Prove a + d a + d + k a + d + c c c + b c d d + k d + c = a b c

Let △ = a + d a + d + k a + d + c c c + b c d d + k d + c Apply C 2 → C 2 − C 1 , C 3 → C 3 − C 1 = a + d k c c b 0 d k c Apply R 1 → R 1 − R 3 = a 0 0 c b 0 d k c Expanding along R 1 , we get: = a [ b c − 0 ] − 0 + 0 = a b c Hence proved.

Prove 1 x x 2 x 2 1 x x x 2 1 = 1 − x 3 2

We have: 1 x x 2 x 2 1 x x x 2 1 Applying R 1 → R 1 + R 2 + R 3 = 1 + x + x 2 1 + x + x 2 1 + x + x 2 x 2 1 x x x 2 1 Taking 1 + x + x 2 common from R 1 = 1 + x + x 2 1 1 1 x 2 1 x x x 2 1 Applying C 2 → C 2 − C 1 , C 3 → C 3 − C 1 = 1 + x + x 2 1 0 0 x 2 1 - x 2 x - x 2 x x 2 - x 1 - x = 1 + x + x 2 1 0 0 x 2 1 - x 1 + x x 1 - x x x x - 1 1 - x Taking x - 1 common from C 2 , C 3 respectively. = 1 + x + x 2 x - 1 x - 1 1 0 0 x 2 - 1 + x - x x x - 1 = x 3 - 1 x - 1 1 0 0 x 2 - 1 + x - x x x - 1 ∵ 1 + x 2 + x ( x − 1 ) = x 3 − 1 3 = x 3 - 1 x - 1 × 1 - 1 + x - x x - 1 = x 3 - 1 x - 1 x + 1 + x 2 = x 3 - 1 x 3 - 1 ∵ 1 + x 2 + x ( x − 1 ) = x 3 − 1 3 = x 3 - 1 2 or 1 − x 3 2

Prove that the points x 1 , y 1 , x 2 , y 2 , x 3 , y 3 are collinear if x 1 y 1 1 x 2 y 2 1 x 3 y 3 1 = 0

Let the points be A x 1 , y 1 , B x 2 , y 2 , C x 3 , y 3 . If the points are collinear then the area of triangle formed by the these points is zero i.e. Area of △ A B C = 0 i.e. △ = 1 2 x 1 y 1 1 x 2 y 2 1 x 3 y 3 1 = 0 ⇒ x 1 y 1 1 x 2 y 2 1 x 3 y 3 1 = 0 Hence proved.

If A + B + C = π , prove that sin 2 ⁡ A cot ⁡ A 1 sin 2 ⁡ B cot ⁡ B 1 sin 2 ⁡ C cot ⁡ C 1 = 0 .

Let △ = sin 2 ⁡ A cot ⁡ A 1 sin 2 ⁡ B cot ⁡ B 1 sin 2 ⁡ C cot ⁡ C 1 Apply R 2 → R 2 − R 1 , R 3 → R 3 − R 1 = sin 2 ⁡ A cot ⁡ A 1 sin 2 ⁡ B − sin 2 ⁡ A cot ⁡ B − cot ⁡ A 0 sin 2 ⁡ C − sin 2 ⁡ A cot ⁡ C − cot ⁡ A 0 = sin 2 ⁡ A cot ⁡ A 1 sin ⁡ ( B + A ) sin ⁡ ( B − A ) cos ⁡ B sin ⁡ B − cos ⁡ A sin ⁡ A 0 sin ⁡ ( C + A ) sin ⁡ ( C − A ) cos ⁡ C sin ⁡ C − cos ⁡ A sin ⁡ A 0 = sin 2 ⁡ A cot A 1 sin ⁡ ( B + A ) sin ⁡ ( B − A ) sin ⁡ A cos ⁡ B − cos ⁡ A sin ⁡ B sin ⁡ A sin ⁡ B 0 sin ⁡ ( C + A ) sin ⁡ ( C − A ) sin ⁡ A cos ⁡ C − cos ⁡ A sin ⁡ C sin ⁡ A sin ⁡ C 0 = sin 2 ⁡ A cot ⁡ A 1 sin ⁡ ( π − C ) sin ⁡ ( B − A ) sin ⁡ ( A − B ) sin ⁡ A sin ⁡ B 0 sin ⁡ ( π − B ) sin ⁡ ( C − A ) sin ⁡ ( A − C ) sinA ⁡ sin ⁡ C 0 ∵ A + B + C = π and ∵ sin 2 ⁡ x − sin 2 ⁡ y = sin ⁡ ( x + y ) sin ⁡ ( x − y ) = sin 2 ⁡ A cot A 1 sin ⁡ ( C ) sin ⁡ ( B − A ) − sin ⁡ ( B − A ) sin ⁡ A sin ⁡ B 0 sin ⁡ ( B ) sin ⁡ ( C − A ) − sin ⁡ ( C − A ) sin ⁡ A sin ⁡ C 0 ∵ sin ⁡ ( − x ) = − sin ⁡ x , sin π - x = sin x = 1 sin ⁡ ( C ) sin ⁡ ( B − A ) × − sin ⁡ ( C − A ) sin ⁡ A sin ⁡ C − sin ⁡ ( B ) sin ⁡ ( C − A ) × − sin ⁡ ( B − A ) sin ⁡ A sin ⁡ B + 0 + 0 [Expanding along 3 column] = − sin ⁡ ( C − A ) sin ⁡ ( B − A ) sin ⁡ A + sin ⁡ ( C − A ) sin ⁡ ( B − A ) sin ⁡ A = 0 Hence proved.

Eliminate x , y , z from a = x y − z , b = y z − x , c = z x − y

We have, a = x y − z , b = y z − x , c = z x − y i.e. we have a y - a z = x ⇒ x - a y + a z = 0 b z - b x = y ⇒ b x + y - b z = 0 c x - c y = z ⇒ c x - c y - z = 0 Now, D = 1 - a a b 1 - b c - c - 1 = 1 - 1 - b c + a - b + b c + a - b c - c = - 1 - b c - a b + a b c - a b c - a c = - 1 - ( a b + b c + c a ) For eliminating x , y , z put D = 0 ⇒ − 1 − ( a b + b c + c a ) = 0 ⇒ 1 + a b + b c + c a = 0

E M B I B E

Elements of Mathematics Class 12>Chapter 5 - Determinants>EXERCISES 5 (a)

Odisha Board Solutions for Chapter: Determinants, Exercise 1: EXERCISES 5 (a)

Author:Odisha Board

Odisha Board Mathematics Solutions for Exercise - Odisha Board Solutions for Chapter: Determinants, Exercise 1: EXERCISES 5 (a)

Attempt the practice questions on Chapter 5: Determinants, Exercise 1: EXERCISES 5 (a) with hints and solutions to strengthen your understanding. Elements of Mathematics Class 12 solutions are prepared by Experienced Embibe Experts.

Questions from Odisha Board Solutions for Chapter: Determinants, Exercise 1: EXERCISES 5 (a) with Hints & Solutions

EASY

12th Odisha Board

IMPORTANT

Elements of Mathematics Class 12>Chapter 5 - Determinants>EXERCISES 5 (a)>Q 9

Prove $|\begin{array}{ccc} a + d & a + d + k & a + d + c \\ c & c + b & c \\ d & d + k & d + c \end{array}| = a b c$

EASY

12th Odisha Board

IMPORTANT

Elements of Mathematics Class 12>Chapter 5 - Determinants>EXERCISES 5 (a)>Q 11

Show that by eliminating $α$ and $β$ from the equations $a_{i} α + b_{i} β + c_{i} = 0, i = 1, 2, 3$ , we get $|\begin{array}{lll} a_{1} b_{1} c_{1} \\ a_{2} b_{2} c_{2} \\ a_{3} b_{3} c_{3} \end{array}| = 0$

HARD

12th Odisha Board

IMPORTANT

Elements of Mathematics Class 12>Chapter 5 - Determinants>EXERCISES 5 (a)>Q 12

Prove $|\begin{array}{lll} 1 x x^{2} \\ x^{2} 1 x \\ x x^{2} 1 \end{array}| = {(1 - x^{3})}^{2}$

EASY

12th Odisha Board

IMPORTANT

Elements of Mathematics Class 12>Chapter 5 - Determinants>EXERCISES 5 (a)>Q 13

Prove that the points $(x_{1}, y_{1}), (x_{2}, y_{2}), (x_{3}, y_{3})$ are collinear if $|\begin{array}{lll} x_{1} y_{1} 1 \\ x_{2} y_{2} 1 \\ x_{3} y_{3} 1 \end{array}| = 0$

HARD

12th Odisha Board

IMPORTANT

Elements of Mathematics Class 12>Chapter 5 - Determinants>EXERCISES 5 (a)>Q 14

If $A + B + C = π$ , prove that $|\begin{array}{lll} \sin^{2} A \cot A 1 \\ \sin^{2} B \cot B 1 \\ \sin^{2} C \cot C 1 \end{array}| = 0$ .

EASY

12th Odisha Board

IMPORTANT

Elements of Mathematics Class 12>Chapter 5 - Determinants>EXERCISES 5 (a)>Q 15

Eliminate $x, y, z$ from $a = \frac{x}{y - z}, b = \frac{y}{z - x}, c = \frac{z}{x - y}$

EASY

12th Odisha Board

IMPORTANT

Elements of Mathematics Class 12>Chapter 5 - Determinants>EXERCISES 5 (a)>Q 16

Given the equations $x = c y + b z, y = a z + c x and z = b x + a y$ where $x, y, z$ are not all zero, prove that $a^{2} + b^{2} + c^{2} + 2 a b c = 1$ by determinant method.