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

Let L.H.S = a 2 2 a b b 2 b 2 a 2 2 a b 2 a b b 2 a 2 Applying c 1 → c 1 + c 2 we get, = a 2 + 2 a b + b 2 2 a b b 2 a 2 + 2 a b + b 2 a 2 2 a b a 2 + 2 a b + b 2 b 2 a 2 = a + b 2 2 a b b 2 a + b 2 a 2 2 a b a + b 2 b 2 a 2 Taking a + b 2 common, = a + b 2 1 2 a b b 2 1 a 2 2 a b 1 b 2 a 2 Applying R 1 → R 1 - R 3 and R 2 → R 2 - R 3 we get = a + b 2 0 b 2 a - b b 2 - a 2 0 a 2 - b 2 a 2 b - a 1 b 2 a 2 Expanding ∆ along R 3 = a + b 2 a b 2 a - b 2 b - a + a 2 - b 2 2 = a + b 2 a b 4 a b - 2 b 2 - 2 a 2 + a b + a 4 + b 4 - 2 a 2 b 2 = a + b 2 5 a 2 b 2 - 2 a b 3 - 2 a 3 b + a 4 + b 4 - 2 a 2 b 2 = a + b 2 a 4 + b 4 - 2 a 3 b - 2 b 3 a + 3 a 2 b 2 = a + b 2 a 2 + b 2 2 - 2 a b a 2 + b 2 + a 2 b 2 = a 3 + b 3 2 = R.H.S Hence proved.

Prove that a 2 + 1 a b a c a b b 2 + 1 b c c a c b c 2 + 1 = 1 + a 2 + b 2 + c 2

Let L.H.S= ∆ = a 2 + 1 a b a c a b b 2 + 1 b c a c b 2 c 2 + 1 Applying R 1 a , R 2 b , R 3 c we get, = a + 1 a b c a b + 1 b c a b c + 1 c Applying a C 1 , b C 2 , C C 3 we get, = a 2 + 1 b 2 c 2 a 2 b 2 + 1 c 2 a 2 b 2 c 2 + 1 Applying C 1 → C 1 + C 2 + C 3 we get, = 1 + a 2 + b 2 + c 2 b 2 c 2 1 + a 2 + b 2 + c 2 b 2 + 1 c 2 1 + a 2 + b 2 + c 2 b 2 c 2 + 1 Taking 1 + a 2 + b 2 + c 2 common = 1 + a 2 + b 2 + c 2 1 b 2 c 2 1 b 2 + 1 c 2 1 b 2 c 2 + 1 Applying R 2 → R 2 - R 1 and R 3 → R 3 - R 1 we get ∆ = 1 + a 2 + b 2 + c 2 1 b 2 c 2 0 1 0 0 0 1 = 1 + a 2 + b 2 + c 2 1 0 0 1 = 1 + a 2 + b 2 + c 2 =R.H.S Hence proved.

Show that a + b + c - c - b - c a + b + c - a - b - a a + b + c = 2 ( a + b ) ( b + c ) ( c + a )

Let, LHS = ∆ = a + b + c - c - b - c a + b + c - a - b - a a + b + c Applying R 3 → R 3 + R 2 and R 1 → R 1 + R 2 we get, = a + b a + b - a + b - c a + b + c - a - b + c b + c b + c Taking a + b and b + c common from R 1 and R 3 = a + b b + c 1 1 - 1 - c a + b + c - a - 1 1 1 Applying R 1 → R 1 + R 3 we get = a + b b + c 0 2 0 - c a + b + c - a - 1 1 1 Expanding along R 1 = a + b b + c - 2 - a - c = 2 a + b b + c c + a = R.H.S Hence proved.

If a , b , c are real numbers such that b + c c + a a + b c + a a + b b + c a + b b + c c + a = 0 , then show that either a + b + c = 0 or, a = b = c

Let ∆ = b + c c + a a + b c + a a + b b + c a + b b + c c + a ⇒ ∆ = 2 ( a + b + c ) 2 ( c + a + b ) 2 ( a + b + c ) c + a a + b b + c a + b b + c c + a [ Applying R 1 → R 1 + R 2 + R 3 ] ⇒ ∆ = 2 ( a + b + c ) 1 1 1 c + a a + b b + c a + b b + c c + a ⇒ ∆ = 2 ( a + b + c ) 1 0 0 c + a b - c b - a a + b c - a c - b [ Applying C 2 → C 2 - C 1 and C 3 → C 3 - C 1 ] = 2 ( a + b + c ) 1 × b - c b - a c - a c - b = 2 ( a + b + c ) { ( b - c ) ( c - b ) - ( b - a ) ( c - a ) } = 2 ( a + b + c ) { c 2 - b 2 - b c + a b + a c + a 2 } = - ( a + b + c ) 2 a 2 + 2 b 2 + 2 c 2 - 2 a b - 2 b c - 2 c a = - ( a + b + c ) ( a - b ) 2 + ( b - c ) 2 + ( c - a ) 2 Given that Δ = 0 ⇒ - ( a + b + c ) ( a - b ) 2 + ( b - c ) 2 + ( c - a ) 2 = 0 ⇒ ( a + b + c ) = 0 or ( a - b ) 2 + ( b - c ) 2 + ( c - a ) 2 = 0 ⇒ ( a + b + c ) = 0 or a = b = c Hence proved.

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

Assume ∆ = 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 ⇒ ∆ = 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 Hence, proved.

E M B I B E

MATHEMATICS CLASS XII VOLUME-1>Chapter 6 - Determinants>EXERCISE

R D Sharma Solutions for Chapter: Determinants, Exercise 1: EXERCISE

Author:R D Sharma

R D Sharma Mathematics Solutions for Exercise - R D Sharma Solutions for Chapter: Determinants, Exercise 1: EXERCISE

Attempt the free practice questions on Chapter 6: Determinants, Exercise 1: EXERCISE with hints and solutions to strengthen your understanding. MATHEMATICS CLASS XII VOLUME-1 solutions are prepared by Experienced Embibe Experts.

Questions from R D Sharma Solutions for Chapter: Determinants, Exercise 1: EXERCISE with Hints & Solutions

EASY

12th CBSE

IMPORTANT

MATHEMATICS CLASS XII VOLUME-1>Chapter 6 - Determinants>EXERCISE>Q 1

Evaluate the following determinant:

|\begin{matrix} 1 & 3 & 9 & 27 \\ 3 & 9 & 27 & 1 \\ 9 & 27 & 1 & 3 \\ 27 & 1 & 3 & 9 \end{matrix}|

EASY

12th CBSE

IMPORTANT

MATHEMATICS CLASS XII VOLUME-1>Chapter 6 - Determinants>EXERCISE>Q 1

Evaluate the following determinant:

$|\begin{matrix} 102 & 18 & 36 \\ 1 & 3 & 4 \\ 17 & 3 & 6 \end{matrix}|$

HARD

12th CBSE

IMPORTANT

MATHEMATICS CLASS XII VOLUME-1>Chapter 6 - Determinants>EXERCISE>Q 28

Prove that

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

HARD

12th CBSE

IMPORTANT

MATHEMATICS CLASS XII VOLUME-1>Chapter 6 - Determinants>EXERCISE>Q 29

Prove that

$|\begin{matrix} a^{2} + 1 & a b & a c \\ a b & b^{2} + 1 & b c \\ c a & c b & c^{2} + 1 \end{matrix}| = 1 + a^{2} + b^{2} + c^{2}$

HARD

12th CBSE

IMPORTANT

MATHEMATICS CLASS XII VOLUME-1>Chapter 6 - Determinants>EXERCISE>Q 31

Show that

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

HARD

12th CBSE

IMPORTANT

MATHEMATICS CLASS XII VOLUME-1>Chapter 6 - Determinants>EXERCISE>Q 49

If $a, b, c$ are real numbers such that $|\begin{array}{l} b + c & c + a & a + b \\ c + a & a + b & b + c \\ a + b & b + c & c + a \end{array}| = 0,$ then show that either $a + b + c = 0$ or, $a = b = c$

HARD

12th CBSE

IMPORTANT

MATHEMATICS CLASS XII VOLUME-1>Chapter 6 - Determinants>EXERCISE>Q 50

If $|\begin{array}{l} p & b & c \\ a & q & c \\ a & b & r \end{array}| = 0$ find the value of $\frac{p}{p - a} + \frac{q}{q - b} + \frac{r}{r - c}, p \neq a, q \neq b, r \neq c$

EASY

12th CBSE

IMPORTANT

MATHEMATICS CLASS XII VOLUME-1>Chapter 6 - Determinants>EXERCISE>Q 55

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

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