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

Given determinant is b + c a - b a c + a b - c b a + b c - a c Applying C 2 → C 2 - C 3 = b + c - b a c + a - c b a + b - a c C 1 → C + C 2 = c - b a a - c b b - a c Taking - 1 common from C 2 = - 1 c b a a c b b a c Expanding along R 1 = - 1 c c 2 - a b - b a c - b 2 + a a 2 - b c = - 1 c 3 - a b c - b a c + b 3 + a 3 - a b c ∴ b + c a - b a c + a b - c b a + b c - a c = 3 a b c - a 3 - b 3 - c 3 Hence proved.

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

Given determinant is b + c a - b a c + a b - c b a + b c - a c Applying C 2 → C 2 - C 3 = b + c - b a c + a - c b a + b - a c C 1 → C + C 2 = c - b a a - c b b - a c Taking - 1 common from C 2 = - 1 c b a a c b b a c Expanding along R 1 = - 1 c c 2 - a b - b a c - b 2 + a a 2 - b c = - 1 c 3 - a b c - b a c + b 3 + a 3 - a b c ∴ b + c a - b a c + a b - c b a + b c - a c = 3 a b c - a 3 - b 3 - c 3 Hence proved.

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

Let Δ = a + b b + c c + a b + c c + a a + b c + a a + b b + c We know that if each element of a row or column is expressed as the sum of two or more quantities, the determinant is expressed as the sum of two or more determinants. ∴ Δ = a b c b c a c a b + b c a c a b a b c Applying C 1 ↔ C 3 in second determinant to get negative value of the deteminant = a b c b c a c a b + - 1 a c b b a c c b a = a b c b c a c a b + - 1 - 1 a b c b c a c a b Applying C 2 ↔ C 3 = 2 a b c b c a c a b = RHS

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

L.H.S = a + b + 2 c a b c b + c + 2 a b c a c + a + 2 b Applying C 1 → C 1 + C 2 + C 3 , we have = 2 ( a + b + c ) a b 2 ( a + b + c ) b + c + 2 a b 2 ( a + b + c ) a c + a + 2 b Taking, 2 ( a + b + c ) common we get, = 2 ( a + b + c ) 1 a b 1 b + c + 2 a b 1 a c + a + 2 b Now, applying R 2 → R 2 - R 1 and R 3 → R 3 - R 1 we get, = 2 ( a + b + c ) 1 a b 0 b + c + a 0 0 0 c + a + b Expanding along C 1 = 2 ( a + b + c ) ( a + b + c ) 2 = 2 ( a + b + c ) 3 = R · H · S Hence proved.

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

Consider, L.H.S = a - b - c 2 a 2 a 2 b b - c - a 2 b 2 c 2 c c - a - b Now by applying, R 1 → R 1 + R 2 + R 3 , we get, = a + b + c a + b + c a + b + c 2 b b - c - a 2 b 2 c 2 c c - a - b Taking ( a + b + c ) common we get, = ( a + b + c ) 1 1 1 2 b b - c - a 2 b 2 c 2 c c - a - b Applying C 2 → C 2 - C 1 and C 3 → C 3 - C 1 we get, = ( a + b + c ) 1 0 0 2 b - b - c - a 0 2 c 0 - c - a - b = ( a + b + c ) 1 0 0 2 b b + c + a 0 2 c 0 b + c + a = ( a + b + c ) 3 = R · H · S Hence proved.

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

Consider, L.H.S = 1 b + c b 2 + c 2 1 c + a c 2 + a 2 1 a + b a 2 + b 2 Now by applying, R 2 → R 2 - R 1 and R 3 → R 3 - R 1 , we get, = 1 b + c b 2 + c 2 0 a - b a 2 - b 2 0 a - c a 2 - c 2 = ( a - b ) ( a - c ) 1 b + c b 2 + c 2 0 1 a + b 0 1 a + c Again by applying R 3 → R 3 - R 2 , we get, = ( a - b ) ( a - c ) 1 b + c b 2 + c 2 0 1 a + b 0 0 c - a = ( a - b ) ( a - c ) ( b - c ) = R · H · S Hence Proved.

Prove a a + b a + 2 b a + 2 b a a + b a + b a + 2 b a = 9 a + b b 2

Consider, L.H.S = a a + b a + 2 b a + 2 b a a + b a + b a + 2 b a Applying R 1 → R 1 + R 2 + R 3 we get, = 3 a + 3 b 3 a + 3 b 3 a + 3 b a + 2 b a a + b a + b a + 2 b a Taking, ( 3 a + 3 b ) common we get, = ( 3 a + 3 b ) 1 1 1 a + 2 b a a + b a + b a + 2 b a Applying, C 1 → C 1 - C 2 and C 3 → C 3 - C 2 we get, = ( 3 a + 3 b ) 0 1 0 2 b a b - b a + 2 b - 2 b = ( 3 a + 3 b ) b 2 0 1 0 2 a 1 - 1 a + 2 b - 2 = 3 ( a + b ) b 2 ( 3 ) = 9 ( a + b ) b 2 = R . H . S Hence proved.

Prove 1 a b c 1 b c a 1 c a b = 1 a a 2 1 b b 2 1 c c 2

Consider, L. H. S = 1 a b c 1 b c a 1 c a b Now by applying, R 1 → a R 1 , R 2 → b R 2 , R 3 → c R 3 We get, = 1 a b c a a 2 a b c b b 2 c a b c c 2 a b c = a b c a b c a a 2 1 b b 2 1 c c 2 1 = - a 1 a 2 b 1 b 2 c 1 c 2 = 1 a a 2 1 b b 2 1 c c 2 Hence proved.

Prove z x y z 2 x 2 y 2 z 4 x 4 y 4 = x y z x 2 y 2 z 2 x 4 y 4 z 4 = x 2 y 2 z 2 x 4 y 4 z 4 x y z = x y z x - y y - z z - x x + y + z

Given, z x y z 2 x 2 y 2 z 4 x 4 y 4 = x y z x 2 y 2 z 2 x 4 y 4 z 4 = x 2 y 2 z 2 x 4 y 4 z 4 x y z = x y z ( x - y ) ( y - z ) ( z - x ) ( x + y + z ) Consider, x y z x 2 y 2 z 2 x 4 y 4 z 4 By taking x y z common = x y z 1 1 1 x y z x 3 y 3 z 3 By applying c 1 → c 1 - c 2 and c 3 → c 3 - c 2 we get, = x y z 0 1 0 x - y y z - y x 3 - y 3 y 3 z 3 - y 3 = x y z ( x - y ) ( z - y ) 0 1 0 1 y 1 x 2 + y 2 + x y y 3 z 2 + y 2 + z y = - x y z ( x - y ) ( z - y ) z 2 + y 2 + z y - x 2 - y 2 - x y = - x y z ( x - y ) ( z - y ) [ ( z - x ) ( z + x 0 + y ( z - x ) ] = x y z ( x - y ) ( y - z ) ( z - x ) ( x + y + z ) = R . H . S Hence proved.

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

Consider, L . H . S = ( b + c ) 2 a 2 b c ( c + a ) 2 b 2 c a ( a + b ) 2 c 2 a b Applying, C 1 → C 1 + C 2 - 2 C 3 = ( b + c ) 2 - a 2 - 2 b c a 2 b c ( c + a ) 2 - b 2 - 2 c a b 2 c a ( a + b ) 2 - c 2 - 2 a b c 2 a b = a 2 + b 2 + c 2 a 2 b c a 2 + b 2 + c 2 b 2 c a a 2 + b 2 + c 2 c 2 a b Taking a 2 + b 2 + c 2 , common, we get = 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 get = 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 ( b - a ) ( c - a ) 1 a 2 b c 0 b + a - c 0 c + a - b = a 2 + b 2 + c 2 ( b - a ) ( c - a ) [ ( b + a ) ( - b ) - ( - c ) ( c + a ) ] = a 2 + b 2 + c 2 ( a - b ) ( c - a ) ( b - c ) ( a + b + c ) = R . H . S Hence proved.

E M B I B E

MATHEMATICS CLASS XII VOLUME-1>Chapter 6 - Determinants>EXERCISE>Q 12

MEDIUM

12th CBSE

IMPORTANT

Earn 100

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

Important Questions on Determinants

HARD

12th CBSE

IMPORTANT

MATHEMATICS CLASS XII VOLUME-1>Chapter 6 - Determinants>EXERCISE>Q 13

Prove that:

$= - 1 [c^{3} - a b c - b a c + b^{3} + a^{3} - a b c]$

MEDIUM

12th CBSE

IMPORTANT

MATHEMATICS CLASS XII VOLUME-1>Chapter 6 - Determinants>EXERCISE>Q 14

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

MEDIUM

12th CBSE

IMPORTANT

MATHEMATICS CLASS XII VOLUME-1>Chapter 6 - Determinants>EXERCISE>Q 15

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

EASY

12th CBSE

IMPORTANT

MATHEMATICS CLASS XII VOLUME-1>Chapter 6 - Determinants>EXERCISE>Q 16

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

MEDIUM

12th CBSE

IMPORTANT

MATHEMATICS CLASS XII VOLUME-1>Chapter 6 - Determinants>EXERCISE>Q 17

Prove $|\begin{matrix} a & a + b & a + 2 b \\ a + 2 b & a & a + b \\ a + b & a + 2 b & a \end{matrix}| = 9 (a + b) b^{2}$

MEDIUM

12th CBSE

IMPORTANT

MATHEMATICS CLASS XII VOLUME-1>Chapter 6 - Determinants>EXERCISE>Q 18

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

MEDIUM

12th CBSE

IMPORTANT

MATHEMATICS CLASS XII VOLUME-1>Chapter 6 - Determinants>EXERCISE>Q 19

Prove $|\begin{matrix} z & x & y \\ z^{2} & x^{2} & y^{2} \\ z^{4} & x^{4} & y^{4} \end{matrix}| = |\begin{matrix} x & y & z \\ x^{2} & y^{2} & z^{2} \\ x^{4} & y^{4} & z^{4} \end{matrix}| = |\begin{matrix} x^{2} & y^{2} & z^{2} \\ x^{4} & y^{4} & z^{4} \\ x & y & z \end{matrix}| = x y z (x - y) (y - z) (z - x) (x + y + z)$

MEDIUM

12th CBSE

IMPORTANT

MATHEMATICS CLASS XII VOLUME-1>Chapter 6 - Determinants>EXERCISE>Q 20

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