If A = 1 - 1 5 3 2 - 4 7 8 9 , then find | A | .

A = 1 - 1 5 3 2 - 4 7 8 9 Hence, determinant will be : A = 1 18 + 32 + 1 27 + 28 + 5 24 - 14 = 50 + 55 + 50 = 155

Solve the following 3 - x - 6 3 - 6 3 - x 3 3 3 - 6 - x = 0

Given, 3 - x - 6 3 - 6 3 - x 3 3 3 - 6 - x = 0 Apply, C 1 → C 1 + C 2 + C 3 ⇒ - x - 6 3 - x 3 - x 3 - x 3 - 6 - x = 0 Apply, R 2 → R 2 - R 1 , R 3 → R 3 - R 1 ⇒ - x - 6 3 0 9 - x 0 0 9 - 9 - x = 0 ⇒ - x 9 - x - 9 - x = 0 ⇒ x = 0 , 9 , - 9

If x , y , z are all different and if x x 2 1 + x 3 y y 2 1 + y 3 z z 2 1 + z 3 = 0 , prove that x y z = - 1 .

Given, x x 2 1 + x 3 y y 2 1 + y 3 z z 2 1 + z 3 = 0 It can be written as, ⇒ x x 2 1 y y 2 1 z z 2 1 + x x 2 x 3 y y 2 y 3 z z 2 z 3 = 0 Changing row/column changes the sign of the determinant. ⇒ - 1 2 1 1 1 x x 2 y y 2 z z 2 + x y z 1 1 1 x x 2 y y 2 z z 2 = 0 ⇒ 1 + x y z 1 1 1 x x 2 y y 2 z z 2 = 0 ∴ x y z = - 1

Prove that x - a 2 b 2 c 2 a 2 x - b 2 c 2 a 2 b 2 x - c 2 = x 2 x - 2 a x - 2 b x - 2 c x + a 2 x - 2 a + b 2 x - 2 b + c 2 x - 2 c .

Given, ( x - a ) 2 b 2 c 2 a 2 ( x - b ) 2 c 2 a 2 b 2 ( x - c ) 2 ⇒ x 2 + a 2 - 2 a x b 2 c 2 a 2 x 2 + b 2 - 2 b x c 2 a 2 b 2 x 2 + c 2 - 2 c x R 2 → R 2 - R 1 , R 3 → R 3 - R 1 ⇒ x 2 + a 2 - 2 a x b 2 c 2 - x x - 2 a x x - 2 b 0 - x x - 2 a 0 x x - 2 c Taking - x common from R 2 & R 3 ⇒ - x 2 x 2 + a 2 - 2 a x b 2 c 2 x - 2 a - x - 2 b 0 x - 2 a 0 - x - 2 c ⇒ - x 2 x 2 + a 2 - 2 a x x - 2 b x - 2 c + b 2 x - 2 a x - 2 c + c 2 x - 2 a x - 2 b ⇒ x 2 x - 2 a x - 2 b x - 2 c x + a 2 x - 2 a + b 2 x - 2 b + c 2 x - 2 c

If p + q + r = 0 , prove that p a q b r c q c r a p b r b p c q a = - p q r a b c b c a c a b .

Given: p + q + r = 0 . . . . i Now, L . H . S . = p a q b r c q c r a p b r b p c q a Expanding along R 1 , we have = p a q r a 2 - p 2 b c - q b q 2 a c - p r b 2 + r c p q c 2 - r 2 a b = p q r a 3 + b 3 + c 3 - a b c p 3 + q 3 + r 3 . . . ( 1 ) Since p + q + r = 0 ⇒ p 3 + q 3 + r 3 = 3 p q r , hence from ( 1 ) , we have L . H . S . = p q r a 3 + b 3 + c 3 - a b c 3 p q r ⇒ L . H . S . = p q r a 3 + b 3 + c 3 - 3 a b c . . . ( 2 ) Now, a b c b c a c a b Applying R 1 → R 1 + R 2 + R 3 , we get = a + b + c b + c + a c + a + b b c a c a b Taking out ( a + b + c ) common from first row, we get = a + b + c 1 1 1 b c a c a b Applying C 2 → C 2 - C 1 and C 3 → C 3 - C 1 = a + b + c 1 0 0 b c - b a - b c a - c b - c Expanding along first row, we get = a + b + c c - b a - b a - c b - c = a + b + c ( c - b ) ( b - c ) - ( a - b ) ( a - c ) = a + b + c c b - c 2 - b 2 + b c - a 2 + a c + a b - b c = a + b + c a b + b c + c a - a 2 - b 2 - c 2 = - a 3 + b 3 + c 3 - 3 a b c . . . . . ( 3 ) ∵ l 3 + m 3 + n 3 - 3 l m n = ( l + m + n ) ( l 2 + m 2 + n 2 - l m - m n - n l ) From ( 2 ) and ( 3 ) , we get p a q b r c q c r a p b r b p c q a = - p q r a b c b c a c a b Hence, proved.

Without expanding the determinant at any stage, show that x 2 + x x + 1 x - 2 2 x 2 + 3 x - 1 3 x 3 x - 3 x 2 + 2 x + 3 2 x - 1 2 x - 1 = x A + B where A and B are determinants of 3 rd order not involving x .

Apply R 2 → R 2 − ( R 1 + R 3 ) ⇒ x 2 + x x + 1 x − 2 - 4 0 0 x 2 + 2 x + 3 2 x - 1 2 x - 1 Apply R 1 + 1 4 x 2 R 2 , R 3 + 1 4 x 2 R 2 ⇒ x x + 1 x − 2 - 4 0 0 2 x + 3 2 x - 1 2 x - 1 Apply R 3 − 2 R 1 ⇒ x + 0 x + 1 x − 2 - 4 0 0 3 - 3 3 ⇒ x x x - 4 0 0 3 - 3 3 + 0 1 - 2 - 4 0 0 3 - 3 3 ⇒ x 1 1 1 - 4 0 0 3 - 3 3 + 0 1 - 2 - 4 0 0 3 - 3 3 Therefore, we got the form, x A + B .

Prove that 2 b c - a 2 c 2 b 2 c 2 2 b c - b 2 a 2 b 2 a 2 2 a b - c 2 = a 3 + b 3 + c 3 - 3 a b c 2

Given, 2 b c - a 2 c 2 b 2 c 2 2 b c - b 2 a 2 b 2 a 2 2 a b - c 2 = a b c b c a c a b × - a - b - c c a b b c a = - 1 a b c b c a c a b × a b c c a b b c a Now expanding, a b c b c a c a b = a b c - a 2 - b b 2 - a c + c a b - c 2 = 3 a b c - a 3 + b 3 + c 3 & a b c c a b b c a = a a 2 - b c - b a c - b 2 + c c 2 - a b = a 3 + b 3 + c 3 - 3 a b c ∴ - 1 a b c b c a c a b × a b c c a b b c a = - 1 3 a b c - a 3 + b 3 + c 3 a 3 + b 3 + c 3 - 3 a b c = - 1 2 a 3 + b 3 + c 3 - 3 a b c a 3 + b 3 + c 3 - 3 a b c = a 3 + b 3 + c 3 - 3 a b c 2 hence proved.

E M B I B E

Gamma Question Bank for Engineering Mathematics>Class XII>Chapter 20 - Determinants>Assignment

Embibe Experts Solutions for Exercise 3: Assignment

Author:Embibe Experts

Embibe Experts Mathematics Solutions for Exercise - Embibe Experts Solutions for Exercise 3: Assignment

Attempt the practice questions from Exercise 3: Assignment with hints and solutions to strengthen your understanding. Gamma Question Bank for Engineering Mathematics solutions are prepared by Experienced Embibe Experts.

Questions from Embibe Experts Solutions for Exercise 3: Assignment with Hints & Solutions

EASY

JEE Main/Advance

IMPORTANT

Gamma Question Bank for Engineering Mathematics>Chapter 20 - Determinants>Assignment>Q 8

Find the area of the triangle whose vertices are $(3, 8), (- 4, 2)$ and $(5, 1)$ .

EASY

JEE Main/Advance

IMPORTANT

Gamma Question Bank for Engineering Mathematics>Chapter 20 - Determinants>Assignment>Q 14

If $A = [\begin{array}{r} 1 & - 1 & 5 \\ 3 & 2 & - 4 \\ 7 & 8 & 9 \end{array}]$ , then find $| A |$ .

EASY

JEE Main/Advance

IMPORTANT

Gamma Question Bank for Engineering Mathematics>Chapter 20 - Determinants>Assignment>Q 15

Solve the following $|\begin{matrix} 3 - x & - 6 & 3 \\ - 6 & 3 - x & 3 \\ 3 & 3 & - 6 - x \end{matrix}| = 0$

EASY

JEE Main/Advance

IMPORTANT

Gamma Question Bank for Engineering Mathematics>Chapter 20 - Determinants>Assignment>Q 16

If $x, y, z$ are all different and if $|\begin{array}{l} x & x^{2} & 1 + x^{3} \\ y & y^{2} & 1 + y^{3} \\ z & z^{2} & 1 + z^{3} \end{array}| = 0$ , prove that $x y z = - 1$ .

EASY

JEE Main/Advance

IMPORTANT

Gamma Question Bank for Engineering Mathematics>Chapter 20 - Determinants>Assignment>Q 17

Prove that $|\begin{matrix} {(x - a)}^{2} & b^{2} & c^{2} \\ a^{2} & {(x - b)}^{2} & c^{2} \\ a^{2} & b^{2} & {(x - c)}^{2} \end{matrix}| = x^{2} (x - 2 a) (x - 2 b) (x - 2 c) (x + \frac{a^{2}}{x - 2 a} + \frac{b^{2}}{x - 2 b} + \frac{c^{2}}{x - 2 c})$ .

HARD

JEE Main/Advance

IMPORTANT

Gamma Question Bank for Engineering Mathematics>Chapter 20 - Determinants>Assignment>Q 18

If $p + q + r = 0$ , prove that $|\begin{array}{l} p a & q b & r c \\ q c & r a & p b \\ r b & p c & q a \end{array}| = - p q r |\begin{array}{l} a & b & c \\ b & c & a \\ c & a & b \end{array}|$ .

EASY

JEE Main/Advance

IMPORTANT

Gamma Question Bank for Engineering Mathematics>Chapter 20 - Determinants>Assignment>Q 19

Without expanding the determinant at any stage, show that $|\begin{matrix} x^{2} + x & x + 1 & x - 2 \\ 2 x^{2} + 3 x - 1 & 3 x & 3 x - 3 \\ x^{2} + 2 x + 3 & 2 x - 1 & 2 x - 1 \end{matrix}| = x A + B$ where $A$ and $B$ are determinants of $3^{rd}$ order not involving $x$ .

MEDIUM

JEE Main/Advance

IMPORTANT

Gamma Question Bank for Engineering Mathematics>Chapter 20 - Determinants>Assignment>Q 20

Prove that $|\begin{matrix} 2 b c - a^{2} & c^{2} & b^{2} \\ c^{2} & 2 b c - b^{2} & a^{2} \\ b^{2} & a^{2} & 2 a b - c^{2} \end{matrix}| = {(a^{3} + b^{3} + c^{3} - 3 a b c)}^{2}$

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