类 RD Sharma 解——第 8 章联立线性方程组的解——练习 8.1 |设置 1

问题 1. 用矩阵法求解下列方程组：

(i) 5x + 7y + 2 = 0

4x + 6y + 3 = 0

解决方案：

The given system of equations can be written in matrix form as,

$\begin{bmatrix}5 & 7 \\ 4 & 6\end{bmatrix} \binom{x}{y} = \binom{ - 2}{ - 3}$

AX = B

Here,

A = $\begin{bmatrix}5 & 7 \\ 4 & 6\end{bmatrix}$, X = $\binom{x}{y}$and B = $\binom{ - 2}{ - 3}$

Now,

|A| = $\begin{bmatrix}5 & 7 \\ 4 & 6\end{bmatrix}$

= 30 – 28

= 2

The given system has a unique solution given by, X = A^-1 B.

Let C_ij be the cofactor of the elements a_ij in A.

C₁₁ = (-1)¹⁺¹ (6) = 6, C₁₂ = (-1)¹⁺² (4) = -4, C₂₁ = -1²⁺¹ (7) = -7 and C₂₂ = (-1)²⁺² (5) = 5

$adj A = \begin{bmatrix}6 & - 4 \\ - 7 & 5\end{bmatrix}^T$

= $\begin{bmatrix}6 & - 7 \\ - 4 & 5\end{bmatrix}$

A^-1 = $\frac{1}{\left| A \right|}adj A$

A^-1 = $\frac{1}{2}\begin{bmatrix}6 & - 7 \\ - 4 & 5\end{bmatrix}\binom{ - 2}{ - 3}$

So, X = A^-1 B

= $\frac{1}{2}\begin{bmatrix}6 & - 7 \\ - 4 & 5\end{bmatrix}$

= $\frac{1}{2}\binom{ - 12 + 21}{8 - 15}$

=> $\binom{x}{y} = \binom{\frac{9}{2}}{\frac{- 7}{2}}$

Therefore, x = 9/2 and y = -7/2.

(ii) 5x + 2y = 3

3x + 2y = 5

解决方案：

The given system of equations can be written in matrix form as,

$\begin{bmatrix}5 & 2 \\ 3 & 2\end{bmatrix} \binom{x}{y} = \binom{3}{5}$

AX = B

Here,

A = $\begin{bmatrix}5 & 2 \\ 3 & 2\end{bmatrix}$, X = $\binom{x}{y}$and B = $\binom{3}{5}$

Now,  

|A| = $\begin{bmatrix}5 & 2 \\ 3 & 2\end{bmatrix}$

= 10 – 6

= 4

The given system has a unique solution given by, X = A^-1 B

Let C_ij be the cofactor of the elements a_ij in A.

C₁₁ = -1¹⁺¹ (2) = 2, C₁₂ = (-1)¹⁺² (3) = – 3, C₂₁ = (-1)²⁺¹ (2) = – 2 and C₂₂ = (-1)²⁺² (5) = 5

$adj A = \begin{bmatrix}2 & - 3 \\ - 2 & 5\end{bmatrix}^T$

= $\begin{bmatrix}2 & - 2 \\ - 3 & 5\end{bmatrix}$

A^-1 = $\frac{1}{\left| A \right|}adj A$

= $\frac{1}{4}\begin{bmatrix}2 & - 2 \\ - 3 & 5\end{bmatrix}$

Now, X = A^-1 B

= $\frac{1}{4}\begin{bmatrix}2 & - 2 \\ - 3 & 5\end{bmatrix}\binom{3}{5}$

= $\frac{1}{4}\binom{6 - 10}{ - 9 + 25}$

=> $\binom{x}{y} = \binom{\frac{- 4}{4}}{\frac{16}{4}}$

Therefore, x = – 1 and y = 4.

(iii) 3x + 4y - 5 = 0

x - y + 3 = 0

解决方案：

 The given system of equations can be written in matrix form as,

$\begin{bmatrix}3 & 4 \\ 1 & - 1\end{bmatrix} \binom{x}{y} = \binom{5}{ - 3}$

AX = B

Here,

A = $\begin{bmatrix}3 & 4 \\ 1 & - 1\end{bmatrix}$, X = $\binom{x}{y}$and B = $\binom{5}{ - 3}$

Now,  

|A| = $\begin{bmatrix}3 & 4 \\ 1 & - 1\end{bmatrix}$

= – 3 – 4

= -7 

So, the given system has a unique solution given by, X = A^-1 B 

Let C_ij be the cofactors of the elements a_ij in A.

C₁₁ = (-1)¹⁺¹ (1) = -1, C₁₂ = (-1)¹⁺² (-1) = 1, C₂₁ = (-1)²⁺¹ (4) = -4 and C₂₂ = (-1)²⁺² (3) = 3

$adj A = \begin{bmatrix}- 1 & - 1 \\ - 4 & 3\end{bmatrix}^T = \begin{bmatrix}- 1 & - 4 \\ - 1 & 3\end{bmatrix}$

$A^{- 1} = \frac{1}{\left| A \right|}adj A$

= $\frac{1}{- 7}\begin{bmatrix}- 1 & - 4 \\ - 1 & 3\end{bmatrix}$

Now, X = A^-1 B

= $\frac{1}{- 7}\begin{bmatrix}- 1 & - 4 \\ - 1 & 3\end{bmatrix}\binom{5}{ - 3}$

= $\frac{1}{- 7}\binom{ - 5 + 12}{ - 5 - 9}$

=> $\binom{x}{y} = \binom{\frac{7}{- 7}}{\frac{- 14}{- 7}}$

Therefore, x = -1 and y = 2.

(iv) 3x + y = 19

3x - y = 23

解决方案：

The given system of equations can be written in matrix form as,

$\begin{bmatrix}3 & 1 \\ 3 & - 1\end{bmatrix} \binom{x}{y} = \binom{19}{23}$

AX = B

Here,  

A = $\begin{bmatrix}3 & 1 \\ 3 & - 1\end{bmatrix}$, X = $\binom{x}{y}$and B = $\binom{19}{23}$

Now,

|A| = $\begin{bmatrix}3 & 1 \\ 3 & - 1\end{bmatrix}$

= – 3 – 3

= -6

So, the given system has a unique solution given by X = A^-1 B.

Let C_ij be the cofactors of the elements a_ij in A.

C₁₁ = (-1)¹⁺¹ (-1) = -1, C₁₂ = (-1)¹⁺² (3) = -3, C₂₁ = (-1)²⁺¹ (1) = -4 and C₂₂ = (-1)²⁺² (3) = 3

$adj A = \begin{bmatrix}- 1 & - 3 \\ - 1 & 3\end{bmatrix}^T$

= $\begin{bmatrix}- 1 & - 1 \\ - 3 & 3\end{bmatrix}$

$A^{- 1} = \frac{1}{\left| A \right|}adj A$

= $\frac{1}{- 6}\begin{bmatrix}- 1 & - 1 \\ - 3 & 3\end{bmatrix}$

Now, X = A^-1 B

= $\frac{1}{- 6}\begin{bmatrix}- 1 & - 1 \\ - 3 & 3\end{bmatrix}\binom{19}{23}$

= $\frac{1}{- 6}\binom{ - 19 - 23}{ - 57 + 69}$

=> $\binom{x}{y} = \binom{\frac{- 42}{- 6}}{\frac{12}{- 6}}$

Therefore, x = 7 and y = -2.

(v) 3x + 7y = 4

x + 2y = -1

解决方案：

The given system of equations can be written in matrix form as,

$\begin{bmatrix}3 & 7 \\ 1 & 2\end{bmatrix} \binom{x}{y} = \binom{4}{ - 1}$

AX = B

Here,

A = $\begin{bmatrix}3 & 7 \\ 1 & 2\end{bmatrix}$, X = $\binom{x}{y}$and B = $\binom{4}{ - 1}$

Now,  

|A| = $\begin{bmatrix}3 & 7 \\ 1 & 2\end{bmatrix}$

= 6 – 7

= -1

So, the given system has a unique solution given by X = A^-1 B.

Let C_ij be the cofactors of the elements a_ij in A.

C₁₁ = (-1)¹⁺¹ (2) = 2, C₁₂ = (-1)¹⁺² (1) = -1, C₂₁ = (-1)²⁺¹ (7) = -7 and C₂₂ = (-1)²⁺² (3) = 3

$adj A = \begin{bmatrix}2 & - 1 \\ - 7 & 3\end{bmatrix}^T$

= $\begin{bmatrix}2 & - 7 \\ - 1 & 3\end{bmatrix}$

$A^{- 1} = \frac{1}{\left| A \right|}adj A$

= $\frac{1}{- 1}\begin{bmatrix}2 & - 7 \\ - 1 & 3\end{bmatrix}$

X = A^-1 B

= $\frac{1}{- 1}\begin{bmatrix}2 & - 7 \\ - 1 & 3\end{bmatrix}\binom{4}{ - 1}$

= $\frac{1}{- 1}\binom{8 + 7}{ - 4 - 3}$

=> $\binom{x}{y} = \binom{\frac{15}{- 1}}{\frac{- 7}{- 1}}$

Therefore x = – 15 and y = 7.

(vi) 3x + y = 7

5x + 3y = 12

解决方案：

The given system of equations can be written in matrix form as,

$\begin{bmatrix}3 & 1 \\ 5 & 3\end{bmatrix} \binom{x}{y} = \binom{7}{12}$

AX = B

Here,

A = $\begin{bmatrix}3 & 1 \\ 5 & 3\end{bmatrix}$, X = $\binom{x}{y}$and B = $\binom{7}{12}$

Now,

|A| = $\begin{bmatrix}3 & 1 \\ 5 & 3\end{bmatrix}$

= 9 – 5

= 4

So, the given system has a unique solution given by X = A^-1 B.

Let C_ij be the cofactors of the elements a_ij in A.

C₁₁ = (-1)¹⁺¹ (3) = 3, C₁₂ = (-1)¹⁺² (5) = -5, C₂₁ = (-1)²⁺¹ (1) = -1 and C₂₂ = (-1)²⁺² (3) = 3

$adj A = \begin{bmatrix}3 & - 5 \\ - 1 & 3\end{bmatrix}^T$

= $\begin{bmatrix}3 & - 1 \\ - 5 & 3\end{bmatrix}$

$A^{- 1} = \frac{1}{\left| A \right|}adj A$

= $\frac{1}{4}\begin{bmatrix}3 & - 1 \\ - 5 & 3\end{bmatrix}$

X = A^-1 B

= $\frac{1}{4}\begin{bmatrix}3 & - 1 \\ - 5 & 3\end{bmatrix}\binom{7}{12}$

= $\frac{1}{4}\binom{21 - 12}{ - 35 + 36}$

=> $\binom{x}{y} = \binom{\frac{9}{4}}{\frac{1}{4}}$

Therefore x = 9/4 and y = 1/4.

问题 2. 用矩阵法求解下列方程组：

(i) x + y - z = 3

2x + 3y + z = 10

3x - y - 7z = 1

解决方案：

A = $\begin{bmatrix}1 & 1 & - 1 \\ 2 & 3 & 1 \\ 3 & - 1 & - 7\end{bmatrix}$

|A| = $\begin{vmatrix}1 & 1 & - 1 \\ 2 & 3 & 1 \\ 3 & - 1 & - 7\end{vmatrix}$

= 1 (- 21 + 1) – 1(-14 – 3) – 1(-2 – 9)

= – 20 + 17 + 11

= 8

So, the given system has a unique solution given by X = A^-1 B.

Let C_ij be the cofactor of the elements a_ij in A.

$C_{11} = \left( - 1 \right)^{1 + 1} \begin{vmatrix}3 & 1 \\ - 1 & - 7\end{vmatrix} = - 20 , C_{12} = \left( - 1 \right)^{1 + 2} \begin{vmatrix}2 & 1 \\ 3 & - 7\end{vmatrix} = 17, C_{13} = \left( - 1 \right)^{1 + 3} \begin{vmatrix}2 & 3 \\ 3 & - 1\end{vmatrix} = - 11$

$C_{21} = \left( - 1 \right)^{2 + 1} \begin{vmatrix}1 & - 1 \\ - 1 & - 7\end{vmatrix} = 8 , C_{22} = \left( - 1 \right)^{2 + 2} \begin{vmatrix}1 & - 1 \\ 3 & - 7\end{vmatrix} = - 4, C_{23} = \left( - 1 \right)^{2 + 3} \begin{vmatrix}1 & 1 \\ 3 & - 1\end{vmatrix} = 4$

$C_{31} = \left( - 1 \right)^{3 + 1} \begin{vmatrix}1 & - 1 \\ 3 & 1\end{vmatrix} = 4 , C_{32} = \left( - 1 \right)^{3 + 2} \begin{vmatrix}1 & - 1 \\ 2 & 1\end{vmatrix} = - 3, C_{33} = \left( - 1 \right)^{3 + 3} \begin{vmatrix}1 & 1 \\ 2 & 3\end{vmatrix} = 1$

$adj A = \begin{bmatrix}- 20 & 17 & - 11 \\ 8 & - 4 & 4 \\ 4 & - 3 & 1\end{bmatrix}^T$

= $\begin{bmatrix}- 20 & 8 & 4 \\ 17 & - 4 & - 3 \\ - 11 & 4 & 1\end{bmatrix}$

$A^{- 1} = \frac{1}{\left| A \right|}adj A$

= $\frac{1}{8}\begin{bmatrix}- 20 & 8 & 4 \\ 17 & - 4 & - 3 \\ - 11 & 4 & 1\end{bmatrix}$

X = A^-1 B

$\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{8}\begin{bmatrix}- 20 & 8 & 4 \\ 17 & - 4 & - 3 \\ - 11 & 4 & 1\end{bmatrix}\begin{bmatrix}3 \\ 10 \\ 1\end{bmatrix}$

$\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{8}\begin{bmatrix}- 60 + 80 + 4 \\ 51 - 40 - 3 \\ - 33 + 40 + 1\end{bmatrix}$

$\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{8}\begin{bmatrix}24 \\ 8 \\ 8\end{bmatrix}$

=> x = 24/8, y = 8/8 and z = 8/8

Therefore, x = 3, y = 1 and z = 1.

(ii) x + y + z = 3

2x - y + z = - 1

2x + y - 3z = - 9

解决方案：

A = $\begin{bmatrix}1 & 1 & 1 \\ 2 & - 1 & 1 \\ 2 & 1 & - 3\end{bmatrix}$

|A| = $\begin{vmatrix}1 & 1 & 1 \\ 2 & - 1 & 1 \\ 2 & 1 & - 3\end{vmatrix}$

= 1 (3 – 1) – 1 (-6 – 2) + 1 (2 + 2)

= 2 + 8 + 4

= 14

So, the given system has a unique solution given by X = A^-1 B.

Let C_ij be the cofactors of the elements a_ij in A.

$C_{11} = \left( - 1 \right)^{1 + 1} \begin{vmatrix}- 1 & 1 \\ 1 & - 3\end{vmatrix} = 2 , C_{12} = \left( - 1 \right)^{1 + 2} \begin{vmatrix}2 & 1 \\ 2 & - 3\end{vmatrix} = 8, C_{13} = \left( - 1 \right)^{1 + 3} \begin{vmatrix}2 & - 1 \\ 2 & 1\end{vmatrix} = 4$

$C_{21} = \left( - 1 \right)^{2 + 1} \begin{vmatrix}1 & 1 \\ 1 & - 3\end{vmatrix} = 4, C_{22} = \left( - 1 \right)^{2 + 2} \begin{vmatrix}1 & 1 \\ 2 & - 3\end{vmatrix} = - 5, C_{23} = \left( - 1 \right)^{2 + 3} \begin{vmatrix}1 & 1 \\ 2 & 1\end{vmatrix} = 1$

$C_{31} = \left( - 1 \right)^{3 + 1} \begin{vmatrix}1 & 1 \\ - 1 & 1\end{vmatrix} = 2 , C_{32} = \left( - 1 \right)^{3 + 2} \begin{vmatrix}1 & 1 \\ 2 & 1\end{vmatrix} = 1, C_{33} = \left( - 1 \right)^{3 + 3} \begin{vmatrix}1 & 1 \\ 2 & - 1\end{vmatrix} = - 3$

$adj A = \begin{bmatrix}2 & 8 & 4 \\ 4 & - 5 & 1 \\ 2 & 1 & - 3\end{bmatrix}^T$

= $\begin{bmatrix}2 & 4 & 2 \\ 8 & - 5 & 1 \\ 4 & 1 & - 3\end{bmatrix}$

$A^{- 1} = \frac{1}{\left| A \right|}adj A$

= $\frac{1}{14}\begin{bmatrix}2 & 4 & 2 \\ 8 & - 5 & 1 \\ 4 & 1 & - 3\end{bmatrix}$

Now, X = A^-1 B

$\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{14}\begin{bmatrix}2 & 4 & 2 \\ 8 & - 5 & 1 \\ 4 & 1 & - 3\end{bmatrix}\begin{bmatrix}3 \\ - 1 \\ - 9\end{bmatrix}$

$\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{14}\begin{bmatrix}6 - 4 - 18 \\ 24 + 5 - 9 \\ 12 - 1 + 27\end{bmatrix}$

$\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{14}\begin{bmatrix}- 16 \\ 20 \\ 38\end{bmatrix}$

=> x = -16/14, y = 20/14 and z = 38/14

Therefore, x = -8/7, y = 10/7 and z = 19/7.

(iii) 6x - 12y + 25z = 4

4x + 15y - 20z = 3

2x + 18y + 15z = 10

解决方案：

A = $\begin{bmatrix}6 & - 12 & 25 \\ 4 & 15 & - 20 \\ 2 & 18 & 15\end{bmatrix}$

|A| = $\begin{vmatrix}6 & - 12 & 25 \\ 4 & 15 & - 20 \\ 2 & 18 & 15\end{vmatrix}$

= 6 (225 + 360) + 12 (60 + 40) + 25 (72 – 30)

= 3510 + 1200 + 1050

= 5760

So, the given system has a unique solution given by X = A^-1 B.

Let C_ij be the cofactors of the elements a_ij in A.

$C_{11} = \left( - 1 \right)^{1 + 1} \begin{vmatrix}15 & - 20 \\ 18 & 15\end{vmatrix} = 585, C_{12} = \left( - 1 \right)^{1 + 2} \begin{vmatrix}4 & - 20 \\ 2 & 15\end{vmatrix} = - 100 , C_{13} = \left( - 1 \right)^{1 + 3} \begin{vmatrix}4 & 15 \\ 2 & 18\end{vmatrix} = 42$

$C_{21} = \left( - 1 \right)^{2 + 1} \begin{vmatrix}- 12 & 25 \\ 18 & 15\end{vmatrix} = 630, C_{22} = \left( - 1 \right)^{2 + 2} \begin{vmatrix}6 & 25 \\ 2 & 15\end{vmatrix} = 40, C_{23} = \left( - 1 \right)^{2 + 3} \begin{vmatrix}6 & - 12 \\ 2 & 18\end{vmatrix} = - 132$

$C_{31} = \left( - 1 \right)^{3 + 1} \begin{vmatrix}- 12 & 25 \\ 15 & - 20\end{vmatrix} = - 135, C_{32} = \left( - 1 \right)^{3 + 2} \begin{vmatrix}6 & 25 \\ 4 & - 20\end{vmatrix} = 220, C_{33} = \left( - 1 \right)^{3 + 3} \begin{vmatrix}6 & - 12 \\ 4 & 15\end{vmatrix} = 138$

$adj A = \begin{bmatrix}585 & - 100 & 42 \\ 630 & 40 & - 132 \\ - 135 & 220 & 138\end{bmatrix}^T$

= $\begin{bmatrix}585 & 630 & - 135 \\ - 100 & 40 & 220 \\ 42 & - 132 & 138\end{bmatrix}$

$A^{- 1} = \frac{1}{\left| A \right|}adj A$

= $\frac{1}{5760}\begin{bmatrix}585 & 630 & - 135 \\ - 100 & 40 & 220 \\ 42 & - 132 & 138\end{bmatrix}$

Now, X = A^-1 B

$\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{5760}\begin{bmatrix}585 & 630 & - 135 \\ - 100 & 40 & 220 \\ 42 & - 132 & 138\end{bmatrix}\begin{bmatrix}4 \\ 3 \\ 10\end{bmatrix}$

$\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{5760}\begin{bmatrix}2340 + 1890 - 1350 \\ - 400 + 120 + 2200 \\ 168 - 396 + 1380\end{bmatrix}$

$\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{5760}\begin{bmatrix}2880 \\ 1920 \\ 1152\end{bmatrix}$

=> x = 2880/5760, y = 1920/5760 and z = 1152/5760

Therefore x = 1/2, y = 1/3 and z = 1/5.

(iv) 3x + 4y + 7z = 14

2x - y + 3z = 4

x + 2y - 3z = 0

解决方案：

A = $\begin{bmatrix}3 & 4 & 7 \\ 2 & - 1 & 3 \\ 2 & 1 & - 3\end{bmatrix}$

|A| = $\begin{vmatrix}3 & 4 & 7 \\ 2 & - 1 & 3 \\ 2 & 1 & - 3\end{vmatrix}$

= 3 (3 – 3) – 4 (- 6 – 6) + 7 (2 + 2)

= 0 + 48 + 28

= 76

So, the given system has a unique solution given by X = A^-1 B.

Let C_ij be the cofactors of the elements a_ij in A.

$C_{11} = \left( - 1 \right)^{1 + 1} \begin{vmatrix}- 1 & 3 \\ 1 & - 3\end{vmatrix} = 0, C_{12} = \left( - 1 \right)^{1 + 2} \begin{vmatrix}2 & 3 \\ 2 & - 3\end{vmatrix} = 12, C_{13} = \left( - 1 \right)^{1 + 3} \begin{vmatrix}2 & - 1 \\ 2 & 1\end{vmatrix} = 4$

$C_{21} = \left( - 1 \right)^{2 + 1} \begin{vmatrix}4 & 7 \\ 1 & - 3\end{vmatrix} = 19 , C_{22} = \left( - 1 \right)^{2 + 2} \begin{vmatrix}3 & 7 \\ 2 & - 3\end{vmatrix} = - 23 , C_{23} = \left( - 1 \right)^{2 + 3} \begin{vmatrix}3 & 4 \\ 2 & 1\end{vmatrix} = 5$

$C_{31} = \left( - 1 \right)^{3 + 1} \begin{vmatrix}4 & 7 \\ - 1 & 3\end{vmatrix} = 19, C_{32} = \left( - 1 \right)^{3 + 2} \begin{vmatrix}3 & 7 \\ 2 & 3\end{vmatrix} = 5, C_{33} = \left( - 1 \right)^{3 + 3} \begin{vmatrix}3 & 4 \\ 2 & - 1\end{vmatrix} = - 11$

$adj A = \begin{bmatrix}0 & 12 & 4 \\ 19 & - 23 & 5 \\ 19 & 5 & - 11\end{bmatrix}^T$

= $\begin{bmatrix}0 & 19 & 19 \\ 12 & - 23 & 5 \\ 4 & 5 & - 11\end{bmatrix}$

$A^{- 1} = \frac{1}{\left| A \right|}adj A$

= $\frac{1}{76}\begin{bmatrix}0 & 19 & 19 \\ 12 & - 23 & 5 \\ 4 & 5 & - 11\end{bmatrix}$

Now, X = A^-1 B

$\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{76}\begin{bmatrix}0 & 19 & 19 \\ 12 & - 23 & 5 \\ 4 & 5 & - 11\end{bmatrix}\begin{bmatrix}14 \\ 4 \\ 0\end{bmatrix}$

$\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{76}\begin{bmatrix}0 + 76 + 0 \\ 168 - 92 + 0 \\ 56 + 20 + 0\end{bmatrix}$

$\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{76}\begin{bmatrix}76 \\ 76 \\ 76\end{bmatrix}$

=> x = 76/76, y = 76/76 and z = 76/76

Therefore x = 1, y = 1 and z = 1.

(五) $\frac{2}{x} - \frac{3}{y} + \frac{3}{z} = 10\\ \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 10\\ \frac{3}{x} - \frac{1}{y} + \frac{2}{z} = 13$

解决方案：

Let 1/x be a, 1/y be b and 1/z be c.

Here,

A = $\begin{bmatrix}2 & - 3 & 3 \\ 1 & 1 & 1 \\ 3 & - 1 & 2\end{bmatrix}$

|A| = $\begin{vmatrix}2 & - 3 & 3 \\ 1 & 1 & 1 \\ 3 & - 1 & 2\end{vmatrix}$

= 2 (2 + 1) + 3 (2 – 3) + 3 (-1 – 3)

= 6 – 3 – 12

= -9

So, the given system has a unique solution given by X = A^-1 B.

Let C_ij be the cofactors of the elements a_ij in A.

$C_{11} = \left( - 1 \right)^{1 + 1} \begin{vmatrix}1 & 1 \\ - 1 & 2\end{vmatrix} = 3, C_{12} = \left( - 1 \right)^{1 + 2} \begin{vmatrix}1 & 1 \\ 3 & 2\end{vmatrix} = 1 , C_{13} = \left( - 1 \right)^{1 + 3} \begin{vmatrix}1 & 1 \\ 3 & - 1\end{vmatrix} = - 4$

$C_{21} = \left( - 1 \right)^{2 + 1} \begin{vmatrix}- 3 & 3 \\ - 1 & 2\end{vmatrix} = 3 , C_{22} = \left( - 1 \right)^{2 + 2} \begin{vmatrix}2 & 3 \\ 3 & 2\end{vmatrix} = - 5, C_{23} = \left( - 1 \right)^{2 + 3} \begin{vmatrix}2 & - 3 \\ 3 & - 1\end{vmatrix} = -7$

$C_{31} = \left( - 1 \right)^{3 + 1} \begin{vmatrix}- 3 & 3 \\ 1 & 1\end{vmatrix} = - 6 , C_{32} = \left( - 1 \right)^{3 + 2} \begin{vmatrix}2 & 3 \\ 1 & 1\end{vmatrix} = 1, C_{33} = \left( - 1 \right)^{3 + 3} \begin{vmatrix}2 & - 3 \\ 1 & 1\end{vmatrix} = 5$

$adj A = \begin{bmatrix}3 & 1 & - 4 \\ 3 & - 5 & - 7 \\ - 6 & 1 & 5\end{bmatrix}^T$

= $\begin{bmatrix}3 & 3 & - 6 \\ 1 & - 5 & 1 \\ - 4 & - 7 & 5\end{bmatrix}$

$A^{- 1} = \frac{1}{\left| A \right|}adj A$

= $\frac{1}{- 9}\begin{bmatrix}3 & 3 & - 6 \\ 1 & - 5 & 1 \\ - 4 & - 7 & 5\end{bmatrix}$

X = A^-1 B

$\begin{bmatrix}a \\ b \\ c\end{bmatrix} = \frac{1}{- 9}\begin{bmatrix}3 & 3 & - 6 \\ 1 & - 5 & 1 \\ - 4 & - 7 & 5\end{bmatrix}\begin{bmatrix}10 \\ 10 \\ 13\end{bmatrix}$

$\begin{bmatrix}a \\ b \\ c\end{bmatrix} = \frac{1}{- 9}\begin{bmatrix}30 + 30 - 78 \\ 10 - 50 + 13 \\ - 40 - 70 + 65\end{bmatrix}$

$\begin{bmatrix}a \\ b \\ c\end{bmatrix} = \frac{1}{- 9}\begin{bmatrix}- 18 \\ - 27 \\ - 45\end{bmatrix}$

=> x = 1/a = – 9/-18, y = 1/b = – 9/- 27 and z = 1/c = -9/-45

Therefore x = 1/2, y = 1/3 and z = 1/5.

(vi) 5x + 3y + z = 16

2x + y + 3z = 19

x + 2y + 4z = 25

解决方案：

A = $\begin{bmatrix}5 & 3 & 1 \\ 2 & 1 & 3 \\ 1 & 2 & 4\end{bmatrix}$

|A| = $\begin{vmatrix}5 & 3 & 1 \\ 2 & 1 & 3 \\ 1 & 2 & 4\end{vmatrix}$

= 5 (4 – 6) – 3 (8 – 3) + 1 (4 – 1)

= -10 – 15 + 3

= – 22

So, the given system has a unique solution given by X = A^-1 B.

Let C_ij be the cofactors of the elements a_ij in A.

$C_{11} = \left( - 1 \right)^{1 + 1} \begin{vmatrix}1 & 3 \\ 2 & 4\end{vmatrix} = - 2, C_{12} = \left( - 1 \right)^{1 + 2} \begin{vmatrix}2 & 3 \\ 1 & 4\end{vmatrix} = - 5 , C_{13} = \left( - 1 \right)^{1 + 3} \begin{vmatrix}2 & 1 \\ 1 & 2\end{vmatrix} = 3$

$C_{21} = \left( - 1 \right)^{2 + 1} \begin{vmatrix}3 & 1 \\ 2 & 4\end{vmatrix} = - 10 , C_{22} = \left( - 1 \right)^{2 + 2} \begin{vmatrix}5 & 1 \\ 1 & 4\end{vmatrix} = 19, C_{23} = \left( - 1 \right)^{2 + 3} \begin{vmatrix}5 & 3 \\ 1 & 2\end{vmatrix} = -7$

$C_{31} = \left( - 1 \right)^{3 + 1} \begin{vmatrix}3 & 1 \\ 1 & 3\end{vmatrix} = 8 , C_{32} = \left( - 1 \right)^{3 + 2} \begin{vmatrix}5 & 1 \\ 2 & 3\end{vmatrix} = - 13 , C_{33} = \left( - 1 \right)^{3 + 3} \begin{vmatrix}5 & 3 \\ 2 & 1\end{vmatrix} = - 1$

$adj A = \begin{bmatrix}- 2 & - 5 & 3 \\ - 10 & 19 & - 7 \\ 8 & - 13 & - 1\end{bmatrix}^T$

= $\begin{bmatrix}- 2 & - 10 & 8 \\ - 5 & 19 & - 13 \\ 3 & - 7 & - 1\end{bmatrix}$

$A^{- 1} = \frac{1}{\left| A \right|}adj A$

= $\frac{1}{- 22}\begin{bmatrix}- 2 & - 10 & 8 \\ - 5 & 19 & - 13 \\ 3 & - 7 & - 1\end{bmatrix}$

X = A^-1 B

$\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{- 22}\begin{bmatrix}- 2 & - 10 & 8 \\ - 5 & 19 & - 13 \\ 3 & - 7 & - 1\end{bmatrix}\begin{bmatrix}16 \\ 19 \\ 25\end{bmatrix}$

$\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{- 22}\begin{bmatrix}- 32 - 190 + 200 \\ - 80 + 361 - 325 \\ 48 - 133 - 25\end{bmatrix}$

$\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{- 22}\begin{bmatrix}- 22 \\ - 44 \\ - 110\end{bmatrix}$

=> x = – 22/- 22, y = – 44/- 22 and z = -110/-22

Therefore x = 1, y = 2 and z = 5.

(vii) 3x + 4y + 2z = 8

2y - 3z = 3

x - 2y + 6z = -2

解决方案：

A = $\begin{bmatrix}3 & 4 & 2 \\ 0 & 2 & - 3 \\ 1 & - 2 & 6\end{bmatrix}$

|A| = $\begin{vmatrix}3 & 4 & 2 \\ 0 & 2 & - 3 \\ 1 & - 2 & 6\end{vmatrix}$

= 3 (12 – 6) – 4 (0 + 3) + 2 (0 – 2)

= 18 – 12 – 4

= 2

Let C_ij be the cofactors of the elements a_ij in A.

$C_{11} = \left( - 1 \right)^{1 + 1} \begin{vmatrix}2 & - 3 \\ - 2 & 6\end{vmatrix} = 6, C_{12} = \left( - 1 \right)^{1 + 2} \begin{vmatrix}0 & - 3 \\ 1 & 6\end{vmatrix} = - 3 , C_{13} = \left( - 1 \right)^{1 + 3} \begin{vmatrix}0 & 2 \\ 1 & - 2\end{vmatrix} = - 2$

$C_{21} = \left( - 1 \right)^{2 + 1} \begin{vmatrix}4 & 2 \\ - 2 & 6\end{vmatrix} = - 28 , C_{22} = \left( - 1 \right)^{2 + 2} \begin{vmatrix}3 & 2 \\ 1 & 6\end{vmatrix} = 16 , C_{23} = \left( - 1 \right)^{2 + 3} \begin{vmatrix}3 & 4 \\ 1 & - 2\end{vmatrix} = 10$

$C_{31} = \left( - 1 \right)^{3 + 1} \begin{vmatrix}4 & 2 \\ 2 & - 3\end{vmatrix} = - 16, C_{32} = \left( - 1 \right)^{3 + 2} \begin{vmatrix}3 & 2 \\ 0 & - 3\end{vmatrix} = 9, C_{33} = \left( - 1 \right)^{3 + 3} \begin{vmatrix}3 & 4 \\ 0 & 2\end{vmatrix} = 6$

$adj A = \begin{bmatrix}6 & - 3 & - 2 \\ - 28 & 16 & 10 \\ - 16 & 9 & 6\end{bmatrix}^T$

= $\begin{bmatrix}6 & - 28 & - 16 \\ - 3 & 16 & 9 \\ - 2 & 10 & 6\end{bmatrix}$

$A^{- 1} = \frac{1}{\left| A \right|}adj A$

= $\frac{1}{2}\begin{bmatrix}6 & - 28 & - 16 \\ - 3 & 16 & 9 \\ - 2 & 10 & 6\end{bmatrix}$

Now X = A^-1 B

$\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{2}\begin{bmatrix}6 & - 28 & - 16 \\ - 3 & 16 & 9 \\ - 2 & 10 & 6\end{bmatrix}\begin{bmatrix}8 \\ 3 \\ - 2\end{bmatrix}$

$\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{2}\begin{bmatrix}48 - 84 + 32 \\ - 24 + 48 - 18 \\ - 16 + 30 - 12\end{bmatrix}$

$\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{2}\begin{bmatrix}- 4 \\ 6 \\ 2\end{bmatrix}$

=> x = -4/2, y = 6/2 and z = 2/2

Therefore x = -2, y = 3 and z = 1.

(viii) 2x + y + z = 2

x + 3y - z = 5

3x + y - 2z = 6

解决方案：

Here,

A = $\begin{bmatrix}2 & 1 & 1 \\ 1 & 3 & - 1 \\ 3 & 1 & - 2\end{bmatrix}$

|A| = $\begin{vmatrix}2 & 1 & 1 \\ 1 & 3 & - 1 \\ 3 & 1 & - 2\end{vmatrix}$

= – 10 – 1 – 8

= -19

Let C_ij be the cofactors of the elements a_ij in A.

$C_{11} = \left( - 1 \right)^{1 + 1} \begin{vmatrix}3 & - 1 \\ 1 & - 2\end{vmatrix} = - 5, C_{12} = \left( - 1 \right)^{1 + 2} \begin{vmatrix}1 & - 1 \\ 3 & - 2\end{vmatrix} = - 1, C_{13} = \left( - 1 \right)^{1 + 3} \begin{vmatrix}1 & 3 \\ 3 & 1\end{vmatrix} = - 8$

$C_{21} = \left( - 1 \right)^{2 + 1} \begin{vmatrix}1 & 1 \\ 1 & - 2\end{vmatrix} = 3, C_{22} = \left( - 1 \right)^{2 + 2} \begin{vmatrix}2 & - 1 \\ 3 & - 2\end{vmatrix} = - 7, C_{23} = \left( - 1 \right)^{2 + 3} \begin{vmatrix}2 & 1 \\ 3 & 1\end{vmatrix} = 1$

$C_{31} = \left( - 1 \right)^{3 + 1} \begin{vmatrix}1 & 1 \\ 3 & - 1\end{vmatrix} = - 4, C_{32} = \left( - 1 \right)^{3 + 2} \begin{vmatrix}2 & 1 \\ 1 & - 1\end{vmatrix} = 3, C_{33} = \left( - 1 \right)^{3 + 3} \begin{vmatrix}2 & 1 \\ 1 & 3\end{vmatrix} = 5$

adj A = $\begin{bmatrix}- 5 & - 1 & - 8 \\ 3 & - 7 & 1 \\ - 4 & 3 & 5\end{bmatrix}^T$

= $\begin{bmatrix}- 5 & 3 & - 4 \\ - 1 & - 7 & 3 \\ - 8 & 1 & 5\end{bmatrix}$

$A^{- 1} = \frac{1}{\left| A \right|}adj A$

= $\frac{1}{- 19}\begin{bmatrix}- 5 & 3 & - 4 \\ - 1 & - 7 & 3 \\ - 8 & 1 & 5\end{bmatrix}$

X = A^-1 B

$\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{- 19}\begin{bmatrix}- 5 & 3 & - 4 \\ - 1 & - 7 & 3 \\ - 8 & 1 & 5\end{bmatrix}\begin{bmatrix}2 \\ 5 \\ 6\end{bmatrix}$

$\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{- 19}\begin{bmatrix}- 10 + 15 - 24 \\ - 2 - 35 + 18 \\ - 16 + 5 + 30\end{bmatrix}$

$\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{- 19}\begin{bmatrix}- 19 \\ -19 \\ 19\end{bmatrix}$

x = -19/-19, y = -19/-19 and z = 19/-19

Therefore x = 1, y = 1 and z = – 1.

(ix) 2x + 6y = 2

3x - z = -8

2x - y + z = -3

解决方案：

A = $\begin{bmatrix}2 & 6 & 0 \\ 3 & 0 & - 1 \\ 2 & - 1 & 1\end{bmatrix}$

|A| = $\begin{vmatrix}2 & 6 & 0 \\ 3 & 0 & - 1 \\ 2 & - 1 & 1\end{vmatrix}$

= 2 (0 – 1) – 6 (3 + 2) + 0 (-3 + 0)

= -2 – 30

= – 32

Let C_ij be the cofactors of the elements a_ij in A.

$C_{11} = \left( - 1 \right)^{1 + 1} \begin{vmatrix}0 & - 1 \\ - 1 & 1\end{vmatrix} = - 1 , C_{12} = \left( - 1 \right)^{1 + 2} \begin{vmatrix}3 & - 1 \\ 2 & 1\end{vmatrix} = - 5 , C_{13} = \left( - 1 \right)^{1 + 3} \begin{vmatrix}3 & 0 \\ 2 & - 1\end{vmatrix} = - 3$

$C_{21} = \left( - 1 \right)^{2 + 1} \begin{vmatrix}6 & 0 \\ - 1 & 1\end{vmatrix} = - 6 , C_{22} = \left( - 1 \right)^{2 + 2} \begin{vmatrix}2 & 0 \\ 2 & 1\end{vmatrix} = 2 , C_{23} = \left( - 1 \right)^{2 + 3} \begin{vmatrix}2 & 6 \\ 2 & - 1\end{vmatrix} = 14$

$C_{31} = \left( - 1 \right)^{3 + 1} \begin{vmatrix}6 & 0 \\ 0 & - 1\end{vmatrix} = - 6 , C_{32} = \left( - 1 \right)^{3 + 2} \begin{vmatrix}2 & 0 \\ 3 & - 1\end{vmatrix} = 2 , C_{33} = \left( - 1 \right)^{3 + 3} \begin{vmatrix}2 & 6 \\ 3 & 0\end{vmatrix} = - 18$

$adj A = \begin{bmatrix}- 1 & - 5 & - 3 \\ - 6 & 2 & 14 \\ - 6 & 2 & - 18\end{bmatrix}^T$

= $\begin{bmatrix}- 1 & - 6 & - 6 \\ - 5 & 2 & 2 \\ - 3 & 14 & - 18\end{bmatrix}$

$A^{- 1} = \frac{1}{\left| A \right|}adj A$

= $\frac{1}{- 32}\begin{bmatrix}- 1 & - 6 & - 6 \\ - 5 & 2 & 2 \\ - 3 & 14 & - 18\end{bmatrix}$

X = A^-1 B

$\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{- 32}\begin{bmatrix}- 1 & - 6 & - 6 \\ - 5 & 2 & 2 \\ - 3 & 14 & - 18\end{bmatrix}\begin{bmatrix}2 \\ - 8 \\ - 3\end{bmatrix}$

$\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{- 32}\begin{bmatrix}- 2 + 48 + 18 \\ - 10 - 16 - 6 \\ - 6 - 112 + 54\end{bmatrix}$

$\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{- 32}\begin{bmatrix}64 \\ - 32 \\ - 64\end{bmatrix}$

=> x = 64/-32, y = -32/-32 and z = -64/-32

Therefore x = – 2, y = 1 and z = 2.

(x) x - y + z = 2

2x - y = 0

2y - z = 1

解决方案：

A = $\begin{bmatrix}1 & - 1 & 1 \\ 2 & - 1 & 0 \\ 0 & 2 & - 1\end{bmatrix}$

|A| = $\begin{vmatrix}1 & - 1 & 1 \\ 2 & - 1 & 0 \\ 0 & 2 & - 1\end{vmatrix}$

= 1 (1 – 0) + 1 (-2 – 0) + 1(4 – 0)

= 1 – 2 + 4

= 3

Let C_ij be the cofactors of the elements a_ij in A.

$C_{11} = \left( - 1 \right)^{1 + 1} \begin{vmatrix}- 1 & 0 \\ 2 & - 1\end{vmatrix} = 1, C_{12} = \left( - 1 \right)^{1 + 2} \begin{vmatrix}2 & 0 \\ 0 & - 1\end{vmatrix} = 2, C_{13} = \left( - 1 \right)^{1 + 3} \begin{vmatrix}2 & - 1 \\ 0 & 2\end{vmatrix} = 4$

$C_{21} = \left( - 1 \right)^{2 + 1} \begin{vmatrix}- 1 & 1 \\ 2 & - 1\end{vmatrix} = 1, C_{22} = \left( - 1 \right)^{2 + 2} \begin{vmatrix}1 & 1 \\ 0 & - 1\end{vmatrix} = - 1, C_{23} = \left( - 1 \right)^{2 + 3} \begin{vmatrix}1 & - 1 \\ 0 & 2\end{vmatrix} = - 2$

$C_{31} = \left( - 1 \right)^{3 + 1} \begin{vmatrix}- 1 & 1 \\ - 1 & 0\end{vmatrix} = 1, C_{32} = \left( - 1 \right)^{3 + 2} \begin{vmatrix}1 & 1 \\ 2 & 0\end{vmatrix} = 2, C_{33} = \left( - 1 \right)^{3 + 3} \begin{vmatrix}1 & - 1 \\ 2 & - 1\end{vmatrix} = 1$

$adj A = \begin{bmatrix}1 & 2 & 4 \\ 1 & - 1 & - 2 \\ 1 & 2 & 1\end{bmatrix}^T$

= $\begin{bmatrix}1 & 1 & 1 \\ 2 & - 1 & 2 \\ 4 & - 2 & 1\end{bmatrix}$

$A^{- 1} = \frac{1}{\left| A \right|}adj A$

= $\frac{1}{1}\begin{bmatrix}1 & 1 & 1 \\ 2 & - 1 & 2 \\ 4 & - 2 & 1\end{bmatrix}$

X = A^-1 B

$\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{3}\begin{bmatrix}1 & 1 & 1 \\ 2 & - 1 & 2 \\ 4 & - 2 & 1\end{bmatrix}\begin{bmatrix}2 \\ 0 \\ 1\end{bmatrix}$

$\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{3}\begin{bmatrix}2 + 1 \\ 4 + 2 \\ 8 + 1\end{bmatrix}$

$\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{1}\begin{bmatrix}3 \\ 6 \\ 9\end{bmatrix}$

=> x = 3/3, y = 6/3 and z = 9/3

Therefore x = 1, y = 2 and z = 3.

(xi) 8x + 4y + 3z = 18

2x + y +z = 5

x + 2y + z = 5

解决方案：

A = $\begin{bmatrix}8 & 4 & 3 \\ 2 & 1 & 1 \\ 1 & 2 & 1\end{bmatrix}$

|A| = $\begin{vmatrix}8 & 4 & 3 \\ 2 & 1 & 1 \\ 1 & 2 & 1\end{vmatrix}$

= 8 (1 – 2) – 4 (2 – 1) + 3(4 – 1)

= – 8 – 4 + 9

= -3

Let C_ij be the cofactors of the elements a_ij in A.

$C_{11} = \left( - 1 \right)^{1 + 1} \begin{vmatrix}1 & 1 \\ 2 & 1\end{vmatrix} = - 1, C_{12} = \left( - 1 \right)^{1 + 2} \begin{vmatrix}2 & 1 \\ 1 & 1\end{vmatrix} = - 1, C_{13} = \left( - 1 \right)^{1 + 3} \begin{vmatrix}2 & 1 \\ 1 & 2\end{vmatrix} = 3$

$C_{21} = \left( - 1 \right)^{2 + 1} \begin{vmatrix}4 & 3 \\ 2 & 1\end{vmatrix} = 2, C_{22} = \left( - 1 \right)^{2 + 2} \begin{vmatrix}8 & 3 \\ 1 & 1\end{vmatrix} = 5, C_{23} = \left( - 1 \right)^{2 + 3} \begin{vmatrix}8 & 4 \\ 1 & 2\end{vmatrix} = - 12$

$C_{31} = \left( - 1 \right)^{3 + 1} \begin{vmatrix}4 & 3 \\ 1 & 1\end{vmatrix} = 1, C_{32} = \left( - 1 \right)^{3 + 2} \begin{vmatrix}8 & 3 \\ 2 & 1\end{vmatrix} = - 2 , C_{33} = \left( - 1 \right)^{3 + 3} \begin{vmatrix}8 & 4 \\ 2 & 1\end{vmatrix} = 0$

$adj A = \begin{bmatrix}- 1 & - 1 & 3 \\ 2 & 5 & - 12 \\ 1 & - 2 & 0\end{bmatrix}^T$

= $\begin{bmatrix}- 1 & 2 & 1 \\ - 1 & 5 & - 2 \\ 3 & - 12 & 0\end{bmatrix}$

$A^{- 1} = \frac{1}{\left| A \right|}adj A$

= $\frac{1}{- 3}\begin{bmatrix}- 1 & 2 & 1 \\ - 1 & 5 & - 2 \\ 3 & - 12 & 0\end{bmatrix}$

X = A^-1 B

$\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{- 3}\begin{bmatrix}- 1 & 2 & 1 \\ - 1 & 5 & - 2 \\ 3 & - 12 & 0\end{bmatrix}\begin{bmatrix}18 \\ 5 \\ 5\end{bmatrix}$

$\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{- 3}\begin{bmatrix}- 18 + 10 + 5 \\ - 18 + 25 - 10 \\ 54 - 60\end{bmatrix}$

$\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{- 3}\begin{bmatrix}- 3 \\ - 3 \\ - 6\end{bmatrix}$

=> x = -3/-3, y = -3/-3 and z = -6/-3

Therefore x = 1, y = 1 and z = 2.

(xii) x + y + z = 6

x + 2z = 7

3x + y + z = 12

解决方案：

A = $\begin{bmatrix}1 & 1 & 1 \\ 1 & 0 & 2 \\ 3 & 1 & 1\end{bmatrix}$

|A| = $\begin{vmatrix}1 & 1 & 1 \\ 1 & 0 & 2 \\ 3 & 1 & 1\end{vmatrix}$

= 1 (0 – 2) – 1 (1 – 6) + 1(1 – 0)

= – 2 + 5 + 1

= 4

Let C_ij be the cofactors of the elements a_ij in A.

$C_{11} = \left( - 1 \right)^{1 + 1} \begin{vmatrix}0 & 2 \\ 1 & 1\end{vmatrix} = - 2, C_{12} = \left( - 1 \right)^{1 + 2} \begin{vmatrix}1 & 2 \\ 3 & 1\end{vmatrix} = 5, C_{13} = \left( - 1 \right)^{1 + 3} \begin{vmatrix}1 & 0 \\ 3 & 1\end{vmatrix} = 1$

$C_{21} = \left( - 1 \right)^{2 + 1} \begin{vmatrix}1 & 1 \\ 1 & 1\end{vmatrix} = 0, C_{22} = \left( - 1 \right)^{2 + 2} \begin{vmatrix}1 & 1 \\ 3 & 1\end{vmatrix} = - 2, C_{23} = \left( - 1 \right)^{2 + 3} \begin{vmatrix}1 & 1 \\ 3 & 1\end{vmatrix} = 2$

$C_{31} = \left( - 1 \right)^{3 + 1} \begin{vmatrix}1 & 1 \\ 0 & 2\end{vmatrix} = 2, C_{32} = \left( - 1 \right)^{3 + 2} \begin{vmatrix}1 & 1 \\ 1 & 2\end{vmatrix} = - 1, C_{33} = \left( - 1 \right)^{3 + 3} \begin{vmatrix}1 & 1 \\ 1 & 0\end{vmatrix} = - 1$

$adj A = \begin{bmatrix}- 2 & 5 & 1 \\ 0 & - 2 & 2 \\ 2 & - 1 & - 1\end{bmatrix}^T$

= $\begin{bmatrix}- 2 & 0 & 2 \\ 5 & - 2 & - 1 \\ 1 & 2 & - 1\end{bmatrix}$

$A^{- 1} = \frac{1}{\left| A \right|}adj A$

= $\frac{1}{4}\begin{bmatrix}- 2 & 0 & 2 \\ 5 & - 2 & - 1 \\ 1 & 2 & - 1\end{bmatrix}$

X = A^-1 B

$\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{4}\begin{bmatrix}- 2 & 0 & 2 \\ 5 & - 2 & - 1 \\ 1 & 2 & - 1\end{bmatrix}\begin{bmatrix}6 \\ 7 \\ 12\end{bmatrix}$

$\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{4}\begin{bmatrix}- 12 + 0 + 24 \\ 30 - 14 - 12 \\ 6 - 14 - 12\end{bmatrix}$

$\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{4}\begin{bmatrix}12 \\ 4 \\ - 20\end{bmatrix}$

=> x = 12/4, y = 4/4 and z = -20/4

Therefore x = 3, y = 1 and z = – 5.

(十三) $\frac{2}{x} + \frac{3}{y} + \frac{10}{z} = 4\\\frac{4}{x} - \frac{6}{y} + \frac{5}{z} = 1\\\frac{6}{x} + \frac{9}{y} - \frac{20}{z} = 2$ , x, y, z ≠ 0

解决方案：

Let 1/x be a, 1/y be b and 1/z be c.

Here,

A = $\begin{bmatrix}2 & 3 & 10 \\ 4 & - 6 & 5 \\ 6 & 9 & - 20\end{bmatrix}$

|A| = $\begin{vmatrix}2 & 3 & 10 \\ 4 & - 6 & 5 \\ 6 & 9 & - 20\end{vmatrix}$

= 2 (120 – 45) – 3 (-80 – 30) + 10 (36 + 36)

= 150 + 330 + 720

= 1200

Let C_ij be the cofactors of the elements a_ij in A.

$C_{11} = \left( - 1 \right)^{1 + 1} \begin{vmatrix}- 6 & 5 \\ 9 & - 20\end{vmatrix} = 75, C_{12} = \left( - 1 \right)^{1 + 2} \begin{vmatrix}4 & 5 \\ 6 & - 20\end{vmatrix} = 110, C_{13} = \left( - 1 \right)^{1 + 3} \begin{vmatrix}4 & - 6 \\ 6 & 9\end{vmatrix} = 72$

$C_{21} = \left( - 1 \right)^{2 + 1} \begin{vmatrix}3 & 10 \\ 9 & - 20\end{vmatrix} = 150, C_{22} = \left( - 1 \right)^{2 + 2} \begin{vmatrix}2 & 10 \\ 6 & - 20\end{vmatrix} = - 100, C_{23} = \left( - 1 \right)^{2 + 3} \begin{vmatrix}2 & 3 \\ 6 & 9\end{vmatrix} = 0$

$C_{31} = \left( - 1 \right)^{3 + 1} \begin{vmatrix}3 & 10 \\ - 6 & 5\end{vmatrix} = 75, C_{32} = \left( - 1 \right)^{3 + 2} \begin{vmatrix}2 & 10 \\ 4 & 5\end{vmatrix} = 30, C_{33} = \left( - 1 \right)^{3 + 3} \begin{vmatrix}2 & 3 \\ 4 & - 6\end{vmatrix} = - 24$

$adj A = \begin{bmatrix}75 & 110 & 72 \\ 150 & - 100 & 0 \\ 75 & 30 & - 24\end{bmatrix}^T$

= $\begin{bmatrix}75 & 150 & 75 \\ 110 & - 100 & 30 \\ 72 & 0 & - 24\end{bmatrix}$

$A^{- 1} = \frac{1}{\left| A \right|}adj A$

= $\frac{1}{1200} \begin{bmatrix}75 & 150 & 75 \\ 110 & - 100 & 30 \\ 72 & 0 & - 24\end{bmatrix}$

X = A^-1 B

$\begin{bmatrix}a \\ b \\ c\end{bmatrix} = \frac{1}{1200} \begin{bmatrix}75 & 150 & 75 \\ 110 & - 100 & 30 \\ 72 & 0 & - 24\end{bmatrix}\begin{bmatrix}4 \\ 1 \\ 2\end{bmatrix}$

$\begin{bmatrix}a \\ b \\ c\end{bmatrix} = \frac{1}{1200}\begin{bmatrix}300 + 150 + 150 \\ 440 - 100 + 60 \\ 288 - 48\end{bmatrix}$

$\begin{bmatrix}a \\ b \\ c\end{bmatrix} = \frac{1}{1200}\begin{bmatrix}600 \\ 400 \\ 240\end{bmatrix}$

=> x = 1/a = 1200/600, y = 1/b = 1200/400 and z = 1/c = 1200/240

Therefore x = 2, y = 3 and z = 5.

(xiv) x - y + 2z = 7

3x + 4y - 5z = -5

2x - y + 3z = 12

解决方案：

A = $\begin{bmatrix}1 & - 1 & 2 \\ 3 & 4 & - 5 \\ 2 & - 1 & 3\end{bmatrix}$

|A| = $\begin{vmatrix}1 & - 1 & 2 \\ 3 & 4 & - 5 \\ 2 & - 1 & 3\end{vmatrix}$

= 1 (12 – 5) + 1 (9 + 10) + 2 (-3 – 8)

= 7 + 19 – 22

= 4

Let C_ij be the cofactors of the elements a_ij in A.

$C_{11} = \left( - 1 \right)^{1 + 1} \begin{vmatrix}4 & - 5 \\ - 1 & 3\end{vmatrix} = 7, C_{12} = \left( - 1 \right)^{1 + 2} \begin{vmatrix}3 & - 5 \\ 2 & 3\end{vmatrix} = - 19, C_{13} = \left( - 1 \right)^{1 + 3} \begin{vmatrix}3 & 4 \\ 2 & - 1\end{vmatrix} = - 11$

$C_{21} = \left( - 1 \right)^{2 + 1} \begin{vmatrix}- 1 & 2 \\ - 1 & 3\end{vmatrix} = 1 , C_{22} = \left( - 1 \right)^{2 + 2} \begin{vmatrix}1 & 2 \\ 2 & 3\end{vmatrix} = - 1 , C_{23} = \left( - 1 \right)^{2 + 3} \begin{vmatrix}1 & - 1 \\ 2 & - 1\end{vmatrix} = - 1$

$C_{31} = \left( - 1 \right)^{3 + 1} \begin{vmatrix}- 1 & 2 \\ 4 & - 5\end{vmatrix} = - 3, C_{32} = \left( - 1 \right)^{3 + 2} \begin{vmatrix}1 & 2 \\ 3 & - 5\end{vmatrix} = 11, C_{33} = \left( - 1 \right)^{3 + 3} \begin{vmatrix}1 & - 1 \\ 3 & 4\end{vmatrix} = 7$

$adj A = \begin{bmatrix}7 & - 19 & - 11 \\ 1 & - 1 & - 1 \\ - 3 & 11 & 7\end{bmatrix}^T$

= $\begin{bmatrix}7 & 1 & - 3 \\ - 19 & - 1 & 11 \\ - 11 & - 1 & 7\end{bmatrix}$

$A^{- 1} = \frac{1}{\left| A \right|}adj A$

= $\frac{1}{4}\begin{bmatrix}7 & 1 & - 3 \\ - 19 & - 1 & 11 \\ - 11 & - 1 & 7\end{bmatrix}$

X = A^-1 B

$\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{4}\begin{bmatrix}7 & 1 & - 3 \\ - 19 & - 1 & 11 \\ - 11 & - 1 & 7\end{bmatrix}\begin{bmatrix}7 \\ - 5 \\ 12\end{bmatrix}$

$\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{4}\begin{bmatrix}49 - 5 - 36 \\ - 133 + 5 + 132 \\ - 77 + 5 + 84\end{bmatrix}$

$\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{4}\begin{bmatrix}8 \\ 4 \\ 12\end{bmatrix}$

=> x = 8/4, y = 4/4 and z = 12/4

Therefore x = 2, y = 1 and z = 3.

问题 3. 证明下列线性方程组是一致的：

(i) 6x + 4y = 2

9x + 6y = 3

解决方案：

 Here,

6x + 4y = 2

9x + 6y = 3

We know, AX = B

A = $\begin{bmatrix}6 & 4 \\ 9 & 6\end{bmatrix}$, X = $\binom{x}{y}$ and B = $\binom{2}{3}$

$\begin{bmatrix}6 & 4 \\ 9 & 6\end{bmatrix}\binom{x}{y} = \binom{2}{3}$

|A| = $\begin{vmatrix}6 & 4 \\ 9 & 6\end{vmatrix}$

= 36 – 36

= 0

Let C_ij be the cofactors of the elements a_ij in A.

C₁₁ = 6, C₁₂ = -9, C₂₁ = -4 and C₂₂ = 6

$adj A = \begin{bmatrix}6 & - 9 \\ - 4 & 6\end{bmatrix}^T$

= $\begin{bmatrix}6 & - 4 \\ - 9 & 6\end{bmatrix}$

(adj A) B = $\begin{bmatrix}6 & - 4 \\ - 9 & 6\end{bmatrix}\binom{2}{3}$

= $\binom{12 - 12}{ - 18 + 18}$

= $\binom{0}{0}$

 Therefore, the system is consistent and has infinitely many solutions.

Hence proved.

(ii) 2x + 3y = 5

6x + 9y = 15

解决方案：

Here,

2x + 3y = 5

6x + 9y = 15

We know, AX = B

A = $\begin{bmatrix}2 & 3 \\ 6 & 9\end{bmatrix}$, X = $\binom{x}{y}$ and B = $\binom{5}{15}$

$\begin{bmatrix}2 & 3 \\ 6 & 9\end{bmatrix}\binom{x}{y} = \binom{5}{15}$

|A| = $\begin{vmatrix}2 & 3 \\ 6 & 9\end{vmatrix}$

= 18 – 18

= 0

Let C_ij be the cofactors of the elements a_ij in A.

C₁₁ = 9, C₁₂ = -6, C₂₁ = -3 and C₂₂ = 2

$adj A = \begin{bmatrix}9 & - 6 \\ - 3 & 2\end{bmatrix}^T$

= $\begin{bmatrix}2 & - 3 \\ - 6 & 9\end{bmatrix}$

(adj A) B = $\begin{bmatrix}9 & - 3 \\ - 6 & 2\end{bmatrix}\binom{5}{15}$

= $\binom{45 - 45}{ - 30 + 30}$

= $\binom{0}{0}$

Therefore, the system is consistent and has infinitely many solutions.

Hence proved.

(iii) 5x + 3y + 7z = 4

3x + 26y + 2z = 9

7x + 2y + 10z = 5

解决方案：

 Here,

5x + 3y + 7z = 4

3x + 26y + 2z = 9

7x + 2y + 10z = 5

We know, AX = B

A = $\begin{bmatrix}5 & 3 & 7 \\ 3 & 26 & 2 \\ 7 & 2 & 10\end{bmatrix}$, X = $\begin{bmatrix}x \\ y \\ z\end{bmatrix}$ and B = $\begin{bmatrix}4 \\ 9 \\ 5\end{bmatrix}$

$\begin{bmatrix}5 & 3 & 7 \\ 3 & 26 & 2 \\ 7 & 2 & 10\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \begin{bmatrix}4 \\ 9 \\ 5\end{bmatrix}$

|A| = $\begin{vmatrix}5 & 3 & 7 \\ 3 & 26 & 2 \\ 7 & 2 & 10\end{vmatrix}$

= 1280 – 48 – 1232\]

= 0

Let C_ij be the cofactors of the elements a_ij in A.

$C_{11} = \left( - 1 \right)^{1 + 1} \begin{vmatrix}26 & 2 \\ 2 & 10\end{vmatrix} = 256 , C_{12} = \left( - 1 \right)^{1 + 2} \begin{vmatrix}3 & 2 \\ 7 & 10\end{vmatrix} = - 16, C_{13} = \left( - 1 \right)^{1 + 3} \begin{vmatrix}3 & 26 \\ 7 & 2\end{vmatrix} = - 176$

$C_{21} = \left( - 1 \right)^{2 + 1} \begin{vmatrix}3 & 7 \\ 2 & 10\end{vmatrix} = - 16 , C_{22} = \left( - 1 \right)^{2 + 2} \begin{vmatrix}5 & 7 \\ 7 & 10\end{vmatrix} = 1, C_{23} = \left( - 1 \right)^{2 + 3} \begin{vmatrix}5 & 3 \\ 7 & 2\end{vmatrix} = 11$

$C_{31} = \left( - 1 \right)^{3 + 1} \begin{vmatrix}3 & 7 \\ 26 & 2\end{vmatrix} = - 176, C_{32} = \left( - 1 \right)^{3 + 2} \begin{vmatrix}5 & 7 \\ 3 & 2\end{vmatrix} = 11, C_{33} = \left( - 1 \right)^{3 + 3} \begin{vmatrix}5 & 3 \\ 3 & 26\end{vmatrix} = 121$

$adj A = \begin{bmatrix}256 & - 16 & - 176 \\ - 16 & 1 & 11 \\ - 176 & 11 & 121\end{bmatrix}^T$

= $\begin{bmatrix}256 & - 16 & - 176 \\ - 16 & 1 & 11 \\ - 176 & 11 & 121\end{bmatrix}$

(adj A)B = $\begin{bmatrix}256 & - 16 & - 176 \\ - 16 & 1 & 11 \\ - 176 & 11 & 121\end{bmatrix}\begin{bmatrix}4 \\ 9 \\ 5\end{bmatrix}$

= $\begin{bmatrix}1024 - 144 - 880 \\ - 64 + 9 + 55 \\ - 704 + 99 + 605\end{bmatrix}$

= $\begin{bmatrix}0 \\ 0 \\ 0\end{bmatrix}$

Therefore, the system is consistent and has infinitely many solutions.

Hence proved.

(iv) x - y + z = 3

2x + y - z = 2

-x -2y + 2z = 1

解决方案：

 Here,

x − y + z = 3

2x + y − z = 2

−x −2y + 2z = 1

We know, AX = B

A = $\begin{bmatrix}1 & - 1 & 1 \\ 2 & 1 & - 1 \\ - 1 & - 2 & 2\end{bmatrix}$, X = $\begin{bmatrix}x \\ y \\ z\end{bmatrix}$ and B = $\begin{bmatrix}3 \\ 2 \\ 1\end{bmatrix}$

$\begin{bmatrix}1 & - 1 & 1 \\ 2 & 1 & - 1 \\ - 1 & - 2 & 2\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \begin{bmatrix}3 \\ 2 \\ 1\end{bmatrix}$

|A| = $\begin{vmatrix}1 & - 1 & 1 \\ 2 & 1 & - 1 \\ - 1 & - 2 & 2\end{vmatrix}$

= 1\left( 2 – 2 \right) + 1\left( 4 – 1 \right) + 1( – 4 + 1)\]

= 0 + 3 – 3

= 0

Let C_ij be the cofactors of the elements a_ij in A.

$C_{11} = \left( - 1 \right)^{1 + 1} \begin{vmatrix}1 & - 1 \\ - 2 & 2\end{vmatrix} = 0, C_{12} = \left( - 1 \right)^{1 + 2} \begin{vmatrix}2 & - 1 \\ - 1 & 2\end{vmatrix} = - 3, C_{13} = \left( - 1 \right)^{1 + 3} \begin{vmatrix}2 & 1 \\ - 1 & - 2\end{vmatrix} = - 3$

$C_{21} = \left( - 1 \right)^{2 + 1} \begin{vmatrix}- 1 & 1 \\ - 2 & 2\end{vmatrix} = 0, C_{22} = \left( - 1 \right)^{2 + 2} \begin{vmatrix}1 & 1 \\ - 1 & 2\end{vmatrix} = 3, C_{23} = \left( - 1 \right)^{2 + 3} \begin{vmatrix}1 & - 1 \\ - 1 & - 2\end{vmatrix} = 3$

$C_{31} = \left( - 1 \right)^{3 + 1} \begin{vmatrix}- 1 & 1 \\ 1 & - 1\end{vmatrix} = 0, C_{32} = \left( - 1 \right)^{3 + 2} \begin{vmatrix}1 & 1 \\ 2 & - 1\end{vmatrix} = 3 , C_{33} = \left( - 1 \right)^{3 + 3} \begin{vmatrix}1 & - 1 \\ 2 & 1\end{vmatrix} = 3$

$adj A = \begin{bmatrix}0 & - 3 & - 3 \\ 0 & 3 & 3 \\ 0 & 3 & 3\end{bmatrix}^T$

= $\begin{bmatrix}0 & 0 & 0 \\ - 3 & 3 & 3 \\ - 3 & 3 & 3\end{bmatrix}$

(adj A) B = $\begin{bmatrix}0 & 0 & 0 \\ - 3 & 3 & 3 \\ - 3 & 3 & 3\end{bmatrix}\begin{bmatrix}3 \\ 2 \\ 1\end{bmatrix}$

= $\begin{bmatrix}0 \\ - 9 + 6 + 3 \\ - 9 + 6 + 3\end{bmatrix}$

= $\begin{bmatrix}0 \\ 0 \\ 0\end{bmatrix}$

Therefore, the system is consistent and has infinitely many solutions.

Hence proved.

(v) x + y + z = 6

x + 2y + 3z = 14

x + 4y + 7z = 30

解决方案：

Here,

x + y + z = 6

x + 2y + 3z = 14

x + 4y + 7z = 30

A = $\begin{bmatrix}1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 4 & 7\end{bmatrix}$, X = $\begin{bmatrix}x \\ y \\ z\end{bmatrix}$ and B = $\begin{bmatrix}6 \\ 14 \\ 30\end{bmatrix}$

$\begin{bmatrix}1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 4 & 7\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \begin{bmatrix}6 \\ 14 \\ 30\end{bmatrix}$

|A| = $\begin{vmatrix}1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 4 & 7\end{vmatrix}$

= 2 – 4 + 2

= 0

Let C_ij be the cofactors of the elements a_ij in A.

$C_{11} = \left( - 1 \right)^{1 + 1} \begin{vmatrix}2 & 3 \\ 4 & 7\end{vmatrix} = 2, C_{12} = \left( - 1 \right)^{1 + 2} \begin{vmatrix}1 & 3 \\ 1 & 7\end{vmatrix} = - 4 , C_{13} = \left( - 1 \right)^{1 + 3} \begin{vmatrix}1 & 2 \\ 1 & 4\end{vmatrix} = 2$

$C_{21} = \left( - 1 \right)^{2 + 1} \begin{vmatrix}1 & 1 \\ 4 & 7\end{vmatrix} = - 3, C_{22} = \left( - 1 \right)^{2 + 2} \begin{vmatrix}1 & 1 \\ 1 & 7\end{vmatrix} = 6, C_{23} = \left( - 1 \right)^{2 + 3} \begin{vmatrix}1 & 1 \\ 1 & 4\end{vmatrix} = - 3$

$C_{31} = \left( - 1 \right)^{3 + 1} \begin{vmatrix}1 & 1 \\ 2 & 3\end{vmatrix} = 1 , C_{32} = \left( - 1 \right)^{3 + 2} \begin{vmatrix}1 & 1 \\ 1 & 3\end{vmatrix} = - 2 , C_{33} = \left( - 1 \right)^{3 + 3} \begin{vmatrix}1 & 1 \\ 1 & 2\end{vmatrix} = 1$

$adj A = \begin{bmatrix}2 & - 4 & 2 \\ - 3 & 6 & - 3 \\ 1 & - 2 & 1\end{bmatrix}^T$

= $\begin{bmatrix}2 & - 3 & 1 \\ - 4 & 6 & - 2 \\ 2 & - 3 & 1\end{bmatrix}$

$\left( adj A \right)B = \begin{bmatrix}2 & - 3 & 1 \\ - 4 & 6 & - 2 \\ 2 & - 3 & 1\end{bmatrix}\begin{bmatrix}6 \\ 14 \\ 30\end{bmatrix}$

= $\begin{bmatrix}12 - 42 + 30 \\ - 24 + 84 - 60 \\ 12 - 42 + 30\end{bmatrix}$

= $\begin{bmatrix}0 \\ 0 \\ 0\end{bmatrix}$

Therefore, the system is consistent and has infinitely many solutions.

Hence proved.

(vi) 2x + 2y - 2z = 1

4x + 4y - z = 2

6x + 6y + 2z = 3

解决方案：

 Here,

2x + 2y − 2z = 1

4x + 4y − z = 2

6x + 6y + 2z = 3

A = $\begin{bmatrix}2 & 2 & - 2 \\ 4 & 4 & - 1 \\ 6 & 6 & 2\end{bmatrix}$, X = $\begin{bmatrix}x \\ y \\ z\end{bmatrix}$ and B = $\begin{bmatrix}1 \\ 2 \\ 3\end{bmatrix}$

$\begin{bmatrix}2 & 2 & - 2 \\ 4 & 4 & - 1 \\ 6 & 6 & 2\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \begin{bmatrix}1 \\ 2 \\ 3\end{bmatrix}$

|A| = $\begin{vmatrix}2 & 2 & - 2 \\ 4 & 4 & - 1 \\ 6 & 6 & 2\end{vmatrix}$

= 2 (8 + 6) – 2 (8 + 6) – 2 (24 – 24)

= 28 – 28 – 0

= 0

Let C_ij be the cofactors of the elements a_ij in A.

$C_{11} = \left( - 1 \right)^{1 + 1} \begin{vmatrix}4 & - 1 \\ 6 & 2\end{vmatrix} = 14, C_{12} = \left( - 1 \right)^{1 + 2} \begin{vmatrix}4 & - 1 \\ 6 & 2\end{vmatrix} = - 14, C_{13} = \left( - 1 \right)^{1 + 3} \begin{vmatrix}4 & 4 \\ 6 & 6\end{vmatrix} = 0$

$C_{21} = \left( - 1 \right)^{2 + 1} \begin{vmatrix}2 & - 2 \\ 6 & 2\end{vmatrix} = - 16, C_{22} = \left( - 1 \right)^{2 + 2} \begin{vmatrix}2 & - 2 \\ 6 & 2\end{vmatrix} = 16, C_{23} = \left( - 1 \right)^{2 + 3} \begin{vmatrix}2 & 2 \\ 6 & 6\end{vmatrix} = 0$

$C_{31} = \left( - 1 \right)^{3 + 1} \begin{vmatrix}2 & - 2 \\ 4 & - 1\end{vmatrix} = 6, C_{32} = \left( - 1 \right)^{3 + 2} \begin{vmatrix}2 & - 2 \\ 4 & - 1\end{vmatrix} = - 6, C_{33} = \left( - 1 \right)^{3 + 3} \begin{vmatrix}2 & 2 \\ 4 & 4\end{vmatrix} = 0$

$adj A = \begin{bmatrix}14 & - 14 & 0 \\ - 16 & 16 & 0 \\ 6 & - 6 & 0\end{bmatrix}^T$

= $\begin{bmatrix}14 & - 16 & 6 \\ - 14 & 16 & - 6 \\ 0 & 0 & 0\end{bmatrix}$

(adj A) B = $\begin{bmatrix}14 & - 16 & 6 \\ - 14 & 16 & - 6 \\ 0 & 0 & 0\end{bmatrix}\begin{bmatrix}1 \\ 2 \\ 3\end{bmatrix}$

= $\begin{bmatrix}14 - 32 + 18 \\ - 14 + 32 - 18 \\ 0\end{bmatrix}$

= $\begin{bmatrix}0 \\ 0 \\ 0\end{bmatrix}$

Therefore, the system is consistent and has infinitely many solutions.

Hence proved.

问题 4. 证明下列每一个线性方程组是不一致的：

(i) 2x + 5y = 7

6x + 15y = 13

解决方案：

The given system of equations can be expressed as follows:

AX = B

Here,  

A = $\begin{bmatrix}2 & 5 \\ 6 & 15\end{bmatrix}$, X = $\binom{x}{y}$ and B = $\binom{7}{13}$

Now,  

|A| = $\begin{vmatrix}2 & 5 \\ 6 & 15\end{vmatrix}$

= 30 – 30

= 0

Let C_ij be the cofactors of the elements a_ij in A.

C₁₁ = 15, C₁₂ = -6, C₂₁ = -5 and C₂₂ = 2

adj A = $\begin{bmatrix}15 & - 6 \\ - 5 & 2\end{bmatrix}^T$

= $\begin{bmatrix}15 & - 5 \\ - 6 & 2\end{bmatrix}$

(adj A) B = $\begin{bmatrix}15 & - 5 \\ - 6 & 2\end{bmatrix}\binom{7}{13}$

= $\binom{105 - 65}{ - 42 + 26}$

= $\binom{40}{ - 16}$ ≠ 0

Therefore, the given system of equations is inconsistent.

Hence proved.

(ii) 2x + 3y = 5

6x + 9y = 10

解决方案：

The given system of equations can be expressed as follows:

AX = B

Here,

A = $\begin{bmatrix}2 & 3 \\ 6 & 9\end{bmatrix}$, X = $\binom{x}{y}$ and B = $\binom{5}{10}$

|A| = $\begin{vmatrix}2 & 3 \\ 6 & 9\end{vmatrix}$

= 18 – 18

= 0

Let C_ij be the cofactors of the elements a_ij in A.

C₁₁ = 9, C₁₂ = -6, C₂₁ = -3 and C₂₂ = 2

$adj A = \begin{bmatrix}9 & - 6 \\ - 3 & 2\end{bmatrix}^T$

= $\begin{bmatrix}9 & - 3 \\ - 6 & 2\end{bmatrix}$

(adj A) B = $\begin{bmatrix}9 & - 3 \\ - 6 & 2\end{bmatrix}\binom{5}{10}$

= $\binom{45 - 30}{ - 30 + 20}$

= $\binom{15}{ - 10}$ ≠ 0

Therefore, the given system of equations is inconsistent.

Hence proved.

(iii) 4x - 2y = 3

6x - 3y = 5

解决方案：

The given system of equations can be expressed as,

AX = B

Here,

A = $\begin{bmatrix}4 & - 2 \\ 6 & - 3\end{bmatrix}$, X = $\binom{x}{y}$ and B = $\binom{3}{5}$

|A| = $\begin{vmatrix}4 & - 2 \\ 6 & - 3\end{vmatrix}$

= 12 – 12

= 0

Let C_ij be the cofactors of the elements a_ij in A.

C₁₁ = -3, C₁₂ = -6, C₂₁ = 2 and C₂₂ = 4

$adj A = \begin{bmatrix}- 3 & - 6 \\ 2 & 4\end{bmatrix}^T$

= $\begin{bmatrix}- 3 & 2 \\ - 6 & 4\end{bmatrix}$

(adj A) B = $\begin{bmatrix}- 3 & 2 \\ - 6 & 4\end{bmatrix}\binom{3}{5}$

= $\binom{ - 9 + 10}{ - 18 + 20}$

= $\binom{1}{2}$ ≠ 0

Therefore, the given system of equations is inconsistent.

Hence proved.

(iv) 4x - 5y - 2z = 2

5x - 4y + 2z = -2

2x + 2y + 8z = -1

解决方案：

The given system of equations can be written as,

AX = B  

Here,

A = $\begin{bmatrix}4 & - 5 & - 2 \\ 5 & - 4 & 2 \\ 2 & 2 & 8\end{bmatrix}$, X = $\begin{bmatrix}x \\ y \\ z\end{bmatrix}$ and B = $\begin{bmatrix}2 \\ - 2 \\ - 1\end{bmatrix}$

|A| = $\begin{vmatrix}4 & - 5 & - 2 \\ 5 & - 4 & 2 \\ 2 & 2 & 8\end{vmatrix}$

= -144 + 180 – 36

= 0

Let C_ij be the cofactors of the elements a_ij in A.

$C_{11} = \left( - 1 \right)^{1 + 1} \begin{vmatrix}- 4 & 2 \\ 2 & 8\end{vmatrix} = 28, C_{12} = \left( - 1 \right)^{1 + 2} \begin{vmatrix}5 & 2 \\ 2 & 8\end{vmatrix} = - 36, C_{13} = \left( - 1 \right)^{1 + 3} \begin{vmatrix}5 & - 4 \\ 2 & 2\end{vmatrix} = 18$

$C_{21} = \left( - 1 \right)^{2 + 1} \begin{vmatrix}- 5 & - 2 \\ 2 & 8\end{vmatrix} = 36 , C_{22} = \left( - 1 \right)^{2 + 2} \begin{vmatrix}4 & - 2 \\ 2 & 8\end{vmatrix} = 36 , C_{23} = \left( - 1 \right)^{2 + 3} \begin{vmatrix}4 & - 5 \\ 2 & 2\end{vmatrix} = - 18$

$C_{31} = \left( - 1 \right)^{3 + 1} \begin{vmatrix}- 5 & - 2 \\ - 4 & 2\end{vmatrix} = - 18, C_{32} = \left( - 1 \right)^{3 + 2} \begin{vmatrix}4 & - 2 \\ 5 & 2\end{vmatrix} = - 18, C_{33} = \left( - 1 \right)^{3 + 3} \begin{vmatrix}4 & - 5 \\ 5 & - 4\end{vmatrix} = 9$

$adj A = \begin{bmatrix}28 & - 36 & 18 \\ 36 & 36 & - 18 \\ - 18 & - 18 & 9\end{bmatrix}^T$

= $\begin{bmatrix}28 & 36 & - 18 \\ - 36 & 36 & - 18 \\ 18 & - 18 & 9\end{bmatrix}$

(adj A) B = $\begin{bmatrix}28 & 36 & - 18 \\ - 36 & 36 & - 18 \\ 18 & - 18 & 9\end{bmatrix}\begin{bmatrix}2 \\ - 2 \\ - 1\end{bmatrix}$

= $\begin{bmatrix}56 - 72 + 18 \\ - 72 - 72 + 18 \\ 36 + 36 - 9\end{bmatrix}$

= $\begin{bmatrix}2 \\ - 126 \\ 63\end{bmatrix}$ ≠ 0

Therefore, the given system of equations is inconsistent.

Hence proved.

(v) 3x - y - 2z = 2

2y - z = -1

3x - 5y = 3

解决方案：

The given system of equations can be written as,

AX = B

Here,

A = $\begin{bmatrix}3 & - 1 & - 2 \\ 0 & 2 & - 1 \\ 3 & - 5 & 0\end{bmatrix}$, X = $\begin{bmatrix}x \\ y \\ z\end{bmatrix}$ and B = $\begin{bmatrix}2 \\ - 1 \\ 3\end{bmatrix}$

|A| = $\begin{vmatrix}3 & - 1 & - 2 \\ 0 & 2 & - 1 \\ 3 & - 5 & 0\end{vmatrix}$

= -15 + 3 + 12

= 0

Let C_ij be the cofactors of the elements a_ij in A.

$C_{11} = \left( - 1 \right)^{1 + 1} \begin{vmatrix}2 & - 1 \\ - 5 & 0\end{vmatrix} = - 5, C_{12} = \left( - 1 \right)^{1 + 2} \begin{vmatrix}0 & - 1 \\ 3 & 0\end{vmatrix} = - 3, C_{13} = \left( - 1 \right)^{1 + 3} \begin{vmatrix}0 & 2 \\ 3 & - 5\end{vmatrix} = - 6$

$C_{21} = \left( - 1 \right)^{2 + 1} \begin{vmatrix}- 1 & - 2 \\ - 5 & 0\end{vmatrix} = 10, C_{22} = \left( - 1 \right)^{2 + 2} \begin{vmatrix}3 & - 2 \\ 3 & 0\end{vmatrix} = 6, C_{23} = \left( - 1 \right)^{2 + 3} \begin{vmatrix}3 & - 1 \\ 3 & - 5\end{vmatrix} = 12$

$C_{31} = \left( - 1 \right)^{3 + 1} \begin{vmatrix}- 1 & - 2 \\ 2 & - 1\end{vmatrix} = 5, C_{32} = \left( - 1 \right)^{3 + 2} \begin{vmatrix}3 & - 2 \\ 0 & - 1\end{vmatrix} = 3, C_{33} = \left( - 1 \right)^{3 + 3} \begin{vmatrix}3 & - 1 \\ 0 & 2\end{vmatrix} = 6$

adj A = $\begin{bmatrix}- 5 & - 3 & - 6 \\ 10 & 6 & 12 \\ 5 & 3 & 6\end{bmatrix}^T$

= $\begin{bmatrix}- 5 & 10 & 5 \\ - 3 & 6 & 3 \\ - 6 & 12 & 6\end{bmatrix}$

(adj A) B = $\begin{bmatrix}- 5 & 10 & 5 \\ - 3 & 6 & 3 \\ - 6 & 12 & 6\end{bmatrix}\begin{bmatrix}2 \\ - 1 \\ 3\end{bmatrix}$

= $\begin{bmatrix}- 10 - 10 + 15 \\ - 6 - 6 + 9 \\ - 12 - 12 + 18\end{bmatrix}$

= $\begin{bmatrix}- 5 \\ - 3 \\ - 6\end{bmatrix}$ ≠ 0

Therefore, the given system of equations is inconsistent.

Hence proved.

(vi) x + y - 2z = 5

x - 2y + z = -2

−2x + y + z = 4

解决方案：

The given system of equations can be written as,

AX = B

Here,  

A = $\begin{bmatrix}1 & 1 & - 2 \\ 1 & - 2 & 1 \\ - 2 & 1 & 1\end{bmatrix}$, X = $\begin{bmatrix}x \\ y \\ z\end{bmatrix}$ and B = $\begin{bmatrix}5 \\ - 2 \\ 4\end{bmatrix}$

|A| = $\begin{vmatrix}1 & 1 & - 2 \\ 1 & - 2 & 1 \\ - 2 & 1 & 1\end{vmatrix}$

= – 3 – 3 + 6

= 0

Let C_ij be the cofactors of the elements a_ij in A.

$C_{11} = \left( - 1 \right)^{1 + 1} \begin{vmatrix}- 2 & 1 \\ 1 & 1\end{vmatrix} = - 3 , C_{12} = \left( - 1 \right)^{1 + 2} \begin{vmatrix}1 & 1 \\ - 2 & 1\end{vmatrix} = - 3, C_{13} = \left( - 1 \right)^{1 + 3} \begin{vmatrix}1 & - 2 \\ - 2 & 1\end{vmatrix} = - 3$

$C_{21} = \left( - 1 \right)^{2 + 1} \begin{vmatrix}1 & - 2 \\ 1 & 1\end{vmatrix} = - 3, C_{22} = \left( - 1 \right)^{2 + 2} \begin{vmatrix}1 & - 2 \\ - 2 & 1\end{vmatrix} = - 3 , C_{23} = \left( - 1 \right)^{2 + 3} \begin{vmatrix}1 & 1 \\ - 2 & 1\end{vmatrix} = - 3$

$C_{31} = \left( - 1 \right)^{3 + 1} \begin{vmatrix}1 & - 2 \\ - 2 & 1\end{vmatrix} = - 3, C_{32} = \left( - 1 \right)^{3 + 2} \begin{vmatrix}1 & - 2 \\ 1 & 1\end{vmatrix} = - 3, C_{33} = \left( - 1 \right)^{3 + 3} \begin{vmatrix}1 & 1 \\ 1 & - 2\end{vmatrix} = - 3$

$adj A = \begin{bmatrix}- 3 & - 3 & - 3 \\ - 3 & - 3 & - 3 \\ - 3 & - 3 & - 3\end{bmatrix}^T$

= $\begin{bmatrix}- 3 & - 3 & - 3 \\ - 3 & - 3 & - 3 \\ - 3 & - 3 & - 3\end{bmatrix}$

(adj A) B = $\begin{bmatrix}- 3 & - 3 & - 3 \\ - 3 & - 3 & - 3 \\ - 3 & - 3 & - 3\end{bmatrix}\begin{bmatrix}5 \\ - 2 \\ 4\end{bmatrix}$

= $\begin{bmatrix}- 15 + 6 - 12 \\ - 15 + 6 - 12 \\ - 15 + 6 - 12\end{bmatrix}$

= $\begin{bmatrix}- 21 \\ - 21 \\ - 21\end{bmatrix}$ ≠ 0

Therefore, the given system of equations is inconsistent.

Hence proved.

问题 5. 如果 A = $\begin{bmatrix}1 & - 1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2\end{bmatrix}$和 B = $\begin{bmatrix}2 & 2 & - 4 \\ - 4 & 2 & - 4 \\ 2 & - 1 & 5\end{bmatrix}$是两个方阵，求 AB 并求解线性方程组：x - y = 3, 2x + 3y + 4z = 17, y + 2z = 7。

解决方案：

Here,

A = $\begin{bmatrix}1 & - 1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2\end{bmatrix}$ and B = $\begin{bmatrix}2 & 2 & - 4 \\ - 4 & 2 & - 4 \\ 2 & - 1 & 5\end{bmatrix}$

Now,

AB = $\begin{bmatrix}1 & - 1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2\end{bmatrix} \begin{bmatrix}2 & 2 & - 4 \\ - 4 & 2 & - 4 \\ 2 & - 1 & 5\end{bmatrix}$

AB = $\begin{bmatrix}6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6\end{bmatrix}$

AB = $6\begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}$

AB = 6I

$\frac{1}{6}AB$ = I

$A^{- 1} = \frac{1}{6}B$

$A^{- 1} = \frac{1}{6}\begin{bmatrix}2 & 2 & - 4 \\ - 4 & 2 & - 4 \\ 2 & - 1 & 5\end{bmatrix}$

X = A^-1 B

X = $\frac{1}{6}\begin{bmatrix}2 & 2 & - 4 \\ - 4 & 2 & - 4 \\ 2 & - 1 & 5\end{bmatrix}\begin{bmatrix}3 \\ 17 \\ 7\end{bmatrix}$

$\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{6}\begin{bmatrix}6 + 34 - 28 \\ - 12 + 34 - 28 \\ 6 - 17 + 35\end{bmatrix}$

$\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{6}\begin{bmatrix}12 \\ - 6 \\ 24\end{bmatrix}$

Therefore x = 2, y = -1 and z = 4.

问题 6. 如果 A = $\begin{bmatrix}2 & - 3 & 5 \\ 3 & 2 & - 4 \\ 1 & 1 & - 2\end{bmatrix}$ ，找到 A ^-1并因此求解线性方程组 2x - 3y + 5z = 11, 3x + 2y - 4z = -5, x + y + 2z = -3。

解决方案：

Here,

A = $\begin{bmatrix}2 & - 3 & 5 \\ 3 & 2 & - 4 \\ 1 & 1 & - 2\end{bmatrix}$

|A| = $\begin{vmatrix}2 & - 3 & 5 \\ 3 & 2 & - 4 \\ 1 & 1 & - 2\end{vmatrix}$

= 0 – 6 + 5

= -1

Let C_ij be the cofactors of the elements a_ij in A.

$C_{11} = \left( - 1 \right)^{1 + 1} \begin{vmatrix}2 & - 4 \\ 1 & - 2\end{vmatrix} = 0, C_{12} = \left( - 1 \right)^{1 + 2} \begin{vmatrix}3 & - 4 \\ 1 & - 2\end{vmatrix} = 2, C_{13} = \left( - 1 \right)^{1 + 3} \begin{vmatrix}3 & 2 \\ 1 & 1\end{vmatrix} = 1$

$C_{21} = \left( - 1 \right)^{2 + 1} \begin{vmatrix}- 3 & 5 \\ 1 & - 2\end{vmatrix} = - 1, C_{22} = \left( - 1 \right)^{2 + 2} \begin{vmatrix}2 & 5 \\ 1 & - 2\end{vmatrix} = - 9, C_{23} = \left( - 1 \right)^{2 + 3} \begin{vmatrix}2 & - 3 \\ 1 & 1\end{vmatrix} = - 5$

$C_{31} = \left( - 1 \right)^{3 + 1} \begin{vmatrix}- 3 & 5 \\ 2 & - 4\end{vmatrix} = 2 , C_{32} = \left( - 1 \right)^{3 + 2} \begin{vmatrix}2 & 5 \\ 3 & - 4\end{vmatrix} = 23, C_{33} = \left( - 1 \right)^{3 + 3} \begin{vmatrix}2 & - 3 \\ 3 & 2\end{vmatrix} = 13$

$adj A = \begin{bmatrix}0 & 2 & 1 \\ - 1 & - 9 & - 5 \\ 2 & 23 & 13\end{bmatrix}^T$

= $\begin{bmatrix}0 & - 1 & 2 \\ 2 & - 9 & 23 \\ 1 & - 5 & 13\end{bmatrix}$

$A^{- 1} = \frac{1}{\left| A \right|}adj A$

= $\frac{1}{- 1}\begin{bmatrix}0 & - 1 & 2 \\ 2 & - 9 & 23 \\ 1 & - 5 & 13\end{bmatrix}$

$\begin{bmatrix}2 & - 3 & 5 \\ 3 & 2 & - 4 \\ 1 & 1 & - 2\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \begin{bmatrix}11 \\ - 5 \\ - 3\end{bmatrix}$

X = A^-1 B

$\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{- 1}\begin{bmatrix}0 & - 1 & 2 \\ 2 & - 9 & 23 \\ 1 & - 5 & 13\end{bmatrix}\begin{bmatrix}11 \\ - 5 \\ - 3\end{bmatrix}$

$\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{- 1}\begin{bmatrix}0 + 5 - 6 \\ 22 + 45 - 69 \\ 11 + 25 - 39\end{bmatrix}$

$\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{- 1}\begin{bmatrix}- 1 \\ - 2 \\ - 3\end{bmatrix}$

=> x = – 1/- 1, y = -2/-1\ and z = -3/-1

Therefore x = 1, y = 2 and z = 3.

问题 7. 求 A ⁻¹ ，如果 A = $\begin{bmatrix}1 & 2 & 5 \\ 1 & - 1 & - 1 \\ 2 & 3 & - 1\end{bmatrix}$ .因此求解以下线性方程组：x + 2y + 5z = 10, x - y - z = -2, 2x + 3y - z = -11。

解决方案：

A = $\begin{bmatrix}1 & 2 & 5 \\ 1 & - 1 & - 1 \\ 2 & 3 & - 1\end{bmatrix}$

|A| = $\begin{vmatrix}1 & 2 & 5 \\ 1 & - 1 & - 1 \\ 2 & 3 & - 1\end{vmatrix}$

= 4 – 2 + 25

= 27

Let C_ij be the cofactors of the elements a_ij in A.

$C_{11} = \left( - 1 \right)^{1 + 1} \begin{vmatrix}- 1 & - 1 \\ 3 & - 1\end{vmatrix} = 4, C_{12} = \left( - 1 \right)^{1 + 2} \begin{vmatrix}1 & - 1 \\ 2 & - 1\end{vmatrix} = - 1, C_{13} = \left( - 1 \right)^{1 + 3} \begin{vmatrix}1 & - 1 \\ - 2 & 3\end{vmatrix} = 5$

$C_{21} = \left( - 1 \right)^{2 + 1} \begin{vmatrix}2 & 5 \\ 3 & - 1\end{vmatrix} = 17, C_{22} = \left( - 1 \right)^{2 + 2} \begin{vmatrix}1 & 5 \\ 2 & - 1\end{vmatrix} = - 11, C_{23} = \left( - 1 \right)^{2 + 3} \begin{vmatrix}1 & 2 \\ 2 & 3\end{vmatrix} = 1$

$C_{31} = \left( - 1 \right)^{3 + 1} \begin{vmatrix}2 & 5 \\ - 1 & - 1\end{vmatrix} = 3, C_{32} = \left( - 1 \right)^{3 + 2} \begin{vmatrix}1 & 5 \\ 1 & - 1\end{vmatrix} = 6, C_{33} = \left( - 1 \right)^{3 + 3} \begin{vmatrix}1 & 2 \\ 1 & - 1\end{vmatrix} = - 3$

$adj A = \begin{bmatrix}4 & - 1 & 5 \\ 17 & - 11 & 1 \\ 3 & 6 & - 3\end{bmatrix}^T$

= $\begin{bmatrix}4 & 17 & 3 \\ - 1 & - 11 & 6 \\ 5 & 1 & - 3\end{bmatrix}$

$A^{- 1} = \frac{1}{\left| A \right|}adj A$

= $\frac{1}{27}\begin{bmatrix}4 & 17 & 3 \\ - 1 & - 11 & 6 \\ 5 & 1 & - 3\end{bmatrix}$

$\begin{bmatrix}1 & 2 & 5 \\ 1 & - 1 & - 1 \\ 2 & 3 & - 1\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \begin{bmatrix}10 \\ - 2 \\ - 11\end{bmatrix}$

X = A^-1 B

$\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{27}\begin{bmatrix}4 & 17 & 3 \\ - 1 & - 11 & 6 \\ 5 & 1 & - 3\end{bmatrix}\begin{bmatrix}10 \\ - 2 \\ - 11\end{bmatrix}$

$\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{27}\begin{bmatrix}40 - 34 - 33 \\ - 10 + 22 - 66 \\ 50 - 2 + 33\end{bmatrix}$

$\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \frac{1}{27}\begin{bmatrix}- 27 \\ - 54 \\ 81\end{bmatrix}$

=> x = -27/27, y = -54/27 and z = 81/27

Therefore, x = – 1, y = -2 and z = 3.