第 12 类 RD Sharma 解决方案 - 第 6 章行列式 - 练习 6.2 |设置 1

问题 1. 评估以下决定因素：

（一世） $\begin{vmatrix} 1 & 3 &5 \\ 2 & 6 & 10 \\ 31 & 11 & 38 \end{vmatrix}$

解决方案：

Considering the determinant, we have

$\triangle = \begin{vmatrix} 1 & 3 &5 \\ 2 & 6 & 10 \\ 31 & 11 & 38 \end{vmatrix} \\ \triangle = 2\begin{vmatrix} 1 & 3 &5 \\ 1 & 3 & 5 \\ 31 & 11 & 38 \end{vmatrix}$

As R1 and R2 are identical

Hence, △ = 0

(二) $\begin{vmatrix} 67 & 19 &21 \\ 39 & 13 & 14 \\ 81 & 24 & 26 \end{vmatrix}$

解决方案：

Considering the determinant, we have

$\triangle = \begin{vmatrix} 67 & 19 &21 \\ 39 & 13 & 14 \\ 81 & 24 & 26 \end{vmatrix}$

C1⇢C1 – 3C3

$\triangle = \begin{vmatrix} 4 & 19 &21 \\ -3 & 13 & 14 \\ 3 & 24 & 26 \end{vmatrix}$

R3⇢R3 + R2 and R1⇢R1 + R2

$\triangle = \begin{vmatrix} 1 & 32 &35 \\ -3 & 13 & 14 \\ 0 & 37 & 40 \end{vmatrix}$

R2⇢R2 + 3R1

$\triangle = \begin{vmatrix} 1 & 32 &35 \\ 0 & 109 & 119 \\ 0 & 37 & 40 \end{vmatrix}$

△ = 1(109 × 40 – 119 × 37)

Hence, △ = -43

㈢ $\begin{vmatrix} a & h & g \\ h & b & f \\ g & f & c \end{vmatrix}$

解决方案：

Considering the determinant, we have

$\triangle = \begin{vmatrix} a & h & g \\ h & b & f \\ g & f & c \end{vmatrix}\\ \triangle = a\begin{vmatrix} b & f \\ f & c \end{vmatrix}-h\begin{vmatrix} h & f \\ g & c \end{vmatrix}+g\begin{vmatrix} h & b \\ g & f \end{vmatrix}$

△ = a(bc – f²) – h(hc – fg) + g(hf – gb)

△ = abc – af² – h²c + fgh + fgh – g²b

Hence, △ = abc + 2fgh – af² – ch² – bg²

(四) $\begin{vmatrix} 1 & -3 & 2 \\ 4 & -1 & 2 \\ 3 & 5 & 2 \end{vmatrix}$

解决方案：

Considering the determinant, we have

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

$\triangle = 1\begin{vmatrix} -1 & 2 \\ 5 & 2 \end{vmatrix}-(-3)\begin{vmatrix} 4 & 2 \\ 3 & 2 \end{vmatrix}+2\begin{vmatrix} 4 & -1 \\ 3 & 5 \end{vmatrix}$

△ = 1(-2 – 10) + 3(8 – 6) + 2(20 + 3)

△ = 1(-12) + 3(2) + 2(23)

△ = -12 + 6 + 46

Hence, △ = 40

(五) $\begin{vmatrix} 1 & 4 & 9 \\ 4 & 9 &16 \\ 9 & 16 & 25 \end{vmatrix}$

解决方案：

Considering the determinant, we have

$\triangle = \begin{vmatrix} 1 & 4 & 9 \\ 4 & 9 &16 \\ 9 & 16 & 25 \end{vmatrix}\\ \triangle = 1\begin{vmatrix} 9 & 16 \\ 16 & 25 \end{vmatrix}-4\begin{vmatrix} 4 & 16 \\ 9 & 25 \end{vmatrix}+9\begin{vmatrix} 4 & 9 \\ 9 & 16 \end{vmatrix}$

△ = 1(225-256) + 4(100-144) + 9(64-81)

△ = 1(-31) – 4(-44) + 9(-17)

△ = -31 + 176 – 153

Hence, △ = -8

(六) $\begin{vmatrix} 6 & -3 & 2 \\ 2 & -1 & 2 \\ -10 & 5 & 2 \end{vmatrix}$

解决方案：

Considering the determinant, we have

$\triangle = \begin{vmatrix} 6 & -3 & 2 \\ 2 & -1 & 2 \\ -10 & 5 & 2 \end{vmatrix}$

Taking -2 common from C1, C2 and C3

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

As C1 and C2 are identical

Hence, △ = 0

(七) $\begin{vmatrix} 1 & 3 & 9 & 27 \\ 3 & 9 & 27 & 1 \\ 9 & 27 & 1 & 3 \\ 27 & 1 & 3 & 9 \end{vmatrix}$

解决方案：

Considering the determinant, we have

$\triangle = \begin{vmatrix} 1 & 3 & 9 & 27 \\ 3 & 9 & 27 & 1 \\ 9 & 27 & 1 & 3 \\ 27 & 1 & 3 & 9 \end{vmatrix}\\ \triangle = \begin{vmatrix} 1 & 3 & 3^2 & 3^3 \\ 3 & 3^2 & 3^3 & 1 \\ 3^2 & 3^3 & 1 & 3 \\ 3^3 & 1 & 3 & 3^2 \end{vmatrix}$

C1⇢C1 + C2 + C3

$\triangle = \begin{vmatrix} 1+3 + 3^2 + 3^3 & 3 & 3^2 & 3^3 \\ 1+3 + 3^2 + 3^3 & 3^2 & 3^3 & 1 \\ 1+3 + 3^2 + 3^3 & 3^3 & 1 & 3 \\ 1+3 + 3^2 + 3^3 & 1 & 3 & 3^2 \end{vmatrix}\\ \triangle = 1+3 + 3^2 + 3^3\begin{vmatrix} 1 & 3 & 3^2 & 3^3 \\ 1 & 3^2 & 3^3 & 1 \\ 1& 3^3 & 1 & 3 \\ 1 & 1 & 3 & 3^2 \end{vmatrix}\\$

C2⇢C2 – C1

C3⇢C3 – C1

C4⇢C4 – C1

$\triangle = 1+3 + 3^2 + 3^3\begin{vmatrix} 1 & 3 & 3^2 & 3^3 \\ 0 & 3^2-3 & 3^3-3^2 & 1-3^3 \\ 0 & 3^3-3 & 1-3^2 & 3-3^3 \\ 0 & 1-3 & 3-3^2 & 3^2-3^3 \end{vmatrix}\\ \triangle = (1+3 + 3^2 + 3^3)1\begin{vmatrix} 3^2-3 & 3^3-3^2 & 1-3^3 \\ 3^3-3 & 1-3^2 & 3-3^3 \\ 1-3 & 3-3^2 & 3^2-3^3 \end{vmatrix}\\ \triangle = 1+3 + 3^2 + 3^3\begin{vmatrix} 6 & 18 & -26 \\ 24 & -8 & -24 \\ -2 & -6 & -18 \end{vmatrix}\\$

Taking 2, -2 and -2 common from C1, C2 and C3

$\triangle = (1+3 + 3^2 + 3^3)(2^3)\begin{vmatrix} 3 & -9 & 13\\ 12 & 4 & 12\\ -1 & 3 & 9\end{vmatrix}\\$

Taking 4 common from R2 and R1⇢R1+3R3

$\triangle = (1+3 + 3^2 + 3^3)(2^2)(2^3)\begin{vmatrix} 0 & 0 & 40\\ 3 & 1 & 3\\ -1 & 3 & 9\end{vmatrix}\\$

△ = (1 + 3 + 32 + 33)(4)(8)[40(9 – (-1))]

△ = (40)(4)(8)[40(9 + 1)]

△ = 40 × 4 × 8 × 40 × 10

Hence, △ = 512000

(八) $\begin{vmatrix} 102 & 18 & 36 \\ 1 & 3 & 4 \\ 17 & 3 & 6 \end{vmatrix}$

解决方案：

Considering the determinant, we have

$\triangle = \begin{vmatrix} 102 & 18 & 36 \\ 1 & 3 & 4 \\ 17 & 3 & 6 \end{vmatrix}$

Taking 6 common from R1, we get

$\triangle = 6\begin{vmatrix} 17 & 3 & 6 \\ 1 & 3 & 4 \\ 17 & 3 & 6 \end{vmatrix}$

As R1 and R3 are identical

Hence, △ = 0

问题 2. 在不展开的情况下，证明以下每个行列式的值都为零：

（一世） $\begin{vmatrix} 8 & 2 & 7 \\ 12 & 3 & 5 \\ 16 & 4 & 3 \end{vmatrix}$

解决方案：

Considering the determinant, we have

$\triangle = \begin{vmatrix} 8 & 2 & 7 \\ 12 & 3 & 5 \\ 16 & 4 & 3 \end{vmatrix}$

Taking 4 common from C1, we get

$\triangle = 4\begin{vmatrix} 2 & 2 & 7 \\ 3 & 3 & 5 \\ 4 & 4 & 3 \end{vmatrix}$

As C1 and C2 are identical

Hence, △ = 0

(二) $\begin{vmatrix} 6 & -3 & 2 \\ 2 & -1 & 2 \\ -10 & 5 & 2 \end{vmatrix}$

解决方案：

Considering the determinant, we have

$\triangle = \begin{vmatrix} 6 & -3 & 2 \\ 2 & -1 & 2 \\ -10 & 5 & 2 \end{vmatrix}$

Taking -2 common from C1, we get

$\triangle = (-2)\begin{vmatrix} -3 & -3 & 2 \\ -1 & -1 & 2 \\ 5 & 5 & 2 \end{vmatrix}$

As C1 and C2 are identical

Hence, △ = 0

㈢ $\begin{vmatrix} 2 & 3 & 7 \\ 13 & 17 & 5 \\ 15 & 20 & 12 \end{vmatrix}$

解决方案：

Considering the determinant, we have

$\triangle = \begin{vmatrix} 2 & 3 & 7 \\ 13 & 17 & 5 \\ 15 & 20 & 12 \end{vmatrix}$

R3⇢R3 – R2

$\triangle = \begin{vmatrix} 2 & 3 & 7 \\ 13 & 17 & 5 \\ 2 & 3 & 7 \end{vmatrix}$

As R1 and R3 are identical

Hence, △ = 0

(四) $\begin{vmatrix} \frac{1}{a} & a^2 & bc \\ \frac{1}{b} & b^2 & ac \\ \frac{1}{c} & c^2 & ab \end{vmatrix}$

解决方案：

Considering the determinant, we have

$\triangle = \begin{vmatrix} \frac{1}{a} & a^2 & bc \\ \frac{1}{b} & b^2 & ac \\ \frac{1}{c} & c^2 & ab \end{vmatrix}$

Multiplying and dividing △ by abc, we get

$\triangle = \frac{abc}{abc}\begin{vmatrix} \frac{1}{a} & a^2 & bc \\ \frac{1}{b} & b^2 & ac \\ \frac{1}{c} & c^2 & ab \end{vmatrix}\\$

Multiplying R1, R2 and R3 by a, b and c respectively

$\triangle = \frac{1}{abc}\begin{vmatrix} 1 & a^3 & abc \\ 1 & b^3 & abc \\ 1 & c^3 & abc \end{vmatrix}$

Taking abc common from C3, we get

$\triangle = \frac{abc}{abc}\begin{vmatrix} 1 & a^3 & 1 \\ 1 & b^3 & 1 \\ 1 & c^3 & 1 \end{vmatrix}$

As C2 and C3 are identical

Hence, △ = 0

(五) $\begin{vmatrix} a+b & 2a+b & 3a+b \\ 2a+b & 3a+b & 4a+b \\ 4a+b & 5a+b & 6a+b \end{vmatrix}$

解决方案：

Considering the determinant, we have

$\triangle = \begin{vmatrix} a+b & 2a+b & 3a+b \\ 2a+b & 3a+b & 4a+b \\ 4a+b & 5a+b & 6a+b \end{vmatrix}$

C3⇢C3 – C2 and C2⇢C2 – C1

$\triangle = \begin{vmatrix} a+b & a & a \\ 2a+b & a & a \\ 4a+b & a & a \end{vmatrix}$

As C2 and C3 are identical

Hence, △ = 0

(六) $\begin{vmatrix} 1 & a & a^2-bc \\ 1 & b & b^2-ac \\ 1 & c & c^2-ab \end{vmatrix}$

解决方案：

Considering the determinant, we have

$\triangle = \begin{vmatrix} 1 & a & a^2-bc \\ 1 & b & b^2-ac \\ 1 & c & c^2-ab \end{vmatrix}$

Splitting the determinant, we have

$\triangle = \begin{vmatrix} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{vmatrix}-\begin{vmatrix} 1 & a & bc \\ 1 & b & ac \\ 1 & c & ab \end{vmatrix}$

R2⇢R2-R1 and R3⇢R3-R1

$\triangle = \begin{vmatrix} 1 & a & a^2 \\ 0 & b-a & b^2-a^2 \\ 0 & c-a & c^2-a^2 \end{vmatrix}-\begin{vmatrix} 1 & a & bc \\ 0 & b-a & ac-bc \\ 0 & c-a & ab-bc \end{vmatrix} \\ \triangle = \begin{vmatrix} 1 & a & a^2 \\ 0 & b-a & (b-a)(b+a) \\ 0 & c-a & (c-a)(c+a) \end{vmatrix}-\begin{vmatrix} 1 & a & bc \\ 0 & b-a & -(b-a)c \\ 0 & c-a & -(c-a)b \end{vmatrix}$

Taking (b-a) and (c-a) common from R2 and R3, we have

$\triangle = (b-a)(c-a)\begin{vmatrix} 1 & a & a^2 \\ 0 & 1 & b+a \\ 0 & 1 & c+a \end{vmatrix}-(b-a)(c-a)\begin{vmatrix} 1 & a & bc \\ 0 & 1 & -c \\ 0 & 1 & -b \end{vmatrix}$

△ = (b – a)(c – a)(c + a – (b + a)) – (b – a)(c – a)(-b – (-c))

△ = (b – a)(c – a)(c + a – b – a) – (b – a)(c – a)(-b + c)

△ = (b – a)(c – a)(c – b) – (b – a)(c – a)(c – b)

Hence, △ = 0

(七) $\begin{vmatrix} 49 & 1 & 6 \\ 39 & 7 & 4 \\ 26 & 2 & 3 \end{vmatrix}$

解决方案：

Considering the determinant, we have

$\triangle = \begin{vmatrix} 49 & 1 & 6 \\ 39 & 7 & 4 \\ 26 & 2 & 3 \end{vmatrix}$

C1⇢C1 – 8C3

$\triangle = \begin{vmatrix} 1 & 1 & 6 \\ 7 & 7 & 4 \\ 2 & 2 & 3 \end{vmatrix}$

As C1 and C2 are identical

Hence, △ = 0

(八) $\begin{vmatrix} 0 & x & y \\ -x & 0 & z \\ -y & -z & 0 \end{vmatrix}$

解决方案：

Considering the determinant, we have

$\triangle = \begin{vmatrix} 0 & x & y \\ -x & 0 & z \\ -y & -z & 0 \end{vmatrix}$

Multiplying and dividing by xyz, we have

$\triangle = \frac{xyz}{xyz}\begin{vmatrix} 0 & x & y \\ -x & 0 & z \\ -y & -z & 0 \end{vmatrix}$

Multiplying C1, C2 and C3 by z, y and x respectively

$\triangle = \frac{1}{xyz}\begin{vmatrix} 0 & xy & yx \\ -xz & 0 & zx \\ -yz & -zy & 0 \end{vmatrix}$

Taking y, x and z common in R1, R2 and R3 respectively

$\triangle = \frac{xyz}{xyz}\begin{vmatrix} 0 & x & x \\ -z & 0 & z \\ -y & -y & 0 \end{vmatrix}$

C2⇢C2 – C3

$\triangle = \frac{xyz}{xyz}\begin{vmatrix} 0 & 0 & x \\ -z & -z & z \\ -y & -y & 0 \end{vmatrix}$

As C1 and C2 are identical

Hence, △ = 0

(九) $\begin{vmatrix} 1 & 43 & 6 \\ 7 & 35 & 4 \\ 3 & 17 & 2 \end{vmatrix}$

解决方案：

Considering the determinant, we have

$\triangle = \begin{vmatrix} 1 & 43 & 6 \\ 7 & 35 & 4 \\ 3 & 17 & 2 \end{vmatrix}$

C2⇢C2 – 7C3

$\triangle = \begin{vmatrix} 1 & 1 & 6 \\ 7 & 7 & 4 \\ 3 & 3 & 2 \end{vmatrix}$

As C1 and C2 are identical

Hence, △ = 0

（X） $\begin{vmatrix} 1^2 & 2^2 & 3^2 & 4^2 \\ 2^2 & 3^2 & 4^2 & 5^5 \\ 3^3 & 4^4 & 5^2 & 6^2 \\ 4^2 & 5^2 & 6^2 &7^2 \end{vmatrix}$

解决方案：

Considering the determinant, we have

$\triangle = \begin{vmatrix} 1^2 & 2^2 & 3^2 & 4^2 \\ 2^2 & 3^2 & 4^2 & 5^5 \\ 3^3 & 4^4 & 5^2 & 6^2 \\ 4^2 & 5^2 & 6^2 &7^2 \end{vmatrix}$

C3⇢C3 – C2 and C4⇢C4 – C1

$\triangle = \begin{vmatrix} 1^2 & 2^2 & 3^2-2^2 & 4^2-1^2 \\ 2^2 & 3^2 & 4^2-3^2 & 5^2-2^2 \\ 3^2 & 4^2 & 5^2-4^2 & 6^2-3^2 \\ 4^2 & 5^2 & 6^2-5^2 &7^2-4^2 \end{vmatrix} \\ \triangle = \begin{vmatrix} 1^2 & 2^2 & 5 & 15 \\ 2^2 & 3^2 & 7 & 21 \\ 3^2 & 4^2 & 9 & 27 \\ 4^2 & 5^2 & 11 &33 \end{vmatrix}$

Taking 3 common from C3, we get

$\triangle = 3\begin{vmatrix} 1^2 & 2^2 & 5 & 5 \\ 2^2 & 3^2 & 7 & 7 \\ 3^2 & 4^2 & 9 & 9 \\ 4^2 & 5^2 & 11 &11 \end{vmatrix}$

As C3 and C4 are identical

Hence, △ = 0

(十一) $\begin{vmatrix} a & b & c \\ a+2x & b+2y & c+2z \\ x & y & z \end{vmatrix}$

解决方案：

Considering the determinant, we have

$\triangle = \begin{vmatrix} a & b & c \\ a+2x & b+2y & c+2z \\ x & y & z \end{vmatrix}$

R3⇢R3 + R1 and R2⇢R2 + R1

$\triangle = \begin{vmatrix} a & b & c \\ a+a+2x & b+b+2y & c+c+2z \\ a+x & b+y & c+z \end{vmatrix} \\ \triangle = \begin{vmatrix} a & b & c \\ 2a+2x & 2b+2y & 2c+2z \\ a+x & b+y & c+z \end{vmatrix}$

Taking 2 common from R2, we get

$\triangle = 2\begin{vmatrix} a & b & c \\ a+x & b+y & c+z \\ a+x & b+y & c+z \end{vmatrix}$

As R2 and R3 are identical

Hence, △ = 0

问题 3。 $\begin{vmatrix} a & b+c & a^2 \\ b & c+a & b^2 \\ c & a+b & c^2 \end{vmatrix}$

解决方案：

Considering the determinant, we have

$\triangle = \begin{vmatrix} a & b+c & a^2 \\ b & c+a & b^2 \\ c & a+b & c^2 \end{vmatrix}$

C2⇢C2+C1

$\triangle = \begin{vmatrix} a & b+c+a & a^2 \\ b & c+a+b & b^2 \\ c & a+b+c & c^2 \end{vmatrix}$

Taking (a+b+c) common from C2, we get

$\triangle = (a+b+c)\begin{vmatrix} a & 1 & a^2 \\ b & 1 & b^2 \\ c & 1 & c^2 \end{vmatrix}$

R3⇢R3-R1 and R2⇢R2-R1

$\triangle = (a+b+c)\begin{vmatrix} a & 1 & a^2 \\ b-a & 0 & b^2-a^2 \\ c-a & 0 & c^2-a^2 \end{vmatrix} \\ \triangle = (a+b+c)\begin{vmatrix} a & 1 & a^2 \\ b-a & 0 & (b-a)(b+a) \\ c-a & 0 & (c-a)(c+a) \end{vmatrix}$

Taking (b – a) and (c – a) from R2 and R3, we have

$\triangle = (a+b+c)\begin{vmatrix} a & 1 & a^2 \\ b-a & 0 & b^2-a^2 \\ c-a & 0 & c^2-a^2 \end{vmatrix} \\ \triangle = (a+b+c)(b-a)(c-a)\begin{vmatrix} a & 1 & a^2 \\ 1 & 0 & b+a \\ 1 & 0 & c+a \end{vmatrix}$

△ = (a + b + c)(b – a)(c – a)[1(b + a – (c + a))]

△ = (a + b + c)(b – a)(c – a)(b + a – c – a)

Hence, △ = (a + b + c)(b – a)(c – a)(b – c)

问题 4。 $\begin{vmatrix} 1 & a & bc \\ 1 & b & ca \\ 1 & c & ab \end{vmatrix}$

解决方案：

Considering the determinant, we have

$\triangle = \begin{vmatrix} 1 & a & bc \\ 1 & b & ca \\ 1 & c & ab \end{vmatrix}$

R3⇢R3-R1 and R2⇢R2-R1

$\triangle = \begin{vmatrix} 1 & a & bc \\ 0 & b-a & ca-bc \\ 0 & c-a & ab-bc \end{vmatrix}\\ \triangle = \begin{vmatrix} 1 & a & bc \\ 0 & b-a & -c(b-a) \\ 0 & c-a & -b(c-a) \end{vmatrix}\\$

Taking (b-a) and (c-a) from R2 and R3, we have

$\triangle = (b-a)(c-a)\begin{vmatrix} 1 & a & bc \\ 0 & 1 & -c \\ 0 & 1 & -b \end{vmatrix}\\$

△ = (b – a)(c – a)[1((1)(-b) – (1)(-c))]

△ = (b – a)(c – a)[-b – (-c)]

△ = (b – a)(c – a)[-b + c]

Hence, △ = (a – b)(b – c)(c – a)

问题 5。 $\begin{vmatrix} x+\lambda & x & x \\ x & x+\lambda & x \\ x & x & x+\lambda \end{vmatrix}$

解决方案：

Considering the determinant, we have

$\triangle = \begin{vmatrix} x+\lambda & x & x \\ x & x+\lambda & x \\ x & x & x+\lambda \end{vmatrix}$

C1⇢C1+C2+C3

$\triangle = \begin{vmatrix} 3x+\lambda & x & x \\ 3x+\lambda & x+\lambda & x \\ 3x+\lambda & x & x+\lambda \end{vmatrix}$

Taking (3x+λ) common from C1, we get

$\triangle = (3x+\lambda)\begin{vmatrix} 1 & x & x \\ 1 & x+\lambda & x \\ 1 & x & x+\lambda \end{vmatrix}$

R3⇢R3-R1 and R2⇢R2-R1

$\triangle = (3x+\lambda)\begin{vmatrix} 1 & x & x \\ 0 & x+\lambda-x & x-x \\ 0 & x-x & x+\lambda-x \end{vmatrix} \\ \triangle = (3x+\lambda)\begin{vmatrix} 1 & x & x \\ 0 & \lambda & 0 \\ 0 & 0 & \lambda \end{vmatrix}$

△ = (3x + λ)[λ(λ(1) – 0)]

△ = (3x + λ)[λ(λ)]

Hence, △ = λ²(3x + λ)

问题 6。 $\begin{vmatrix} a & b & c \\ c & a & b \\ b & c & a \end{vmatrix}$

解决方案：

Considering the determinant, we have

$\triangle = \begin{vmatrix} a & b & c \\ c & a & b \\ b & c & a \end{vmatrix}$

C1⇢C1 + C2 + C3

$\triangle = \begin{vmatrix} a+b+c & b & c \\ c+a+b & a & b \\ b+c+a & c & a \end{vmatrix}$

Taking (a + b + c) common from C1, we get

$\triangle = (a+b+c)\begin{vmatrix} 1 & b & c \\ 1 & a & b \\ 1 & c & a \end{vmatrix}$

R3⇢R3 – R1 and R2⇢R2 – R1

$\triangle = (a+b+c)\begin{vmatrix} 1 & b & c \\ 0 & a-b & b-c \\ 0 & c-b & a-c \end{vmatrix}$

△ = (a + b + c)[1((a – b)(a – c) – (c – b)(b – c))]

△ = (a + b + c)[(a² – ac – ab + bc) – (cb – c² – b² + bc)]

△ = (a + b + c)[a² – ac – ab + bc + c² + b² – 2bc]

Henfce, △ = (a + b + c)[a² + b² + c² – ac – ab – bc]

问题 7。 $\begin{vmatrix} sin\hspace{0.1cm}\alpha & cos\hspace{0.1cm}\alpha & cos(\alpha+\delta) \\ sin\hspace{0.1cm}\beta & cos\hspace{0.1cm}\beta & cos(\beta+\delta) \\ sin\hspace{0.1cm}\gamma & cos\hspace{0.1cm}\gamma & cos(\gamma+\delta) \end{vmatrix}$

解决方案：

Considering the determinant, we have

$\triangle = \begin{vmatrix} sin\hspace{0.1cm}\alpha & cos\hspace{0.1cm}\alpha & cos(\alpha+\delta) \\ sin\hspace{0.1cm}\beta & cos\hspace{0.1cm}\beta & cos(\beta+\delta) \\ sin\hspace{0.1cm}\gamma & cos\hspace{0.1cm}\gamma & cos(\gamma+\delta) \end{vmatrix}$

$\triangle = \begin{vmatrix} sin\hspace{0.1cm}\alpha sin\hspace{0.1cm}\delta & cos\hspace{0.1cm}\alpha cos\hspace{0.1cm}\delta & cos(\alpha+\delta) \\ sin\hspace{0.1cm}\beta sin\hspace{0.1cm}\delta & cos\hspace{0.1cm}\beta cos\hspace{0.1cm}\delta & cos(\beta+\delta) \\ sin\hspace{0.1cm}\gamma sin\hspace{0.1cm}\delta & cos\hspace{0.1cm}\gamma cos\hspace{0.1cm}\delta & cos(\gamma+\delta) \end{vmatrix}$

C2⇢C2 – C1

$\triangle = \begin{vmatrix} sin\hspace{0.1cm}\alpha sin\hspace{0.1cm}\delta & cos\hspace{0.1cm}\alpha cos\hspace{0.1cm}\delta - sin\hspace{0.1cm}\alpha sin\hspace{0.1cm}\delta & cos(\alpha+\delta) \\ sin\hspace{0.1cm}\beta sin\hspace{0.1cm}\delta & cos\hspace{0.1cm}\beta cos\hspace{0.1cm}\delta - sin\hspace{0.1cm}\beta sin\hspace{0.1cm}\delta & cos(\beta+\delta) \\ sin\hspace{0.1cm}\gamma sin\hspace{0.1cm}\delta & cos\hspace{0.1cm}\gamma cos\hspace{0.1cm}\delta - sin\hspace{0.1cm}\gamma sin\hspace{0.1cm}\delta & cos(\gamma+\delta) \end{vmatrix}$

Using the trigonometric identity,

cos a cos b – sin a sin b =  cos (a + b)

$\triangle = \begin{vmatrix} sin\hspace{0.1cm}\alpha sin\hspace{0.1cm}\delta & cos(\alpha+\delta) & cos(\alpha+\delta) \\ sin\hspace{0.1cm}\beta sin\hspace{0.1cm}\delta & cos(\beta+\delta) & cos(\beta+\delta) \\ sin\hspace{0.1cm}\gamma sin\hspace{0.1cm}\delta & cos(\gamma+\delta) & cos(\gamma+\delta) \end{vmatrix}$

As C2 and C3 are identical

Hence, △ = 0

证明下列恒等式：

问题 8。 $\begin{vmatrix} a & b & c \\ a-b & b-c & c-a \\ b+c & c+a & a+b \end{vmatrix}$ = a ³ + b ² + c ³ – 3abc

解决方案：

Considering the determinant, we have

$\triangle = \begin{vmatrix} a & b & c \\ a-b & b-c & c-a \\ b+c & c+a & a+b \end{vmatrix}$

R3⇢R3 + R1 and R2⇢R2 + R1

$\triangle = \begin{vmatrix} a & b & c \\ a-b & b-c & c-a \\ b+c+a & c+a+b & a+b+c \end{vmatrix}$

Taking (a + b + c) common from R3, we get

$\triangle = (a+b+c)\begin{vmatrix} a & b & c \\ a-b & b-c & c-a \\ 1 & 1 & 1 \end{vmatrix}$

R2⇢R2 – R1

$\triangle = (a+b+c)\begin{vmatrix} a & b & c \\ a-b-a & b-c-b & c-a-c \\ 1 & 1 & 1 \end{vmatrix}\\ \triangle = (a+b+c)\begin{vmatrix} a & b & c \\ -b & -c & -a \\ 1 & 1 & 1 \end{vmatrix}$

Taking (-1) common from R2, we get

$\triangle = (a+b+c)(-1)\begin{vmatrix} a & b & c \\ b & c & a \\ 1 & 1 & 1 \end{vmatrix}$

C1⇢C1 – C2 and C2⇢C2 – C3

$\triangle = (a+b+c)(-1)\begin{vmatrix} a-b & b-c & c \\ b-c & c-a & a \\ 0 & 0 & 1 \end{vmatrix}$

△ = (-1)(a + b + c)[1((a – b)(c – a) – (b – c)(b – c))]

△ = (-1)(a + b + c)[(a – b)(c – a) – (b – c)²]

△ = (-1)(a + b + c)[(ac – a² – bc + ab) – (b² – 2cb + c²)]

△ = (-1)(a + b + c)(ac – a² – bc + ab – b² + 2cb – c²)

△ = (a + b + c)(-ac + a² – bc – ab + b² + c²)

△ = (a + b + c)(a² + b² + c² – ac – ab – cb)

△ = a³ + b³ + c³ – 3abc

Hence proved 

问题 9。 $\begin{vmatrix}b+c & a-b & a \\ c+a & b-c & b \\ a+b & c-a & c \end{vmatrix}$ = 3abc – a ³ – b ² – c ³

解决方案：

Considering the determinant, we have

$\triangle = \begin{vmatrix}b+c & a-b & a \\ c+a & b-c & b \\ a+b & c-a & c \end{vmatrix}$

C1⇢C1 + C3

$\triangle = \begin{vmatrix}b+c+a & a-b & a \\ c+a+b & b-c & b \\ a+b+c & c-a & c \end{vmatrix}$

Taking (a + b + c) common from C1, we get

$\triangle = (a+b+c)\begin{vmatrix}1 & a-b & a \\ 1 & b-c & b \\ 1 & c-a & c \end{vmatrix}$

△ = (a + b + c)[1((b – c)c – b(c – a)) – 1((a – b)c – a(c – a)) + 1(b(a – b) – a(b – c))]

△ = (a + b + c)[(b – c)c – b(c – a) – (a – b)c + a(c – a) + b(a – b) – a(b – c)]

△ = (a + b + c)[(bc – c²-bc + ab) – (ac – bc) + ac – a² + ab – b² – (ab – ac)]

△ = (a + b + c)[bc – c² – bc + ab – ac + bc + ac – a² + ab – b² – ab + ac]

△ = (a + b + c)[bc – c² + ab + ac – a² – b²]

△ = (a + b + c)[bc + ab + ac – a² – b² – c²]

△ = 3abc – a³ – b³ – c³

Hence proved 

问题 10。 $\begin{vmatrix}a+b & b+c & c+a \\ b+c & c+a & a+b \\ c+a & a+b& b+c \end{vmatrix} = 2\begin{vmatrix}a & b & c \\ b & c & a \\ c & a& b \end{vmatrix}$

解决方案：

Considering the determinant, we have

$\triangle = \begin{vmatrix}a+b & b+c & c+a \\ b+c & c+a & a+b \\ c+a & a+b& b+c \end{vmatrix}$

C1⇢C1 + C2 + C3

$\triangle = \begin{vmatrix}a+b+b+c+c+a & b+c & c+a \\ b+c+c+a+a+b & c+a & a+b \\ c+a+a+b+b+c & a+b& b+c \end{vmatrix}\\ \triangle = \begin{vmatrix}2(a+b+c) & b+c & c+a \\ 2(a+b+c) & c+a & a+b \\ 2(a+b+c) & a+b& b+c \end{vmatrix}$

Taking 2 common from C1, we get

$\triangle = 2\begin{vmatrix}a+b+c & b+c & c+a \\ a+b+c & c+a & a+b \\ a+b+c & a+b& b+c \end{vmatrix}$

C2⇢C2 – C1 and C3⇢C3 – C1

$\triangle = 2\begin{vmatrix}a+b+c & b+c-(a+b+c) & c+a-(a+b+c) \\ a+b+c & c+a-(a+b+c) & a+b-(a+b+c) \\ a+b+c & a+b-(a+b+c) & b+c-(a+b+c) \end{vmatrix}\\ \triangle = 2\begin{vmatrix}a+b+c & -a & -b \\ a+b+c & -b & -c \\ a+b+c & -c & -a \end{vmatrix}\\$

Taking (-1) and (-1) common from C2 and C3,

$\triangle = 2(-1)(-1)\begin{vmatrix}a+b+c & a & b \\ a+b+c & b & c \\ a+b+c & c & a \end{vmatrix}\\ \triangle = 2\begin{vmatrix}a+b+c & a & b \\ a+b+c & b & c \\ a+b+c & c & a \end{vmatrix}\\$

By splitting the determinant, we get

$\triangle = 2\begin{vmatrix}c & a & b \\ a & b & c \\ b & c & a \end{vmatrix}+2\begin{vmatrix}a & a & b \\ b & b & c \\ c & c & a \end{vmatrix}+2\begin{vmatrix}b & a & b \\ c & b & c \\ a & c & a \end{vmatrix}\\ \triangle = 2\begin{vmatrix}c & a & b \\ a & b & c \\ b & c & a \end{vmatrix}+2(0)+2(0)\\ \triangle = 2\begin{vmatrix}c & a & b \\ a & b & c \\ b & c & a \end{vmatrix}+0+0\\ \triangle = 2\begin{vmatrix}c & a & b \\ a & b & c \\ b & c & a \end{vmatrix}$

Hence proved 

问题 11。 $\begin{vmatrix}a+b+2c & a & b \\ c & b+c+2a & b \\ c & a & c+a+2b \end{vmatrix}$ = 2(a + b + c) ³

Considering the determinant, we have

$\triangle = \begin{vmatrix}a+b+2c & a & b \\ c & b+c+2a & b \\ c & a & c+a+2b \end{vmatrix}$

C1⇢C1 + C2 + C3

$\triangle = \begin{vmatrix}2a+2b+2c & a & b \\ 2b+2c+2a & b+c+2a & b \\ 2a+2b+2c & a & c+a+2b \end{vmatrix}$

Taking (2a + 2b + 2c) common from C1, we get

$\triangle = (2a+2b+2c)\begin{vmatrix}1 & a & b \\ 1 & b+c+2a & b \\ 1 & a & c+a+2b \end{vmatrix}\\ \triangle = 2(a+b+c)\begin{vmatrix}1 & a & b \\ 1 & b+c+2a & b \\ 1 & a & c+a+2b \end{vmatrix}$

R2⇢R2 – R1 and R3⇢R3 – R1

$\triangle = 2(a+b+c)\begin{vmatrix}1 & a & b \\ 0 & b+c+2a-a & b-b \\ 0 & a-a & c+a+2b-b \end{vmatrix}\\ \triangle = 2(a+b+c)\begin{vmatrix}1 & a & b \\ 0 & b+c+a & 0 \\ 0 & 0 & c+a+b \end{vmatrix}$

△ = 2(a + b + c)[1((a + b + b)(a + b + c) – 0)]

△ = 2(a + b + c)[(a + b + b)²]

△ = 2(a + b + c)³

Hence proved 

问题 12。 $\begin{vmatrix}a-b-c & 2a & 2a \\ 2b & b-c-a & 2b \\ 2c & 2c & c-a-b \end{vmatrix}$ = (a + b + c) ³

解决方案：

Considering the determinant, we have

$\triangle = \begin{vmatrix}a-b-c & 2a & 2a \\ 2b & b-c-a & 2b \\ 2c & 2c & c-a-b \end{vmatrix}$

R1⇢R1 + R2 + R3

$\triangle = \begin{vmatrix}a-b-c+2b+2c & 2a+b-c-a+2c & 2a+2b+c-a-b \\ 2b & b-c-a & 2b \\ 2c & 2c & c-a-b \end{vmatrix}\\ \triangle = \begin{vmatrix}a+b+c & a+b+c & a+b+c \\ 2b & b-c-a & 2b \\ 2c & 2c & c-a-b \end{vmatrix}$

Taking (a + b + c) common from R1, we get

$\triangle = (a+b+c)\begin{vmatrix}1 & 1 & 1 \\ 2b & b-c-a & 2b \\ 2c & 2c & c-a-b \end{vmatrix}$

C2⇢C2 – C1 and C3⇢C3 – C1

$\triangle = (a+b+c)\begin{vmatrix}1 & 0 & 0 \\ 2b & b-c-a-2b & 2b-2b \\ 2c & 2c-2c & c-a-b-2c \end{vmatrix}\\ \triangle = (a+b+c)\begin{vmatrix}1 & 0 & 0 \\ 2b & -c-a-b & 0 \\ 2c & 0 & -a-b-c \end{vmatrix}$

△ = (a + b + c)[1((-b – c – a)(-b – c – a) – 0)]

△ = (a + b + c)[(b + c + a)(b + c + a)]

△ = (a + b + c)[(b + c + a)²]

△ = (a + b + c)³

Hence proved 

问题 13。 $\begin{vmatrix}1 & b+c & b^2+c^2 \\ 1 & c+a & c^2+a^2 \\ 1 & a+b & a^2+b^2 \end{vmatrix}$ = (a – b)(b – c)(c – a)

解决方案：

Considering the determinant, we have

$\triangle = \begin{vmatrix}1 & b+c & b^2+c^2 \\ 1 & c+a & c^2+a^2 \\ 1 & a+b & a^2+b^2 \end{vmatrix}$

R2⇢R2 – R1 and R3⇢R3 – R1

$\triangle = \begin{vmatrix}1 & b+c & b^2+c^2 \\ 0 & a-b & a^2-b^2 \\ 0 & a-c & a^2-c^2 \end{vmatrix}\\ \triangle = \begin{vmatrix}1 & b+c & b^2+c^2 \\ 0 & a-b & (a-b)(a+b) \\ 0 & a-c & (a-c)(a+c) \end{vmatrix}$

Taking (a – b) and (a – c) common from R2 and R3 respectively, we get

$\triangle =(a-b)(a-c) \begin{vmatrix}1 & b+c & b^2+c^2 \\ 0 & 1 & a+b \\ 0 & 1 & a+c \end{vmatrix}$

△ = (a – b)(a – c)[1(1(a + c) – 1(a + b))]

△ = (a – b)(a – c)[(a + c) – (a + b)]

△ = (a – b)(a – c)[a + c – a – b]

△ = (a – b)(a – c)

△ = (a – b)(a – c)(c – b)

△ = (a – b)(b – c)(c – a)

Hence proved 

问题 14。 $\begin{vmatrix}a & a+b & a+2b \\ a+2b & a & a+b \\ a+b & a+2b & a \end{vmatrix}$ = 9(a + b)b ²

解决方案：

Considering the determinant, we have

$\triangle = \begin{vmatrix}a & a+b & a+2b \\ a+2b & a & a+b \\ a+b & a+2b & a \end{vmatrix}$

C1⇢C1 + C2 + C3

$\triangle = \begin{vmatrix}3a+3b & 3a+3b & 3a+3b \\ a+2b & a & a+b \\ a+b & a+2b & a \end{vmatrix}$

Taking (3a + 3b) common from R1, we get

$\triangle = (3a+3b)\begin{vmatrix}1 & 1 & 1 \\ a+2b & a & a+b \\ a+b & a+2b & a \end{vmatrix}$

C2⇢C2 – C1 and C3⇢C3 – C2

$\triangle = (3a+3b)\begin{vmatrix}1 & 0 & 0 \\ a+2b & a-(a+2b) & a+b-a \\ a+b & a+2b-(a+b) & a-(a+2b) \end{vmatrix}\\ \triangle = (3a+3b)\begin{vmatrix}1 & 0 & 0 \\ a+2b & -2b & b \\ a+b & b & -2b \end{vmatrix}$

△ = 3(a + b)[1((-2b)(-2b) – b(b))]

△ = 3(a + b)[4b² – b²]

△ = 3(a + b)[3b²]

△ = 9(a + b)b²

Hence proved 

问题 15。 $\begin{vmatrix}1 & a & bc \\ 1 & b & ca \\ 1 & c & ab \end{vmatrix} = \begin{vmatrix}1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{vmatrix}$

解决方案：

Considering the determinant, we have

$\triangle = \begin{vmatrix}1 & a & bc \\ 1 & b & ca \\ 1 & c & ab \end{vmatrix}$

R1⇢aR1, R2⇢bR2 and R3⇢cR3

$\triangle = \frac{1}{abc}\begin{vmatrix}a & a^2 & abc \\ b & b^2 & bca \\ c & c^2 & cab \end{vmatrix}$

Taking (abc) common from C3, we get

$\triangle = \frac{abc}{abc}\begin{vmatrix}a & a^2 & 1 \\ b & b^2 & 1 \\ c & c^2 & 1 \end{vmatrix}\\ \triangle = -\begin{vmatrix}a & 1 & a^2 \\ b & 1 & b^2 \\ c & 1 & c^2 \end{vmatrix}\\ \triangle = \begin{vmatrix}1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{vmatrix}$

Hence proved 

问题 16。 $\begin{vmatrix}z & x & y \\ z^2 & x^2 & y^2 \\ z^4 & x^4 & y^4 \end{vmatrix} = \begin{vmatrix}x & y & z \\ x^2 & y^2 & z^2 \\ x^4 & y^4 & z^4 \end{vmatrix}=\begin{vmatrix} z^2 & x^2 & y^2 \\ x^4 & y^4 & z^4 \\x & y & z \end{vmatrix}$ = xyz(x – y)(y – z)(z – x)(x + y + z)

解决方案：

Considering the determinant, we have

C1↔C2 and then 

$\triangle = \begin{vmatrix}z & x & y \\ z^2 & x^2 & y^2 \\ z^4 & x^4 & y^4 \end{vmatrix}\\ \triangle = - \begin{vmatrix}x & z & y \\ x^2 & z^2 & y^2 \\ x^4 & z^4 & y^4 \end{vmatrix}$

C2↔C3

$\triangle = \begin{vmatrix}x & y & z \\ x^2 & y^2 & z^2 \\ x^4 & y^4 & z^4 \end{vmatrix}$

R1↔R2

$\triangle = -\begin{vmatrix}x^2 & y^2 & z^2 \\ x & y & z \\ x^4 & y^4 & z^4 \end{vmatrix}$

R2↔R3

$\triangle = \begin{vmatrix}x^2 & y^2 & z^2 \\ x^4 & y^4 & z^4 \\ x & y & z \end{vmatrix}$

Taking,

$\triangle = \begin{vmatrix}x & y & z \\ x^2 & y^2 & z^2 \\ x^4 & y^4 & z^4 \end{vmatrix}$

Taking x, y and z common from C1, C2 and C3 respectively

$\triangle = (xyz)\begin{vmatrix}1 & 1 & 1 \\ x & y & z \\ x^3 & y^3 & z^3 \end{vmatrix}$

C1⇢C1 – C2 and C3⇢C3 – C2

$\triangle = (xyz)\begin{vmatrix}0 & 1 & 0 \\ x-y & y & z-y \\ x^3-y^3 & y^3 & z^3-y^3 \end{vmatrix}\\ \triangle = (xyz)\begin{vmatrix}0 & 1 & 0 \\ x-y & y & z-y \\ (x-y)(x^2+xy+y^2) & y^3 & (z-y)(z^2+zy+y^2) \end{vmatrix}$

Taking (x – y) and (z – y) common from C1 and C3 respectively, we get

$\triangle = (xyz)(x-y)(z-y)\begin{vmatrix}0 & 1 & 0 \\ 1 & y & 1 \\ x^2+xy+y^2 & y^3 & z^2+zy+y^2 \end{vmatrix}$

△ = (xyz)(x – y)(z – y)[1(1(z² + zy + y²) – 1(x² + xy + y²))]

△ = (xyz)(x – y)(z – y)[z² + zy + y² – (x² + xy + y²)]

△ = (xyz)(x – y)(z – y)[z² + zy + y² – x² – xy – y²]

△ = (xyz)(x – y)(z – y)[z² + zy – x² – xy]

△ = (xyz)(x – y)(z – y)[z² – x² + zy – xy]

△ = (xyz)(x – y)(z – y)[(z – x)(z + x) + y(z – x)]

△ = (xyz)(x – y)(z – y)(z – x)[z + x + y]

△ = (xyz)(x – y)(z – y)(z – x)(x + y + z)

$\triangle=\begin{vmatrix}z & x & y \\ z^2 & x^2 & y^2 \\ z^4 & x^4 & y^4 \end{vmatrix} = \begin{vmatrix}x & y & z \\ x^2 & y^2 & z^2 \\ x^4 & y^4 & z^4 \end{vmatrix}=\begin{vmatrix} z^2 & x^2 & y^2 \\ x^4 & y^4 & z^4 \\x & y & z \end{vmatrix}=xyz(x-y)(y-z)(z-x)(x+y+z)$

Hence proved 

问题 17。 $\begin{vmatrix}(b+c)^2 & a^2 & bc \\ (c+a)^2 & b^2 & ca \\ (a+b)^2 & c^2 & ab \end{vmatrix}$ = (a – b)(b – c)(c – a)(a + b + c)(a ² + b ² + c ² )

解决方案：

Considering the determinant, we have

$\triangle = \begin{vmatrix}(b+c)^2 & a^2 & bc \\ (c+a)^2 & b^2 & ca \\ (a+b)^2 & c^2 & ab \end{vmatrix}$

C1⇢C1 + C2 – 2C3

$\triangle = \begin{vmatrix}(b+c)^2+a^2-2bc & a^2 & bc \\ (c+a)^2+b^2-2ca & b^2 & ca \\ (a+b)^2+c^2-2ab & c^2 & ab \end{vmatrix}\\ \triangle = \begin{vmatrix}a^2+b^2+c^2 & a^2 & bc \\ a^2+b^2+c^2 & b^2 & ca \\ a^2+b^2+c^2 & c^2 & ab \end{vmatrix}$

Taking (a² + b² + c²) common from C1, we get

$\triangle = (a^2+b^2+c^2)\begin{vmatrix}1 & a^2 & bc \\ 1 & b^2 & ca \\ 1 & c^2 & ab \end{vmatrix}$

C2⇢C2-C1 and C3⇢C3-C1

$\triangle = (a^2+b^2+c^2)\begin{vmatrix}1 & a^2 & bc \\ 0 & b^2-a^2 & ca-bc \\ 0 & c^2-a^2 & ab-bc \end{vmatrix}\\ \triangle = (a^2+b^2+c^2)\begin{vmatrix}1 & a^2 & bc \\ 0 & (b-a)(b+a) & -c(b-a) \\ 0 & (c-a)(c+a) & -b(c-a) \end{vmatrix}$

Taking (b – a) and (c – a) common from R2 and R3, we get

$\triangle = (a^2+b^2+c^2)(b-a)(c-a)\begin{vmatrix}1 & a^2 & bc \\ 0 & b+a & -c \\ 0 & c+a & -b \end{vmatrix}$

△ = (a² + b² + c²)(b – a)(c – a)[1((b + a)(-b) – (c + a)(-c))]

△ = (a² + b² + c²)(b – a)(c – a)[(b + a)(-b) + (c + a)c]

△ = (a² + b² + c²)(b – a)(c – a)[(-b² – ab) + (c² + ac)]

△ = (a² + b² + c²)(b – a)(c – a)

△ = (a² + b² + c²)(b – a)(c – a)[(c – b)(c + b) + a(c – b)]

△ = (a² + b² + c²)(b – a)(c – a)(c – b)

△ = (a² + b² + c²)(a + b + c)(a – b)(b – c)(c – a)

Hence proved