RD Sharma Class 12 Ex 6.2 Solutions Chapter 6 Determinants

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TextbookNCERT
ClassClass 12th
SubjectMaths
Chapter6
Exercise6.2
CategoryRD Sharma Solutions

Table of Contents

RD Sharma Class 12 Ex 6.2 Solutions Chapter 6 Determinants

Question 1. Evaluate the following determinant:

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

Solution:

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

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

Solution:

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

(iii) \begin{vmatrix} a & h & g \\ h & b & f \\ g & f & c \end{vmatrix}

Solution:

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

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

Solution:

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

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

Solution:

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

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

Solution:

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

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

Solution:

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

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

Solution:

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

Question 2. Without expanding, show that the values of each of the following determinants are zero:

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

Solution:

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

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

Solution:

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

(iii) \begin{vmatrix} 2 & 3 & 7 \\ 13 & 17 & 5 \\ 15 & 20 & 12 \end{vmatrix}

Solution:

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

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

Solution:

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

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

Solution:

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

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

Solution:

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

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

Solution:

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

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

Solution:

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

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

Solution:

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}

Solution:

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

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

Solution:

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

Question 3. \begin{vmatrix} a & b+c & a^2 \\ b & c+a & b^2 \\ c & a+b & c^2 \end{vmatrix}

Solution:

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)

Question 4. \begin{vmatrix} 1 & a & bc \\ 1 & b & ca \\ 1 & c & ab \end{vmatrix}

Solution:

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)

Question 5. \begin{vmatrix} x+\lambda & x & x \\ x & x+\lambda & x \\ x & x & x+\lambda \end{vmatrix}

Solution:

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)

Question 6. \begin{vmatrix} a & b & c \\ c & a & b \\ b & c & a \end{vmatrix}

Solution:

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)[(a– ac – ab + bc) – (cb – c– b+ bc)]

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

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

Question 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}

Solution:

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

Prove the following identities:

Question 8. \begin{vmatrix} a & b & c \\ a-b & b-c & c-a \\ b+c & c+a & a+b \end{vmatrix} = a+ b+ c– 3abc

Solution:

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 – a– bc + ab) – (b– 2cb + c2)]

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

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

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

△ = a3 + b3 + c3 – 3abc

Hence proved 

Question 9. \begin{vmatrix}b+c & a-b & a \\ c+a & b-c & b \\ a+b & c-a & c \end{vmatrix} = 3abc – a– b– c3

Solution:

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 – a+ ab – b– (ab – ac)]

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

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

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

△ = 3abc – a3 – b3 – c3

Hence proved 

Question 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}

Solution:

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 

Question 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 

Question 12. \begin{vmatrix}a-b-c & 2a & 2a \\ 2b & b-c-a & 2b \\ 2c & 2c & c-a-b \end{vmatrix} = (a + b + c)3

Solution:

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 

Question 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)

Solution:

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 

Question 14. \begin{vmatrix}a & a+b & a+2b \\ a+2b & a & a+b \\ a+b & a+2b & a \end{vmatrix} = 9(a + b)b2

Solution:

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)[4b– b2]

△ = 3(a + b)[3b2]

△ = 9(a + b)b2

Hence proved 

Question 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}

Solution:

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 

Question 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)

Solution:

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(z+ zy + y2) – 1(x+ xy + y2))]

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

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

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

△ = (xyz)(x – y)(z – y)[z– x+ 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 

Question 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+ b+ c2)

Solution:

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 (a+ b+ 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}

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

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

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

△ = (a+ b+ c2)(b – a)(c – a)

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

△ = (a+ b+ c2)(b – a)(c – a)(c – b)

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

Hence proved 

Prove the following identities:

Question 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

Solution:

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 

Question 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+ b+ c2)

Solution:

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 -(a+ b+ 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}

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

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

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

△ = (a+ b+ c2)(a – b)(c – a)

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

△ = (a+ b+ c2)(a – b)(c – a)(c – b)

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

Hence proved 

Question 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+ b+ c2)

Solution:

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)(c+ a+ ac) – (c + a – b)(b+ a+ ab))]

△ = (b – a)(c – a)(b – c)(a + b + c)

△ = -(a – b)(c – a)(b – c)(a + b + c)

Hence proved 

Question 21. \begin{vmatrix}a^2 & bc & ac+c^2 \\ a^2+ab & b^2 & ac \\ ab & b^2+bc & c^2 \end{vmatrix} = 4a2b2c2

Solution:

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 

Question 22. \begin{vmatrix}x+4 & x & x \\ x & x+4 & x \\ x & x & x+4 \end{vmatrix} = 16(3x + 4)

Solution:

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 

Question 23. \begin{vmatrix}1 & 1+p & 1+p+q \\ 2 & 3+2p & 4+3p+2q \\ 3 & 6+3p & 10+6p+3q \end{vmatrix} = 1

Solution:

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 

Question 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)

Solution:

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 

Question 25. \begin{vmatrix}a^2 & 2ab& b^2 \\ b^2 & a^2 & 2ab \\ 2ab & b^2 & a^2 \end{vmatrix} = (a+ b3)2

Solution:

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((a– b2)(a– b2) – (b– 2ab)(2ab – a2))]

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

△ = (a + b)2 [(a+ b– ab)2]

△ = (a+ b3)2

Hence proved

Question 26. \begin{vmatrix}a^2+1 & ab & ac \\ ab & b^2+1 & bc \\ ca & cb & c^2+1 \end{vmatrix} = 1 + a+ b+ c2

Solution:

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 (a+ b+ c+ 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}

△ = (a+ b+ c+ 1)[1((1)(1) – (0)(0))]

△ = (a+ b+ c+ 1)[1]

△ = (a+ b+ c+ 1)

Hence proved !!

Question 27. \begin{vmatrix}1 & a & a^2 \\ a^2 & 1 & a \\ a & a^2 & 1 \end{vmatrix} = (a– 1)2

Solution:

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 (a+ 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}

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

△ = (a+ a + 1)(1 – a)2[(1 + a) + a2]

△ = (a+ a + 1)(1 – a)2[1 + a + a2]

△ = ((a+ a + 1)(1 – a))2

△ = (a– 1)2

Hence proved 

Question 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)

Solution:

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 

Question 29. \begin{vmatrix}b+c & a & a \\ b & c+a & b \\ c & c & a+b \end{vmatrix} = 4abc

Solution:

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 

Question 30. \begin{vmatrix}b^2+c^2 & ab & ac \\ ba & c^2+a^2 & bc \\ ca & cb & a^2+b^2 \end{vmatrix} = 4a2b2c2

Solution:

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

Question 31. \begin{vmatrix}0 & b^2a & c^2a \\ a^2b & 0 & c^2b \\ a^2c & b^2c & 0 \end{vmatrix}= 2a3b3c3

Solution:

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 

Question 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

Solution:

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 

Question 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

Solution:

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 

Question 34. \begin{vmatrix} x+4 & 2x & 2x \\ 2x & x+4 & 2x \\ 2x & 2x & x+4 \end{vmatrix}= (5x + 4)(4 -x)2

Solution:

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 

Prove the following identities:

Question 35. \begin{vmatrix} y+z & z & y \\ z & z+x & x \\ y & x & x+y \end{vmatrix} = 4xyz

Solution:

Considering the determinant, we have

\triangle = \begin{vmatrix} y+z & z & y \\ z & z+x & x \\ y & x & x+y \end{vmatrix}

R1⇢R1 – R2 – R3

\triangle = \begin{vmatrix} y+z-z-y & z-z-x-x & y-x-x-y \\ z & z+x & x \\ y & x & x+y \end{vmatrix}
\triangle = \begin{vmatrix} 0 & -2x & -2x \\ z & z+x & x \\ y & x & x+y \end{vmatrix}

C2⇢C2 – C3

\triangle = \begin{vmatrix} 0 & -2x-(-2x) & -2x \\ z & z+x-x & x \\ y & x-(x+y) & x+y \end{vmatrix}
\triangle = \begin{vmatrix} 0 & 0 & -2x \\ z & z & x \\ y & -y & x+y \end{vmatrix}

△ = [-2x((z)(-y) – (z)(y))]

△ = [-2x(-zy – zy)]

△ = [-2x(-2zy)]

△ = 4xyz

Hence proved 

Question 36. \begin{vmatrix} -a(b^2+c^2-a^2) & 2b^3 & 2c^3 \\ 2a^3 & -b(c^2+a^2-b^2) & 2c^3 \\ 2a^3 & 2b^3 & -c(a^2+b^2-c^2) \end{vmatrix} = abc(a+ b+ c2)3

Solution:

Considering the determinant, we have

\triangle = \begin{vmatrix} -a(b^2+c^2-a^2) & 2b^3 & 2c^3 \\ 2a^3 & -b(c^2+a^2-b^2) & 2c^3 \\ 2a^3 & 2b^3 & -c(a^2+b^2-c^2) \end{vmatrix}

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

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

R1⇢R1 – R3 and R2⇢R2 – R3

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

Taking (a+ b+ c2) common from R1 and R2, we get

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

C3⇢C3 + C1

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

△ = abc(a+ b+ c2)2[-1((-1)(a+ c– b2) – (1)(2b2))]

△ = abc(a+ b+ c2)2[a+ c– b+ 2b2]

△ = abc(a+ b+ c2)2[a+ c+ b2]

△ = abc(a+ b+ c2)3

Hence proved 

Question 37. \begin{vmatrix} 1+a & 1 & 1 \\ 1 & 1+a & 1 \\ 1 & 1 & 1+a \end{vmatrix} = a+ 3a2

Solution:

Considering the determinant, we have

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

C1⇢C1 + C2 + C3

\triangle = \begin{vmatrix} 3+a & 1 & 1 \\ 3+a & 1+a & 1 \\ 3+a & 1 & 1+a \end{vmatrix}

Taking (3 + a) common from C1, we get

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

R2⇢R2 – R1 and R3⇢R3 – R1

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

△ = (3 + a)[1((a)(a) – (0)(0))]

△ = (3 + a)[a2]

△ = 3a+ a3

Hence proved 

Question 38. \begin{vmatrix} y+z & x & y \\ z+x & z & x \\ x+y & y & z \end{vmatrix} = (x + y + z)(x – z)2

Solution:

Considering the determinant, we have

\triangle = \begin{vmatrix} y+z & x & y \\ z+x & z & x \\ x+y & y & z \end{vmatrix}

R1⇢R1 + R2 + R3

\triangle = \begin{vmatrix} 2(x+y+z) & x+y+z & y+x+z \\ z+x & z & x \\ x+y & y & z \end{vmatrix}

Taking (x + y + z) common from R1, we get

\triangle = (x+y+z)\begin{vmatrix} 2 & 1 & 1 \\ z+x & z & x \\ x+y & y & z \end{vmatrix}

C1⇢C1 – C2 – C3

\triangle = (x+y+z)\begin{vmatrix} 2-1-1 & 1 & 1 \\ z+x-z-x & z & x \\ x+y-y-z & y & z \end{vmatrix}
\triangle = (x+y+z)\begin{vmatrix} 0 & 1 & 1 \\ 0 & z & x \\ x-z & y & z \end{vmatrix}

△ = (x + y + z)[(x – z)(x – z)]

△ = (x + y + z)[(x – z)2]

△ = (x + y + z)(x – z)2

Hence proved 

Question 39. Without expanding, prove that \begin{vmatrix} a & b & c \\ x & y & z \\ p & q & r \end{vmatrix}=\begin{vmatrix} x & y & z \\ p & q & r \\a & b & c \end{vmatrix}=\begin{vmatrix} y & b & q \\ x & z & p \\z & c & r \end{vmatrix}

Solution:

Considering the determinant, we have

\triangle = \begin{vmatrix} a & b & c \\ x & y & z \\ p & q & r \end{vmatrix}

R1↔R2

\triangle = (-1)\begin{vmatrix} x & y & z \\ a & b & c \\ p & q & r \end{vmatrix}

R2↔R3

\triangle = (-1)(-1)\begin{vmatrix} x & y & z \\ p & q & r \\ a & b & c \end{vmatrix}\\ \triangle = \begin{vmatrix} x & y & z \\ p & q & r \\ a & b & c \end{vmatrix}

C1↔C2

\triangle = (-1)\begin{vmatrix} y & x & z \\ q & p & r \\ b & a & c \end{vmatrix}

R2↔R3

\triangle = (-1)(-1)\begin{vmatrix} y & x & z \\ b & a & c \\ q & p & r \end{vmatrix}\\ \triangle = \begin{vmatrix} y & x & z \\ b & a & c \\ q & p & r \end{vmatrix}

Taking transpose, we have

\triangle = \begin{vmatrix} y & b & q \\ x & a & p \\ z & c & r \end{vmatrix}

Hence proved 

\begin{vmatrix} a & b & c \\ x & y & z \\ p & q & r \end{vmatrix}=\begin{vmatrix} x & y & z \\ p & q & r \\a & b & c \end{vmatrix}=\begin{vmatrix} y & b & q \\ x & z & p \\z & c & r \end{vmatrix}

Question 40. Show that \begin{vmatrix} x+1 & x+2 & x+a \\ x+2 & x+3 & x+b \\ x+3 & x+4 & x+c \end{vmatrix}=0       where a, b, c are in AP.

Solution:

Considering the determinant, we have

\triangle = \begin{vmatrix} x+1 & x+2 & x+a \\ x+2 & x+3 & x+b \\ x+3 & x+4 & x+c \end{vmatrix}

C1⇢C1 + C2 + C3

\triangle = \begin{vmatrix} 3x+3+a & x+2 & x+a \\ 3x+5+b & x+3 & x+b \\ 3x+7+c & x+4 & x+c \end{vmatrix}

R2⇢R2 – R1 and R3⇢R3 – R2

\triangle = \begin{vmatrix} 3x+3+a & x+2 & x+a \\ 2+b-a & x+3-x-2 & x+b-x-a \\ 2+c-b & x+4-x-3 & x+c-x-b \end{vmatrix}
\triangle = \begin{vmatrix} 3x+3+a & x+2 & x+a \\ 2+b-a & 1 & b-a \\ 2+c-b & 1 & c-b \end{vmatrix}

As a, b and c are in AP

then, b – a = c – b = λ

\triangle = \begin{vmatrix} 3x+3+a & x+2 & x+a \\ 2+\lambda & 1 & \lambda \\ 2+\lambda & 1 & \lambda \end{vmatrix}

As, R2 and R3 are identical

△ = 0

Hence proved 

Question 41. Show that \begin{vmatrix} x-3 & x-4 & x-\alpha \\ x-2 & x-3 & x-\beta \\ x-1 & x-2 & x-\gamma \end{vmatrix}=0     where α, β, γ are in AP.

Solution:

Considering the determinant, we have

\triangle = \begin{vmatrix} x-3 & x-4 & x-\alpha \\ x-2 & x-3 & x-\beta \\ x-1 & x-2 & x-\gamma \end{vmatrix}

C1⇢C1 + C2 + C3

\triangle = \begin{vmatrix} x-3+x-4+x-\alpha & x-4 & x-\alpha \\ x-2+x-3+x-\beta & x-3 & x-\beta \\ x-1+x-2+x-\gamma & x-2 & x-\gamma \end{vmatrix}
\triangle = \begin{vmatrix} 3x-7-\alpha & x-4 & x-\alpha \\ 3x-5-\beta & x-3 & x-\beta \\ 3x-3-\gamma & x-2 & x-\gamma \end{vmatrix}

R2⇢R2 – R1 and R3⇢R3 – R2

\triangle = \begin{vmatrix} 3x-7-\alpha & x-4 & x-\alpha \\ 3x-5-\beta-(3x-7-\alpha) & x-3-(x-4) & x-\beta-(x-\alpha) \\ 3x-3-\gamma-(3x-5-\beta) & x-2-(x-3) & x-\gamma-(x-\beta) \end{vmatrix}
\triangle = \begin{vmatrix} 3x-7-\alpha & x-4 & x-\alpha \\ 2+\alpha-\beta & 1 & \alpha-\beta \\ 2+\beta-\gamma & 1 & \beta-\gamma \end{vmatrix}

As α, β, γ are in AP

then, β – α = γ – β = λ

\triangle = \begin{vmatrix} 3x-7-\alpha & x-4 & x-\alpha \\ 2+(-\lambda) & 1 & -\lambda \\ 2+(-\lambda) & 1 & -\lambda \end{vmatrix}

As, R2 and R3 are identical

△ = 0

Hence proved 

Question 42. Evaluate \begin{vmatrix} x & 1 & 1 \\ 1 & x & 1 \\ 1 & 1 & x \end{vmatrix}

Solution:

Considering the determinant, we have

\triangle = \begin{vmatrix} x & 1 & 1 \\ 1 & x & 1 \\ 1 & 1 & x \end{vmatrix}

C1⇢C1 + C2 + C3

\triangle = \begin{vmatrix} x+1+1 & 1 & 1 \\ 1+x+1 & x & 1 \\ 1+1+x & 1 & x \end{vmatrix}
\triangle = \begin{vmatrix} x+2 & 1 & 1 \\ x+2 & x & 1 \\ x+2 & 1 & x \end{vmatrix}

Taking (x + 2) common from C1. we get

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

R2⇢R2 – R1 and R3⇢R3 – R2

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

△ = (x + 2)[1((x – 1)(x – 1) – (0)(0))]

△ = (x + 2)(x – 1)2

Hence proved 

Question 43. If a, b, c are real numbers such that \begin{vmatrix} b+c & c+a & a+b \\ c+a & a+b & b+c \\ a+b & b+c & c+a \end{vmatrix}=0     , then show that either a + b + c = 0 or,  a = b = c.

Solution:

Considering the determinant, we have

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

C1⇢C1 + C2 + C3

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

Taking 2(a + b + c) common from C1. We get

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

R2⇢R2 – R1 and R3⇢R3 – R2

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

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

△ = 2(a + b + c)[ba – b– ca + cb – (c – a)2]

△ = 2(a + b + c)[ba – b– ca + cb – (c+ a– 2ac)]

△ = 2(a + b + c)[ba – b– ca + cb – c– a+ 2ac]

△ = 2(a + b + c)[ba + bc + ac – b– c– a2]

As, it is given that

△ = 0

2 (a + b + c)(ba + bc + ac – b– c– a2) = 0

(a + b + c)(ba + bc + ac – b– c– a2) = 0

So, either (a + b + c) = 0 or (ba + bc + ac – b– c– a2) = 0

As, ba + bc + ac – b– c– a2 = 0

On multiplying it by -2, we get

-2ba – 2bc – 2ac + 2b+ 2c+ 2a2 = 0

(a – b)2 + (b – c)2 + (c – a)2= 0

As, square power is always positive

(a – b)2 = (b – c)2 = (c – a)2

(a – b) = (b – c) = (c – a)

a = b = c

Hence proved 

Question 44. Show that x=2 is a root of the equation \begin{vmatrix} x & -6 & -1 \\ 2 & -3x & x-3 \\ -3 & 2x & x+2 \end{vmatrix}=0     and solve it completely.

Solution:

Considering the determinant, we have

\triangle = \begin{vmatrix} x & -6 & -1 \\ 2 & -3x & x-3 \\ -3 & 2x & x+2 \end{vmatrix}

On putting x = 2, we get

\triangle = \begin{vmatrix} 2 & -6 & -1 \\ 2 & -3(2) & 2-3 \\ -3 & 2(2) & 2+2 \end{vmatrix}
\triangle = \begin{vmatrix} 2 & -6 & -1 \\ 2 & -6 & -1 \\ -3 & 4 & 4 \end{vmatrix}=0

As, R1 = R2

△ = 0

\triangle = \begin{vmatrix} x & -6 & -1 \\ 2 & -3x & x-3 \\ -3 & 2x & x+2 \end{vmatrix}

R3⇢R3 – R1

\triangle = \begin{vmatrix} x & -6 & -1 \\ 2 & -3x & x-3 \\ -3-x & 2x-(-6) & x+2-(-1) \end{vmatrix}
\triangle = \begin{vmatrix} x & -6 & -1 \\ 2 & -3x & x-3 \\ -(x+3) & 2(x+3) & x+3 \end{vmatrix}

Taking (x + 3) common from R3, we get

\triangle = (x+3)\begin{vmatrix} x & -6 & -1 \\ 2 & -3x & x-3 \\ -1 & 2 & 1 \end{vmatrix}

R1⇢R1 – R2

\triangle = (x+3)\begin{vmatrix} x-2 & -6-(-3x) & -1-(x-3) \\ 2 & -3x & x-3 \\ -1 & 2 & 1 \end{vmatrix}
\triangle = (x+3)\begin{vmatrix} x-2 & 3x-6 & -x+2 \\ 2 & -3x & x-3 \\ -1 & 2 & 1 \end{vmatrix}

Taking (x – 2) common from R1. We get

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

C3⇢C3 + C1

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

Now, taking (x – 1) common from C3. We get

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

△ = (x + 3)(x – 2)(x – 1)[1((1)(2) – (3)(-1))]

△ = (x + 3)(x – 2)(x – 1)[2 + 3]

△ = 5 (x + 3)(x – 2)(x – 1)

△ = 0

5 (x + 3)(x – 2)(x – 1) = 0

x = 2, 1, -3

Question 45. Solve the following determinant equations:

(i) \begin{vmatrix} x+a & b & c \\ a & x+b & c \\ a & b & x+c \end{vmatrix}=0

Solution:

Considering the determinant, we have

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

C1⇢C1 + C2 + C3

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

Now, taking (x + a + b + c) common from C1. We get

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

R2⇢R2 – R1 and R3⇢R3 – R1

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

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

△ = (x + a + b + c)[x2]

As △ = 0

(x + a + b + c) x2 = 0

x + a + b + c = 0 or x2 = 0

x = -(a + b + c) or x = 0

(ii) \begin{vmatrix} x+a & x & x \\ x & x+a & x \\ x & x & x+a \end{vmatrix}=0, a \neq0

Solution:

Considering the determinant, we have

\triangle = \begin{vmatrix} x+a & x & x \\ x & x+a & x \\ x & x & x+a \end{vmatrix}

C1⇢C1 + C2 + C3

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

Now, taking (3x + a) common from C1. We get

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

R2⇢R2 – R1 and R3⇢R3 – R1

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

△ = (3x + a)[1((a)(a) – (0)(0))]

△ = (3x + a)[a2]

As △ = 0

(3x + a)[a2] = 0

x = -a/3

(iii) \begin{vmatrix} 3x-8 & 3 & 3 \\ 3 & 3x-8 & 3 \\ 3 & 3 & 3x-8 \end{vmatrix}=0

Solution:

Considering the determinant, we have

\triangle = \begin{vmatrix} 3x-8 & 3 & 3 \\ 3 & 3x-8 & 3 \\ 3 & 3 & 3x-8 \end{vmatrix}

C1⇢C1 + C2 + C3

\triangle = \begin{vmatrix} 3x-8+3+3 & 3 & 3 \\ 3+3x-8+3 & 3x-8 & 3 \\ 3+3+3x-8 & 3 & 3x-8 \end{vmatrix}
\triangle = \begin{vmatrix} 3x-2 & 3 & 3 \\ 3x-2 & 3x-8 & 3 \\ 3x-2 & 3 & 3x-8 \end{vmatrix}

Now, taking (3x – 2) common from C1. We get

\triangle = (3x-2)\begin{vmatrix} 1 & 3 & 3 \\ 1 & 3x-8 & 3 \\ 1 & 3 & 3x-8 \end{vmatrix}

R2⇢R2 – R1 and R3⇢R3 – R1

\triangle = (3x-2)\begin{vmatrix} 1 & 3 & 3 \\ 0 & 3x-8-3 & 3-3 \\ 0 & 3-3 & 3x-8-3 \end{vmatrix}
\triangle = (3x-2)\begin{vmatrix} 1 & 3 & 3 \\ 0 & 3x-11 & 0 \\ 0 & 0 & 3x-11 \end{vmatrix}

△ = (3x – 2)[1((3x – 11)(3x – 11) – (0)(0))]

△ = (3x – 2)[(3x – 11)2]

As △ = 0

(3x – 2)(3x – 11)2 = 0

3x – 2 = 0 and 3x – 11 = 0

x = 2/3 and x = 11/3

(iv) \begin{vmatrix} 1 & x & x^2 \\ 1 & a & a^2 \\ 1 & b & b^2 \end{vmatrix}=0, a \neq b

Solution:

Considering the determinant, we have

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

R2⇢R2 – R1 and R3⇢R3 – R1

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

Now, taking (a – x) and (b – x) common from R2 and R3 respectively. We get

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

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

△ = (a – x)(b – x)[b + x – (a + x)]

△ = (a – x)(b – x)[b + x – a – x]

△ = (a – x)(b – x)[b – a]

As △ = 0

(a – x)(b – x)(b – a) = 0

a – x = 0 and b – x = 0

x = a and x = b

(v) \begin{vmatrix} x+1 & 3 & 5 \\ 2 & x+2 & 5 \\ 2 & 3 & x+4 \end{vmatrix}=0

Solution:

Considering the determinant, we have

\triangle = \begin{vmatrix} x+1 & 3 & 5 \\ 2 & x+2 & 5 \\ 2 & 3 & x+4 \end{vmatrix}

C1⇢C1 + C2 + C3

\triangle = \begin{vmatrix} x+1+3+5 & 3 & 5 \\ 2+x+2+5 & x+2 & 5 \\ 2+3+x+4 & 3 & x+4 \end{vmatrix}
\triangle = \begin{vmatrix} x+9 & 3 & 5 \\ x+9 & x+2 & 5 \\ x+9 & 3 & x+4 \end{vmatrix}

Now, taking (x + 9) common from C1. We get

\triangle = (x+9)\begin{vmatrix} 1 & 3 & 5 \\ 1 & x+2 & 5 \\ 1 & 3 & x+4 \end{vmatrix}

R2⇢R2 – R1 and R3⇢R3 – R1

\triangle = (x+9)\begin{vmatrix} 1 & 3 & 5 \\ 0 & x+2-3 & 5-5 \\ 0 & 3-3 & x+4-5 \end{vmatrix}
\triangle = (x+9)\begin{vmatrix} 1 & 3 & 5 \\ 0 & x-1 & 0 \\ 0 & 0 & x-1 \end{vmatrix}

△ = (x + 9)[1((x – 1)(x – 1) – (0)(0))]

△ = (x + 9)(x – 1)2

As △ = 0

(x + 9)(x – 1)2 = 0

x + 9 = 0 or (x – 1)2 = 0

x = -9 or x = 1

(vi) \begin{vmatrix} 1 & x & x^3 \\ 1 & b & b^3 \\ 1 & c & c^3 \end{vmatrix}=0, b \neq c

Solution:

Considering the determinant, we have

\triangle = \begin{vmatrix} 1 & x & x^3 \\ 1 & b & b^3 \\ 1 & c & c^3 \end{vmatrix}

R2⇢R2 – R1 and R3⇢R3 – R1

\triangle = \begin{vmatrix} 1 & x & x^3 \\ 0 & b-x & b^3-x^3 \\ 0 & c-x & c^3-x^3 \end{vmatrix}
\triangle = \begin{vmatrix} 1 & x & x^3 \\ 0 & b-x & (b-x)(b^2+x^2+bx) \\ 0 & c-x & (c-x)(c^2+x^2+cx) \end{vmatrix}

Now, taking (b – x) and (c – x) common from R2 and R3 respectively. We get

\triangle = (b-x)(c-x)\begin{vmatrix} 1 & x & x^3 \\ 0 & 1 & b^2+x^2+bx \\ 0 & 1 & c^2+x^2+cx \end{vmatrix}

△ = (b – x)(c – x)[1((c+ x+ cx)(1) – (b+ x+ bx)(1))]

△ = (b – x)(c – x)[(c+ x+ cx) – (b+ x+ bx)]

△ = (b – x)(c – x)

△ = (b – x)(c – x)

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

△ = (b – x)(c – x)(c – b)

As △ = 0

(b – x)(c – x)(c – b) = 0

b – x = 0 or c – x = 0 or c – b = 0 or c + b + x = 0

x = b or x = c or c = b or x = -(c + b)

(vii) \begin{vmatrix} 15-2x & 11-3x & 7-x \\ 11 & 17 & 14 \\ 10 & 16 & 13 \end{vmatrix}=0

Solution:

Considering the determinant, we have

\triangle = \begin{vmatrix} 15-2x & 11-3x & 7-x \\ 11 & 17 & 14 \\ 10 & 16 & 13 \end{vmatrix}

R2⇢R2 – R3

\triangle = \begin{vmatrix} 15-2x & 11-3x & 7-x \\ 11-10 & 17-16 & 14-13 \\ 10 & 16 & 13 \end{vmatrix}
\triangle = \begin{vmatrix} 15-2x & 11-3x & 7-x \\ 1 & 1 & 1 \\ 10 & 16 & 13 \end{vmatrix}

R2⇢R2 – R1 and R1⇢R1 – R3

\triangle = \begin{vmatrix} 15-2x-(7-x) & 11-3x-(15-2x) & 7-x \\ 0 & 0 & 1 \\ 10-13 & 16-10 & 13 \end{vmatrix}
\triangle = \begin{vmatrix} 8-x & -x-4 & 7-x \\ 0 & 0 & 1 \\ -3 & 6 & 13 \end{vmatrix}

△ = -1[(8 – x)(6) – (-x – 4)(-3)]

△ = -1[(8 – x)(6) – (x + 4)(3)]

△ = [(x + 4)(3) – (8 – x)(6)]

△ = [3x + 12 – (48 – 6x)]

△ = [9x – 36]

As △ = 0

9x – 36 = 0

x = 4

(viii) \begin{vmatrix} 1 & 1 & x \\ p+1 & p+1 & p+x \\ 3 & x+1 & x+2 \end{vmatrix}=0

Solution:

Considering the determinant, we have

\triangle = \begin{vmatrix} 1 & 1 & x \\ p+1 & p+1 & p+x \\ 3 & x+1 & x+2 \end{vmatrix}

R2⇢R2 – R1

\triangle = \begin{vmatrix} 1 & 1 & x \\ p+1-1 & p+1-1 & p+x-x \\ 3 & x+1 & x+2 \end{vmatrix}
\triangle = \begin{vmatrix} 1 & 1 & x \\ p & p & p \\ 3 & x+1 & x+2 \end{vmatrix}

Now, taking p common from R2. We get

\triangle = p\begin{vmatrix} 1 & 1 & x \\ 1 & 1 & 1 \\ 3 & x+1 & x+2 \end{vmatrix}

R2⇢R2 – R1

\triangle = p\begin{vmatrix} 1 & 1 & x \\ 0 & 0 & 1-x \\ 3 & x+1 & x+2 \end{vmatrix}

Now, taking 1 – x common from R2. We get

\triangle = p(1-x)\begin{vmatrix} 1 & 1 & x \\ 0 & 0 & 1 \\ 3 & x+1 & x+2 \end{vmatrix}

△ = p(1 – x)[-1((x + 1)(1) – (3)(1))]

△ = p(x – 1)[x + 1 – 3]

△ = p(x – 1)[x – 2]

As △ = 0

p(x – 1)(x – 2) = 0

x – 1 = 0 or x – 2 = 0

x = 1 or x = 2

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