第 12 类 RD Sharma 解决方案 - 第 6 章行列式 - 练习 6.2 |设置 2

证明下列恒等式：

问题 18。 $\begin{vmatrix}(a+1)(a+2) & a+2 & 1 \\ (a+2)(a+3) & a+3 & 1 \\ (a+3)(a+4) & a+4 & 1 \end{vmatrix}$ = -2

解决方案：

Considering the determinant, we have

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

R2⇢R2 – R1 and R3⇢R3 – R2

$\triangle = \begin{vmatrix}(a+1)(a+2) & a+2 & 1 \\ (a+2)(a+3)-(a+1)(a+2) & a+3-(a+2) & 0 \\ (a+3)(a+4)-(a+2)(a+3) & a+4-(a+3) & 0 \end{vmatrix}\\ \triangle = \begin{vmatrix}(a+1)(a+2) & a+2 & 1 \\ (a+2)[a+3-a-1] & a+3-a-2 & 0 \\ (a+3)[a+4-a-2] & a+4-a-3 & 0 \end{vmatrix}\\ \triangle = \begin{vmatrix}(a+1)(a+2) & a+2 & 1 \\ 2(a+2) & 1 & 0 \\ 2(a+3) & 1 & 0 \end{vmatrix}$

△ = 1[2(a + 2) – 2(a + 3)]

△ = (4a + 4 – (4a + 6))

△ = (4a + 4 – 4a – 6)

△ = -2

Hence proved 

问题 19。 $\begin{vmatrix}a^2 & a^2-(b-c)^2 & bc \\ b^2 & b^2-(c-a)^2 & ca \\ c^2 & c^2-(a-b)^2 & ab \end{vmatrix}$ = (a – b)(b – c)(c – a)(a + b + c)(a ² + b ² + c ² )

解决方案：

Considering the determinant, we have

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

C2⇢C2 – 2C1 – 2C3

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

Taking -(a² + b² + c²) common from C2, we get

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

R2⇢R2 – R1 and R3⇢R3 – R1

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

Taking (b – a) and (c – a) common from R1 and R2, we get

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

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

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

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

△ = (a² + b² + c²)(a – b)(c – a)

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

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

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

Hence proved 

问题 20。 $\begin{vmatrix}1 & a^2+bc & a^3 \\ 1 & b^2+ca & b^3 \\ 1 & c^2+ab & c^3 \end{vmatrix}$ = (a – b)(b – c)(c – a)(a ² + b ² + c ² )

解决方案：

Considering the determinant, we have

$\triangle = \begin{vmatrix}1 & a^2+bc & a^3 \\ 1 & b^2+ca & b^3 \\ 1 & c^2+ab & c^3 \end{vmatrix}$

R2⇢R2 – R1 and R3⇢R3 – R1

$\triangle = \begin{vmatrix}1 & a^2+bc & a^3 \\ 0 & b^2+ca-a^2-bc & b^3-a^3 \\ 0 & c^2+ab-a^2-bc & c^3-a^3 \end{vmatrix}\\ \triangle = \begin{vmatrix}1 & a^2+bc & a^3 \\ 0 & b^2-a^2-c(b-a) & b^3-a^3 \\ 0 & c^2-a^2-b(c-a) & c^3-a^3 \end{vmatrix}\\ \triangle = \begin{vmatrix}1 & a^2+bc & a^3 \\ 0 & (b-a)(b+a-c) & (b-a)(b^2+ba+a^2) \\ 0 & (c-a)(c+a-b) & (c-a)(c^2+ac+c^2) \end{vmatrix}$

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

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

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

△ = (b – a)(c – a)(b – c)(a + b + c)

△ = -(a – b)(c – a)(b – c)(a + b + c)

Hence proved 

问题 21。 $\begin{vmatrix}a^2 & bc & ac+c^2 \\ a^2+ab & b^2 & ac \\ ab & b^2+bc & c^2 \end{vmatrix}$ = 4a ² b ² c ²

解决方案：

Considering the determinant, we have

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

Taking a, b and c common from C1, C2 and C3 we get

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

C1⇢C1 + C2 + C3

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

Taking 2 common from C1, we get

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

C2⇢C2 – C1 and C3⇢C3 – C1

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

C1⇢C1 + C2 + C3

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

Taking c, a and b common from C1, C2 and C3 we get

$\triangle = 2abc(abc)\begin{vmatrix}1 & -1 & 0 \\ 0 & -1 & -1 \\ 1 & 0 & -1 \end{vmatrix}$

R3⇢R3 – R1

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

△ = 2a²b²c²[1((-1)(-1) – (-1)(1))]

△ = 2a²b²c²[1 – (-1)]

△ = 2a²b²c²[1 + 1]

△ = 4a²b²c²

Hence proved 

问题 22。 $\begin{vmatrix}x+4 & x & x \\ x & x+4 & x \\ x & x & x+4 \end{vmatrix}$ = 16(3x + 4)

解决方案：

Considering the determinant, we have

$\triangle = \begin{vmatrix}x+4 & x & x \\ x & x+4 & x \\ x & x & x+4 \end{vmatrix}$

C1⇢C1 + C2 + C3

$\triangle = \begin{vmatrix}3x+4 & x & x \\ 3x+4 & x+4 & x \\ 3x+4 & x & x+4 \end{vmatrix}$

Taking (3x + 4) common from C1, we get

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

R2⇢R2 – R1 and R3⇢R3 – R1

$\triangle = (3x+4)\begin{vmatrix}1 & x & x \\ 0 & x+4-x & x-x \\ 0 & x-(x+4) & x+4-x \end{vmatrix}\\ \triangle = (3x+4)\begin{vmatrix}1 & x & x \\ 0 & 4 & 0 \\ 0 & -4 & 4 \end{vmatrix}$

△ = (3x + 4)[1((4)(4) – (-4)(0))]

△ = (3x + 4)[16 – 0]

△ = 16(3x + 4)

Hence proved 

问题 23。 $\begin{vmatrix}1 & 1+p & 1+p+q \\ 2 & 3+2p & 4+3p+2q \\ 3 & 6+3p & 10+6p+3q \end{vmatrix}$ = 1

解决方案：

Considering the determinant, we have

$\triangle = \begin{vmatrix}1 & 1+p & 1+p+q \\ 2 & 3+2p & 4+3p+2q \\ 3 & 6+3p & 10+6p+3q \end{vmatrix}$

C2⇢C2 – pC1 and C3⇢C3 – qC1

$\triangle = \begin{vmatrix}1 & 1 & 1+p \\ 2 & 3 & 4+3p \\ 3 & 6 & 10+6p \end{vmatrix}$

C3⇢C3 – pC2

$\triangle = \begin{vmatrix}1 & 1 & 1 \\ 2 & 3 & 4 \\ 3 & 6 & 10 \end{vmatrix}$

C2⇢C2 – C1 and C3⇢C3 – C2

$\triangle = \begin{vmatrix}1 & 0 & 0 \\ 2 & 1 & 1 \\ 3 & 3 & 4 \end{vmatrix}$

△ = 1[(1)(4) – (1)(3)]

△ = [4 – 3]

△ = 1

Hence proved 

问题 24。 $\begin{vmatrix}a & b-c & c-b \\ a-c & b & c-a \\ a-b & b-a & c \end{vmatrix}$ = (a + b – c)(b + c – a)(c + a – b)

解决方案：

Considering the determinant, we have

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

R1⇢R1 – R2 – R3

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

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

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

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

C2⇢C2 + C1 and C3⇢C3 + C1

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

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

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

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

△ = (b + c – a)(b + a – c)(c + a – b)

Hence proved 

问题 25。 $\begin{vmatrix}a^2 & 2ab& b^2 \\ b^2 & a^2 & 2ab \\ 2ab & b^2 & a^2 \end{vmatrix}$ = (a ³ + b ³ ) ²

解决方案：

Considering the determinant, we have

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

C1⇢C1 + C2 + C3

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

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

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

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

R2⇢R2 – R1 and R3⇢R3 – R1

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

R2⇢R2 – R3

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

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

△ = (a + b)² [1((a² – b²)(a² – b²) – (b² – 2ab)(2ab – a²))]

△ = (a + b)² [(a² – b²)² + (b² – 2ab)(a² – 2ab)]

△ = (a + b)² [(a² + b² – ab)²]

△ = (a³ + b³)²

Hence proved

问题 26。 $\begin{vmatrix}a^2+1 & ab & ac \\ ab & b^2+1 & bc \\ ca & cb & c^2+1 \end{vmatrix}$ = 1 + a ² + b ² + c ²

解决方案：

Considering the determinant, we have

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

Multiplying a, b and c to R1, R2 and R3 we get

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

Taking a, b and c common from C1, C2 and C3 we get

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

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

R1⇢R1 + R2 + R3

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

Taking (a² + b² + c² + 1) common from R1, we get

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

C2⇢C2-C1 and C3⇢C3-C1

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

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

△ = (a² + b² + c² + 1)[1((1)(1) – (0)(0))]

△ = (a² + b² + c² + 1)[1]

△ = (a² + b² + c² + 1)

Hence proved !!

问题 27。 $\begin{vmatrix}1 & a & a^2 \\ a^2 & 1 & a \\ a & a^2 & 1 \end{vmatrix}$ = (a ³ – 1) ²

解决方案：

Considering the determinant, we have

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

C1⇢C1 + C2 + C3

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

Taking (a² + a + 1) common from C1, we get

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

R2⇢R2 – R1 and R3⇢R3 – R1

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

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

Taking (1 – a) common from R2 and R3, we get

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

△ = (a² + a + 1)(1 – a)²[1((1 + a)(1) – (a)(-a))]

△ = (a² + a + 1)(1 – a)²[(1 + a) + a²]

△ = (a² + a + 1)(1 – a)²[1 + a + a²]

△ = ((a² + a + 1)(1 – a))²

△ = (a³ – 1)²

Hence proved 

问题 28。 $\begin{vmatrix}a+b+c & -c & -b \\ -c & a+b+c & -a \\ -b & -a & a+b+c \end{vmatrix}$ = 2(a + b)(b + c)(c + a)

解决方案：

Considering the determinant, we have

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

C1⇢C1 + C3 and C2⇢C2 + C3

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

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

Taking (c + a) and (b + c) common from C1 and C2, we get

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

R2⇢R2 + R1 and R3⇢R3 + R2

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

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

△ = (c + a)(b + c)[0 + 2(a + b)]

△ = 2(a + b)(c + a)(b + c)

Hence proved 

问题 29。 $\begin{vmatrix}b+c & a & a \\ b & c+a & b \\ c & c & a+b \end{vmatrix}$ = 4abc

解决方案：

Considering the determinant, we have

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

R1⇢R1 + R2 + R3

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

Taking 2 common from R1, we get

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

R2⇢R2 – R1 and R3⇢R3 – R1

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

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

R1⇢R1 + R2 + R3

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

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

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

△ = 2[-c(0 – ab) + b(ac – 0)]

△ = 2[abc + abc]

△ = 2[2abc]

△ = 4abc

Hence proved 

问题 30。 $\begin{vmatrix}b^2+c^2 & ab & ac \\ ba & c^2+a^2 & bc \\ ca & cb & a^2+b^2 \end{vmatrix}$ = 4a ² b ² c ²

解决方案：

Considering the determinant, we have

$\triangle = \begin{vmatrix}b^2+c^2 & ab & ac \\ ba & c^2+a^2 & bc \\ ca & cb & a^2+b^2 \end{vmatrix}$

Multiplying a, b and c to R1, R2 and R3 respectively, we get

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

Taking common a, b and c to C1, C2 and C3 respectively, we get

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

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

R1⇢R1 + R2 + R3

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

Taking 2 common from R1, we get

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

R1⇢R1 – R2

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

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

△ = 2

△ = 2

△ = 2

△ = 2

问题 31。 $\begin{vmatrix}0 & b^2a & c^2a \\ a^2b & 0 & c^2b \\ a^2c & b^2c & 0 \end{vmatrix}$ = 2a ³ b ³ c ³

解决方案：

Considering the determinant, we have

$\triangle = \begin{vmatrix}0 & b^2a & c^2a \\ a^2b & 0 & c^2b \\ a^2c & b^2c & 0 \end{vmatrix}$

Taking a², b² and c² common from C1, C2 and C3. we get

$\triangle = a^2b^2c^2\begin{vmatrix}0 & a & a \\ b & 0 & b \\ c & c & 0 \end{vmatrix}$

Taking a, b and c common from R1, R2 and R3. we get

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

C2⇢C2 – C3

$\triangle = a^3b^3c^3\begin{vmatrix}0 & 0 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & 0 \end{vmatrix}$

△ = a³b³c³[1((1)(1) – (1)(-1))]

△ = a³b³c³[1 + 1]

△ = 2a³b³c³

Hence proved 

问题 32。 $\begin{vmatrix} \frac{a^2+b^2}{c} & c & c \\ a & \frac{b^2+c^2}{a} & a \\ b & b & \frac{c^2+a^2}{b} \end{vmatrix}$ = 4abc

解决方案：

Considering the determinant, we have

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

Multiplying c, a and b to R1, R2 and R3. We get

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

R1⇢R1 – R2 – R3

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

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

Taking -2 common from R1, we get

$\triangle = \frac{-2}{abc}\begin{vmatrix} 0 & b^2 & a^2 \\ a^2 & b^2+c^2 & a^2 \\ b^2 & b^2 & c^2+a^2 \end{vmatrix}$

R2⇢R2 – R1 and R3⇢R3 – R1

$\triangle = \frac{-2}{abc}\begin{vmatrix} 0 & b^2 & a^2 \\ a^2 & c^2 & 0 \\ b^2 & 0 & c^2 \end{vmatrix}$

$\triangle = \frac{-2}{abc}[-b^2((a^2)(c^2)-(0)(b^2))+a^2((a^2)(0)-(b^2)(c^2))]\\ \triangle = \frac{-2}{abc}[-b^2(a^2c^2)+a^2(-b^2c^2)]\\ \triangle = \frac{-2}{abc}[-b^2a^2c^2-a^2b^2c^2]\\ \triangle = \frac{-2}{abc}[-2a^2b^2c^2]$

△ = 4abc

Hence proved 

问题 33。 $\begin{vmatrix} -bc & b^2+bc & c^2+bc \\ a^2+ac & -ac & c^2+ac \\ a^2+ab & b^2+ab & -ab \end{vmatrix}$ = (ab + bc + ca) ³

解决方案：

Considering the determinant, we have

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

Multiplying a, b and c to R1, R2 and R3. We get

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

Taking a, b and c common from C1, C2 and C3. we get

$\triangle = \frac{abc}{abc}\begin{vmatrix} -bc & a(b+c) & a(c+b) \\ b(a+c) & -ac & b(c+a) \\ c(a+b) & c(b+a) & -ab \end{vmatrix}$

R1⇢R1 + R2 + R3

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

$\triangle = \begin{vmatrix} ba+ca+bc & ab+bc+ca & ab+bc+ca \\ ab+bc & -ac & bc+ba \\ ca+cb & cb+ca & -ab \end{vmatrix}$

Taking (ab + bc + ca) common from R1, we get

$\triangle = (ab+bc+ca)\begin{vmatrix} 1 & 1 & 1 \\ ab+bc & -ac & bc+ba \\ ca+cb & cb+ca & -ab \end{vmatrix}$

C1⇢C1 – C2 and C3⇢C3 – C2

$\triangle = (ab+bc+ca)\begin{vmatrix} 0 & 1 & 0 \\ ab+bc+ac & -ac & bc+ba+ac \\ ca+cb-cb-ca & cb+ca & -ab-cb-ca \end{vmatrix}$

$\triangle = (ab+bc+ca)\begin{vmatrix} 0 & 1 & 0 \\ ab+bc+ac & -ac & bc+ba+ac \\ 0 & cb+ca & -(ab+cb+ca) \end{vmatrix}$

Taking (ab + bc + ca) common from C1 and C2, we get

$\triangle = (ab+bc+ca)(ab+bc+ca)(ab+bc+ca)\begin{vmatrix} 0 & 1 & 0 \\ 1 & -ac & 1 \\ 0 & cb+ca & -1 \end{vmatrix}$

△ = (ab + bc + ca)³ [-1((1)(-1) – (1)(0))]

△ = (ab + bc + ca)³ [-1(-1)]

△ = (ab + bc + ca)³

Hence proved 

问题 34。 $\begin{vmatrix} x+4 & 2x & 2x \\ 2x & x+4 & 2x \\ 2x & 2x & x+4 \end{vmatrix}$ = (5x + 4)(4 -x) ²

解决方案：

Considering the determinant, we have

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

C1⇢C1 + C2 + C3

$\triangle = \begin{vmatrix} 5x+4 & 2x & 2x \\ 5x+4 & x+4 & 2x \\ 5x+4 & 2x & x+4 \end{vmatrix}$

Taking (5x + 4) common from C1, we get

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

R2⇢R2 – R1 and R3⇢R3 – R1

$\triangle = (5x+4)\begin{vmatrix} 1 & 2x & 2x \\ 0 & x+4-2x & 2x-2x \\ 0 & 2x-2x & x+4-2x \end{vmatrix}$

$\triangle = (5x+4)\begin{vmatrix} 1 & 2x & 2x \\ 0 & 4-x & 0 \\ 0 & 0 & 4-x \end{vmatrix}$

△ = (5x + 4)[1((4 – x)(4 – x) – (0)(0))]

△ = (5x + 4)[(4 – x)²]

△ = (5x + 4)(4 – x)²

Hence proved