Without expanding, prove that a b c x y z p q r = x y z p q r a b c = y b q x a p z c r

Given that, ∆ = x y z p q r a b c R 2 ↔ R 3 = - x y z a b c p q r R 1 ↔ R 2 = a b c x y z p q r Now take y b q x a p z c r = y x z b a c q p r ( since A = A T ) C 1 ↔ C 2 = - x y z a b c p q r R 1 ↔ R 2 = a b c x y z p q r Hence proved.

Without expanding, prove that a b c x y z p q r = x y z p q r a b c = y b q x a p z c r

Given that, ∆ = x y z p q r a b c R 2 ↔ R 3 = - x y z a b c p q r R 1 ↔ R 2 = a b c x y z p q r Now take y b q x a p z c r = y x z b a c q p r ( since A = A T ) C 1 ↔ C 2 = - x y z a b c p q r R 1 ↔ R 2 = a b c x y z p q r Hence proved.

Show that x + 1 x + 2 x + a x + 2 x + 3 x + b x + 3 x + 4 x + c = 0 where a , b , c are in A.P

Given that a , b , c are in A.P ⇒ b = a + c 2 . Let ∆ = x + 1 x + 2 x + a x + 2 x + 3 x + b x + 3 x + 4 x + c Applying R 2 → 2 R 2 ∆ = 1 2 x + 1 x + 2 x + a 2 x + 4 2 x + 6 2 x + 2 b x + 3 x + 4 x + c Applying R 2 → R 2 - R 1 + R 3 ∆ = 1 2 x + 1 x + 2 x + a 0 0 2 b - a - c x + 3 x + 4 x + c ∆ = 1 2 x + 1 x + 2 x + a 0 0 0 x + 3 x + 4 x + c [ ∵ 2 b = a + c ] ∆ = 0 Thus we get x + 1 x + 2 x + a x + 2 x + 3 x + b x + 3 x + 4 x + c = 0 where a , b , c are in A.P

Show that x - 3 x - 4 x - α x - 2 x - 3 x - β x - 1 x - 2 x - γ = 0 , where α , β , γ are in A.P.

Given that α , β , γ are in A.P ⇒ 2 β = α + γ Let ∆ = x - 3 x - 4 x - α x - 2 x - 3 x - β x - 1 x - 2 x - γ ∆ = x - 3 x - 4 x - α 2 x - 4 2 x - 6 2 x - 2 β x - 1 x - 2 x - γ [ Applying R 2 → 2 R 2 ] ∆ = x - 3 x - 4 x - α 0 0 - 2 β + α + γ x - 1 x - 2 x - γ [ Applying R 2 → R 2 - ( R 1 + R 3 ) ] ∆ = x - 3 x - 4 x - α 0 0 0 x - 1 x - 2 x - γ [ ∵ 2 β = α + γ ] ∆ = 0 Thus we get x - 3 x - 4 x - α x - 2 x - 3 x - β x - 1 x - 2 x - γ = 0 , where α , β , γ are in A.P.

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.

Show that x = 2 is a root of the equation x - 6 - 1 2 - 3 x x - 3 - 3 2 x x + 2 = 0 and solve it completely.

Let ∆ = x - 6 - 1 2 - 3 x x - 3 - 3 2 x x + 2 = x - 6 - 1 2 - 3 x x - 3 - 3 - x 2 x + 6 x + 3 [Applying R 3 → R 3 - R 1 ] = ( x + 3 ) x - 6 - 1 2 - 3 x x - 3 - 1 2 1 = ( x + 3 ) x - 2 3 x - 6 - x + 2 2 - 3 x x - 3 - 1 2 1 [ Applying R 1 → R 1 - R 2 ] = ( x + 3 ) ( x - 2 ) 1 3 - 1 2 - 3 x x - 3 - 1 2 1 = ( x + 3 ) ( x - 2 ) 1 3 0 2 - 3 x x - 1 - 1 2 0 [Applying C 3 → C 3 + C 1 ] = ( x + 3 ) ( x - 2 ) ( x - 1 ) 1 3 - 1 2 [Expanding along C 3 ] = - 5 ( x + 3 ) ( x - 2 ) ( x - 1 ) . We get - 5 ( x + 3 ) ( x - 2 ) ( x - 1 ) = 0 [ Given condition ] ⇒ x = 2 or x = - 3 or x = 1 . Thus we have shown that x = 2 is a root of x - 6 - 1 2 - 3 x x - 3 - 3 2 x x + 2 = 0 .

Solve the following determinant equation: x + a b c a x + b c a b x + c = 0

Let ∆ = x + a b c a x + b c a b x + c = x + a + b + c b c x + a + b + c x + b c x + a + b + c b x + c [Applying C 1 → C 1 + C 2 + C 3 ] = x + a + b + c 1 b c 1 x + b c 1 b x + c = x + a + b + c 1 b c 0 x 0 1 b x + c [Applying R 2 → R 2 - R 1 ] = x + a + b + c 1 b c 0 x 0 0 0 x [Applying R 3 → R 3 - R 1 ] ∆ = 0 + 0 + x + a + b + c x 2 - 0 [ Expanding along R 3 ] Thus we get x + a + b + c x 2 - 0 = 0 ⇒ x 2 = 0 or x + a + b + c = 0 ⇒ x = 0 or x = - a + b + c

Solve the following determinant equation: x + a x x x x + a x x x x - a = 0 , a ≠ 0 .

Let ∆ = x + a x x x x + a x x x x - a = 3 x + a x x 3 x + a x + a x 3 x + a x x - a [Applying C 1 → C 1 + C 2 + C 3 ] = 3 x + a 1 x x 1 x + a x 1 x x - a = 3 x + a 1 x x 0 a 0 1 x x - a [Applying R 2 → R 2 - R 1 ] = 3 x + a 1 x x 0 a 0 0 0 a [Applying R 3 → R 3 - R 1 ] = 3 x + a a 2 - 0 [ Expanding along C 1 ] From the given condition, we get ∆ = ( 3 x + a ) a 2 = 0 Thus, x = - a 3 .

Solve the following determinant equation: 3 x - 8 3 3 3 3 x - 8 3 3 3 3 x - 8 = 0 .

Let ∆ = 3 x - 8 3 3 3 3 x - 8 3 3 3 3 x - 8 = 3 x - 2 3 3 3 x - 2 3 x - 8 3 3 x - 2 3 3 x - 8 [Applying C 1 → C 1 + C 2 + C 3 ] = 3 x - 2 1 3 3 1 3 x - 8 3 1 3 3 x - 8 = 3 x - 2 1 3 3 0 3 x - 11 0 1 3 3 x - 8 [Applying R 2 → R 2 - R 1 ] = 3 x - 2 1 3 3 0 3 x - 11 0 0 0 3 x - 11 [Applying R 3 → R 3 - R 1 ] = 3 x - 2 3 x - 11 2 = 0 [ Expanding along C 1 ] From the given condition, we get ∆ = ( 3 x - 2 ) ( 3 x - 11 ) 2 ⇒ x = 2 3 or x = 11 3 .

E M B I B E

MATHEMATICS CLASS XII VOLUME-1>Chapter 6 - Determinants>EXERCISE>Q 46

EASY

12th CBSE

IMPORTANT

Earn 100

Without expanding, prove that

$|\begin{matrix} a & b & c \\ x & y & z \\ p & q & r \end{matrix}| = |\begin{matrix} x & y & z \\ p & q & r \\ a & b & c \end{matrix}| = |\begin{matrix} y & b & q \\ x & a & p \\ z & c & r \end{matrix}|$

Important Questions on Determinants

MEDIUM

12th CBSE

IMPORTANT

MATHEMATICS CLASS XII VOLUME-1>Chapter 6 - Determinants>EXERCISE>Q 47

Show that

|\begin{matrix} x + 1 & x + 2 & x + a \\ x + 2 & x + 3 & x + b \\ x + 3 & x + 4 & x + c \end{matrix}| = 0

where

a, b, c

are in A.P

MEDIUM

12th CBSE

IMPORTANT

MATHEMATICS CLASS XII VOLUME-1>Chapter 6 - Determinants>EXERCISE>Q 48

Show that

|\begin{matrix} x - 3 & x - 4 & x - α \\ x - 2 & x - 3 & x - β \\ x - 1 & x - 2 & x - γ \end{matrix}| = 0

, where

α, β, γ

are in A.P.

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

MEDIUM

12th CBSE

IMPORTANT

MATHEMATICS CLASS XII VOLUME-1>Chapter 6 - Determinants>EXERCISE>Q 51

Show that $x = 2$ is a root of the equation $|\begin{array}{r} x & - 6 & - 1 \\ 2 & - 3 x & x - 3 \\ - 3 & 2 x & x + 2 \end{array}| = 0$ and solve it completely.