Using properties of determinants, prove that: 1 1 1 + 3 x 1 + 3 y 1 1 1 1 + 3 z 1 = 9 3 x y z + x y + y z + z x .

To prove: ∆ = 1 1 1 + 3 x 1 + 3 y 1 1 1 1 + 3 z 1 = 9 3 x y z + x y + y z + z x Consider LHS as, ∆ = 1 1 1 + 3 x 1 + 3 y 1 1 1 1 + 3 z 1 ⇒ ∆ = 1 1 1 + 3 x 3 y 0 - 3 x 0 3 z - 3 x [Applying R 2 → R 2 - R 1 , R 3 → R 3 - R 1 ] Expanding along R 1 , we get: ∆ = 1 0 + 9 x z - 1 - 9 x y - 0 + 1 + 3 x 9 y z - 0 = 9 x z + 9 x y + 9 y z + 27 x y z = 9 3 x y z + x y + y z + z x = RHS Hence, proved.

Using the properties of determinants, prove that a 2 + 2 a 2 a + 1 1 2 a + 1 a + 2 1 3 3 1 = ( a - 1 ) 3 .

Consider a 2 + 2 a 2 a + 1 1 2 a + 1 a + 2 1 3 3 1 Applying R 1 → R 1 - R 2 and R 2 → R 2 - R 3 = a 2 - 1 a - 1 0 2 a - 2 a - 1 0 0 3 1 = ( a - 1 ) 2 a + 1 1 0 2 1 0 0 3 1 Expanding along C 3 = ( a - 1 ) 2 [ 0 + 0 + ( a + 1 - 2 ) ] = ( a - 1 ) 3 . Hence, proved

If A = 2 0 0 - 1 2 3 3 3 5 then find A ( adj A ) .

Given that, A = 2 0 0 - 1 2 3 3 3 5 We know that, A · adj ( A ) = | A | I ⇒ A · adj ( A ) = 2 0 0 - 1 2 3 3 3 5 1 0 0 0 1 0 0 0 1 ⇒ A · adj ( A ) = 2 1 0 0 0 1 0 0 0 1 ⇒ A · adj ( A ) = 2 I Thus the value of A · adj ( A ) is 2 I .

Using properties of determinants, prove that ( b + c ) 2 a 2 b c ( c + a ) 2 b 2 c a ( a + b ) 2 c 2 a b = ( a 2 + b 2 + c 2 ) ( a + b + c ) ( b - c ) ( c - a ) ( a - b ) .

We have to prove that ( b + c ) 2 a 2 b c ( c + a ) 2 b 2 c a ( a + b ) 2 c 2 a b = ( a 2 + b 2 + c 2 ) ( a + b + c ) ( b - c ) ( c - a ) ( a - b ) L . H . S . = ( b + c ) 2 a 2 b c ( c + a ) 2 b 2 c a ( a + b ) 2 c 2 a b Operating C 1 → C 1 + C 2 - 2 C 3 , we have = b 2 + c 2 + 2 b c + a 2 - 2 b c a 2 b c c 2 + a 2 + a c + b 2 - 2 c a b 2 c a a 2 + b 2 + 2 a b + c 2 - 2 a b c 2 a b = b 2 + c 2 + a 2 a 2 b c c 2 + a 2 + b 2 b 2 c a a 2 + b 2 + c 2 c 2 a b Taking ( a 2 + b 2 + c 2 ) common from first column, we have = ( a 2 + b 2 + c 2 ) 1 a 2 b c 1 b 2 c a 1 c 2 a b Applying R 2 → R 2 - R 1 and R 3 → R 3 - R 1 , we have = ( a 2 + b 2 + c 2 ) 1 a 2 b c 0 b 2 - a 2 c a - b c 0 c 2 - a 2 a b - b c = ( a 2 + b 2 + c 2 ) 1 a 2 b c 0 ( b - a ) ( b + a ) c ( a - b ) 0 ( c - a ) ( c + a ) b ( a - c ) Taking out ( a - b ) common from R 2 and ( c - a ) from R 3 , we have = ( a 2 + b 2 + c 2 ) ( a - b ) ( c - a ) 1 a 2 b c 0 - 1 ( b + a ) c 0 ( c + a ) - b Expanding along first column, we have = ( a 2 + b 2 + c 2 ) ( a - b ) ( c - a ) [ ( a + b ) b - c ( a + c ) ] = ( a 2 + b 2 + c 2 ) ( a - b ) ( c - a ) [ a b + b 2 - a c - c 2 ] = ( a 2 + b 2 + c 2 ) ( a - b ) ( c - a ) [ a ( b - c ) + b 2 - c 2 ] = ( a 2 + b 2 + c 2 ) ( a - b ) ( c - a ) [ a ( b - c ) + ( b - c ) ( b + c ) ] [ ∵ m 2 - n 2 = ( m - n ) ( m + n ) ] = ( a 2 + b 2 + c 2 ) ( a + b + c ) ( b - c ) ( c - a ) ( a - b ) = R . H . S . Hence, the given result is proved.

E M B I B E

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 4 - Determinants>CBSE-2018

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

Author:Embibe Experts

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

Attempt the free practice questions on Chapter 4: Determinants, Exercise 1: CBSE-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: CBSE-2020 with Hints & Solutions

MEDIUM

12th CBSE

IMPORTANT

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

Using properties of determinants, prove that: $|\begin{matrix} 1 & 1 & 1 + 3 x \\ 1 + 3 y & 1 & 1 \\ 1 & 1 + 3 z & 1 \end{matrix}| = 9 (3 x y z + x y + y z + z x)$ .

HARD

12th CBSE

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 4 - Determinants>CBSE-2018>Q 3

If $A = [\begin{matrix} 2 & - 3 & 5 \\ 3 & 2 & - 4 \\ 1 & 1 & - 2 \end{matrix}]$ , find $A^{- 1}$ . Use it to solve the system of equations

$2 x - 3 y + 5 z = 11$

$3 x + 2 y - 4 z = - 5$

$x + y - 2 z = - 3$

EASY

12th CBSE

IMPORTANT

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

If $A$ and $B$ are square matrices of the same order $3$ , such that $| A | = 2$ , and $AB = I$ , write the value of $|B|$ .

MEDIUM

12th CBSE

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 4 - Determinants>CBSE-2019>Q 5

Using the properties of determinants, prove that $|\begin{matrix} a^{2} + 2 a & 2 a + 1 & 1 \\ 2 a + 1 & a + 2 & 1 \\ 3 & 3 & 1 \end{matrix}| = (a - 1)^{3}$ .

EASY

12th CBSE

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 4 - Determinants>CBSE-2020>Q 7

If $A$ is a square matrix of order $3$ and $| A | = 5$ then the value of $|2 A^{'}|$ is

EASY

12th CBSE

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 4 - Determinants>CBSE-2020>Q 8

If $A = [\begin{matrix} 2 & 0 & 0 \\ - 1 & 2 & 3 \\ 3 & 3 & 5 \end{matrix}]$ then find $A (adj A)$ .

HARD

12th CBSE

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 4 - Determinants>CBSE-2020>Q 9

If $A = [\begin{array}{ccc} 1 & 2 & - 3 \\ 3 & 2 & - 2 \\ 2 & - 1 & 1 \end{array}]$ . Find $A^{- 1}$ and use it to solve the following system of equations $x + 2 y - 3 z = 6, 3 x + 2 y - 2 z = 3, 2 x - y + z = 2$ .

HARD

12th CBSE

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 4 - Determinants>CBSE-2020>Q 10

Using properties of determinants, prove that $|\begin{matrix} (b + c)^{2} & a^{2} & b c \\ (c + a)^{2} & b^{2} & c a \\ (a + b)^{2} & c^{2} & a b \end{matrix}| = (a^{2} + b^{2} + c^{2}) (a + b + c) (b - c) (c - a) (a - b)$ .

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