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

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Byju's Answer

Standard XII

Mathematics

Evaluation of a Determinant

Prove that -b...

Question

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

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Solution

$∆ = |\begin{matrix} - b c & b^{2} + b c & c^{2} + b c \\ a^{2} + a c & - a c & c^{2} + a c \\ a^{2} + a b & b^{2} + a b & - a b \end{matrix}| = \frac{1}{a b c} |\begin{matrix} - a b c & a b^{2} + a b c & a c^{2} + a b c \\ a^{2} b + a b c & - a b c & c^{2} b + a b c \\ a^{2} c + a b c & b^{2} c + a b c & - a b c \end{matrix}| [Applying R_{1} \to a R_{1}, R_{2} \to b R_{2} and R_{3} \to c R_{3} and then dividing by a b c] = \frac{a b c}{a b c} |\begin{matrix} - b c & a b + a c & a c + a b \\ a b + b c & - a c & c b + a b \\ a c + b c & b c + a c & - a b \end{matrix}| [Taking out a, b and c common from the three columns] |\begin{matrix} a b + b c + c a & a b + b c + c a & a b + b c + c a \\ a b + b c & - a c & c b + a b \\ a c + b c & b c + a c & - a b \end{matrix}| [Applying R_{1} \to R_{1} + R_{2} + R_{3}] = (a b + b c + c a) |\begin{matrix} 1 & 1 & 1 \\ a b + b c & - a c & c b + a b \\ a c + b c & b c + a c & - a b \end{matrix}| = (a b + b c + c a) |\begin{matrix} 0 & 0 & 1 \\ 0 & - (a b + b c + a c) & c b + a b \\ a c + b c + a b & b c + a c + a b & - a b \end{matrix}| [Applying C_{1} \to C_{1} - C_{3} and C_{2} \to C_{2} - C_{3}] = (a b + b c + c a) |\begin{matrix} 0 & - (a b + b c + a c) \\ a c + b c + a b & b c + a c + a b \end{matrix}| = (a b + b c + c a) (a b + b c + a c)^{2} = (a b + b c + c a)^{3}$

Hence proved.

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Similar questions

Q. If none of

a, b, c

is zero, Whether the given equation

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

is ?

Q. Evaluate:

∣ ∣ ∣ ∣ ∣ \begin{matrix} - b c & b^{2} + b c & c^{2} + b c a^{2} + a c & - a c & c^{2} + a c a^{2} a b & b^{2} + a b & - a b \end{matrix} ∣ ∣ ∣ ∣ ∣

Let

Δ = ∣ ∣ ∣ ∣ ∣ \begin{matrix} - b c & b^{2} + b c & c^{2} + b c a^{2} + a c & - a c & c^{2} + a c a^{2} + a b & b^{2} + a b & - a b \end{matrix} ∣ ∣ ∣ ∣ ∣

and the equation

p x^{3} + q x^{2} + r x + s = 0

has roots

a, b, c

, where

a, b, c \in R^{+}

.The value of

Δ

Q. Let

Δ = ∣ ∣ ∣ ∣ ∣ \begin{matrix} - b c & b^{2} + b c & c^{2} + b c a^{2} + a c & - a c & c^{2} + a c a^{2} + a b & b^{2} + a b & - a b \end{matrix} ∣ ∣ ∣ ∣ ∣

and the equation

p x^{2} + q x^{2} + r x + s = 0

has roots,

a, b, c

where

a, b, c \in R^{+}

Δ = 27

and

a^{2} + b^{2} + c^{2} = 3,

then

Q. Let

a, b, c

be the roots of the equation

p x^{3} + q x^{2} + r x + s = 0

. If

∣ ∣ ∣ ∣ ∣ \begin{matrix} - b c & b^{2} + b c & c^{2} + b c a^{2} + a c & - a c & c^{2} + a c a^{2} + a b & b^{2} + a b & - a b \end{matrix} ∣ ∣ ∣ ∣ ∣ = 27, a + b + c \geq 0

and

a^{2} + b^{2} + c^{2} = 3,

then the value of

3 p + q

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Prove that -bcb2+bcc2+bca2+ac-acc2+aca2+abb2+ab-ab=ab+bc+ca3

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