Class 12 RD Sharma Solutions Chapter 6 Determinants Exercise 6.2 | Set 3 (original) (raw)

Last Updated : 21 Aug, 2024

**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(a2 + b2 + 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 (a2 + b2 + 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(a2 + b2 + c2)2[-1((-1)(a2 + c2 - b2) - (1)(2b2))]

△ = abc(a2 + b2 + c2)2[a2 + c2 - b2 + 2b2]

△ = abc(a2 + b2 + c2)2[a2 + c2 + b2]

**△ = abc(a 2 **+ b 2 **+ c 2 ) 3

Hence proved

Question 37. \begin{vmatrix} 1+a & 1 & 1 \\ 1 & 1+a & 1 \\ 1 & 1 & 1+a \end{vmatrix} = a3 + 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 2 **+ a 3

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 - b2 - ca + cb - (c - a)2]

△ = 2(a + b + c)[ba - b2 - ca + cb - (c2 + a2 - 2ac)]

△ = 2(a + b + c)[ba - b2 - ca + cb - c2 - a2 + 2ac]

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

As, it is given that

△ = 0

2 (a + b + c)(ba + bc + ac - b2 - c2 - a2) = 0

(a + b + c)(ba + bc + ac - b2 - c2 - a2) = 0

So, either (a + b + c) = 0 or (ba + bc + ac - b2 - c2 - a2) = 0

As, ba + bc + ac - b2 - c2 - a2 = 0

On multiplying it by -2, we get

-2ba - 2bc - 2ac + 2b2 + 2c2 + 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((c2 + x2 + cx)(1) - (b2 + x2 + bx)(1))]

△ = (b - x)(c - x)[(c2 + x2 + cx) - (b2 + x2 + bx)]

△ = (b - x)(c - x)[c2 + x2 + cx - b2 - x2 - bx]

△ = (b - x)(c - x)[c2 - b2 + c(c - b)]

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

△ = (b - x)(c - x)(c - b)[c + b + x]

As △ = 0

(b - x)(c - x)(c - b)[c + b + x] = 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

Summary

This set of problems focuses on various aspects of determinants, including: