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Lets take LHS, nbsp; LHS = nbsp;1+a nbsp; amp; 1 amp;1 nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; ...

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Standard XII

Mathematics

Addition of Vectors

Prove that be...

Question

Prove that
$∣ ∣ ∣ ∣ \begin{matrix} 1 + a & 1 & 1 1 & 1 + b & 1 1 & 1 & 1 + c \end{matrix} ∣ ∣ ∣ ∣$ = $a b c (1 + \frac{1}{a} + \frac{1}{b} + \frac{1}{c})$ .

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Solution

Lets take LHS,
LHS = $∣ ∣ ∣ ∣ \begin{matrix} 1 + a & 1 & 1 1 & 1 + b & 1 1 & 1 & 1 + c \end{matrix} ∣ ∣ ∣ ∣$

Take $a$ common from $R_{1}$ , $b$ common from $R_{2}$ and $c$ common from $R_{3}$ ,
we get,
LHS = $a b c ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 + \frac{1}{a} & \frac{1}{a} & \frac{1}{a} \frac{1}{b} & 1 + \frac{1}{b} & \frac{1}{b} \frac{1}{c} & \frac{1}{c} & 1 + \frac{1}{c} \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣$

Now, perform $R_{1} \to R_{1} + (R_{2} + R_{3})$
LHS = $a b c ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 + \frac{1}{a} + \frac{1}{b} + \frac{1}{c} & 1 + \frac{1}{a} + \frac{1}{b} + \frac{1}{c} & 1 + \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \frac{1}{b} & 1 + \frac{1}{b} & \frac{1}{b} \frac{1}{c} & \frac{1}{c} & 1 + \frac{1}{c} \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣$

Take $1 + \frac{1}{a} + \frac{1}{b} + \frac{1}{c}$ common from $R_{1}$
we get,
LHS = $a b c (1 + \frac{1}{a} + \frac{1}{b} + \frac{1}{c}) ∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & 1 & 1 \frac{1}{b} & 1 + \frac{1}{b} & \frac{1}{b} \frac{1}{c} & \frac{1}{c} & 1 + \frac{1}{c} \end{matrix} ∣ ∣ ∣ ∣ ∣$

Now, perform $C_{2} \to C_{2} - C_{1}$ and $C_{3} \to C_{3} - C_{1}$
we get,
LHS = $a b c (1 + \frac{1}{a} + \frac{1}{b} + \frac{1}{c}) ∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & 0 & 0 \frac{1}{b} & 1 & 0 \frac{1}{c} & 0 & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣$

LHS = $a b c (1 + \frac{1}{a} + \frac{1}{b} + \frac{1}{c}) (1)$
LHS = $a b c (1 + \frac{1}{a} + \frac{1}{b} + \frac{1}{c})$
LHS = RHS

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Similar questions

Q. Prove that

∣ ∣ ∣ ∣ \begin{matrix} 1 + a & 1 & 1 1 & 1 + b & 1 1 & 1 & 1 + c \end{matrix} ∣ ∣ ∣ ∣ = a b c (1 + \frac{1}{a} + \frac{1}{b} + \frac{1}{c})

Q. If a, b, c

> 0

then prove

(a b c) {(\frac{1}{a} + \frac{1}{b} + \frac{1}{c})}^{3} \geq 27

$Δ_{1}$ , $∣ ∣ ∣ ∣ \begin{matrix} 1 & 1 & 1 a & b & c a^{2} & b^{2} & c^{2} \end{matrix} ∣ ∣ ∣ ∣$ , $Δ_{2}$ = $∣ ∣ ∣ ∣ \begin{matrix} 1 & b c & a 1 & c a & b 1 & a b & c \end{matrix} ∣ ∣ ∣ ∣$ then

∣ ∣ ∣ ∣ \begin{matrix} 1 + a & 1 & 1 1 & 1 + b & 1 1 & 1 & 1 + c \end{matrix} ∣ ∣ ∣ ∣ = a b c (1 + \frac{1}{a} + \frac{1}{b} + \frac{1}{c})

Q. Show that

∣ ∣ ∣ ∣ \begin{matrix} 1 + a & 1 & 1 1 & 1 + b & 1 1 & 1 & 1 + c \end{matrix} ∣ ∣ ∣ ∣ = a b c (1 + \frac{1}{a} + \frac{1}{b} + \frac{1}{c}) = a b c + b c + c a + a b

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Prove that ∣∣ ∣∣1+a1111+b1111+c∣∣ ∣∣ = abc(1+1a+1b+1c).

Prove that
$∣ ∣ ∣ ∣ \begin{matrix} 1 + a & 1 & 1 1 & 1 + b & 1 1 & 1 & 1 + c \end{matrix} ∣ ∣ ∣ ∣$ = $a b c (1 + \frac{1}{a} + \frac{1}{b} + \frac{1}{c})$ .