Use properties of determinants to solve for x ; x + a b c c x + b a a b x + c = 0 and x ≠ 0 .

Given x + a b c c x + b a a b x + c = 0 Applying R 1 → R 1 - R 3 , we get x + a - a b - b c - ( x + c ) c x + b a a b x + c = 0 x 0 - x c x + b a a b x + c = 0 ⇒ x 1 0 - 1 c x + b a a b x + c = 0 [ taking x common from R 1 ] Apply C 3 → C 3 + 1 C 1 , we get x 1 0 - 1 + 1 c x + b a + c a b x + c + a = 0 x 1 0 0 c x + b a + c a b x + a + c = 0 Expand along R 1 , we get x ( 1 x + b × x + a + c - b × a + c ) = 0 ⇒ x x 2 + a + c x + b x + b a + c - b a + c = 0 ∵ x ≠ 0 ⇒ x 2 + a + b + c x = 0 ⇒ x x + a + b + c = 0 ⇒ x + a + b + c = 0 x ≠ 0 ∴ x = - a + b + c The required value of x is - a + b + c .

Using properties of determinants prove that: x x x 2 + 1 x + 1 y y y 2 + 1 y + 1 z z z 2 + 1 z + 1 = x - y y - z z - x x + y + z .

To prove: x x x 2 + 1 x + 1 y y y 2 + 1 y + 1 z z z 2 + 1 z + 1 = x - y y - z z - x x + y + z Consider, LHS, ∆ = x x x 2 + 1 x + 1 y y y 2 + 1 y + 1 z z z 2 + 1 z + 1 Operating R 2 → R 2 - R 1 and R 3 → R 3 - R 1 , then ∆ = x x x 2 + 1 x + 1 y - x y 3 + y - x 3 - x y - x z - x z 3 + z - x 3 - x z - x ⇒ ∆ = x x 3 + x x + 1 y - x y 3 - x 3 + y - x y - x z - x z 3 - x 3 + z - x z - x ⇒ ∆ = x x 3 + x x + 1 y - x y - x y 2 + x 2 + x y + y - x y - x z - x z - x z 2 + x 2 + x z + z - x z - x ⇒ ∆ = y - x z - x x x 3 + x x + 1 1 y 2 + x 2 + x y + 1 1 1 z 2 + x 2 + x z + 1 1 [Taking y - x z - x From R 2 And R 3 ] Operating R 3 → R 3 - R 2 , then ∆ = - x - y z - x x x 3 + x x + 1 1 y 2 + x 2 + x y + 1 1 0 z 2 - y 2 + x ( z - y ) 0 ⇒ ∆ = - x - y z - x x x 3 + x x + 1 1 y 2 + x 2 + x y + 1 1 0 ( z - y ) ( z + y + x ) 0 ⇒ ∆ = - x - y z - x z - y x + y + z x x 3 + x x + 1 1 y 2 + x 2 + x y + 1 1 0 1 0 Expanding along row three, we obtain as ∆ = ( x - y ) ( y - z ) ( z - x ) ( x + y + z ) ( - x + x + 1 ) ⇒ ∆ = ( x - y ) ( y - z ) ( z - x ) ( x + y + z ) = RHS Hence, proved.

E M B I B E

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 4 - Determinants>ICSE-2019

Embibe Experts Solutions for Chapter: Determinants, Exercise 1: ICSE-2020

Author:Embibe Experts

Embibe Experts Mathematics Solutions for Exercise - Embibe Experts Solutions for Chapter: Determinants, Exercise 1: ICSE-2020

Attempt the practice questions on Chapter 4: Determinants, Exercise 1: ICSE-2020 with hints and solutions to strengthen your understanding. EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS solutions are prepared by Experienced Embibe Experts.

Questions from Embibe Experts Solutions for Chapter: Determinants, Exercise 1: ICSE-2020 with Hints & Solutions

EASY

12th ICSE

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 4 - Determinants>ICSE-2018>Q 1

Without expanding at any stage find the value of $|\begin{matrix} a & b & c \\ a + 2 x & b + 2 y & c + 2 z \\ x & y & z \end{matrix}|$ .

HARD

12th ICSE

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 4 - Determinants>ICSE-2018>Q 2

Use properties of determinants to solve for $x;$

$|\begin{matrix} x + a & b & c \\ c & x + b & a \\ a & b & x + c \end{matrix}| = 0$ and $x \neq 0 .$

EASY

12th ICSE

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 4 - Determinants>ICSE-2019>Q 3

Using determinants, find the values of $k$ , if the area of triangle with vertices $(- 2, 0), (0, 4) and (0, k)$ is $4$ square units.

HARD

12th ICSE

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 4 - Determinants>ICSE-2019>Q 4

Using properties of determinants prove that: $|\begin{array}{l} x & x (x^{2} + 1) & x + 1 \\ y & y (y^{2} + 1) & y + 1 \\ z & z (z^{2} + 1) & z + 1 \end{array}| = (x - y) (y - z) (z - x) (x + y + z)$ .

EASY

12th ICSE

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 4 - Determinants>ICSE-2020>Q 5

Without expanding at any stage, find the value of the determinant: $∆ = |\begin{array}{l} 20 & a & b + c \\ 20 & b & a + c \\ 20 & c & a + b \end{array}|$ .