The value of the Determinant - - loga x-y loga y-z loga z-x- logb y-z logb z-x logb x-y- logc z-x logc x-y logc y-z - is equal to

The correct option is D None of these | [ #160; log_a( xy) amp;log_a( yz) amp;log_a( zx); #160; #160;log_b( yz) amp;log_b( zx) amp;log_ ...

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Byju's Answer

Standard XII

Mathematics

Definition of a Determinant

the value of ...

Question

the value of the Determinant $∣ ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} {log}_{a} (\frac{x}{y}) & {log}_{a} (\frac{y}{z}) & {log}_{a} (\frac{z}{x}) {log}_{b} (\frac{y}{z}) & {log}_{b} (\frac{z}{x}) & {log}_{b} (\frac{x}{y}) {log}_{c} (\frac{z}{x}) & {log}_{c} (\frac{x}{y}) & {log}_{c} (\frac{y}{z}) \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ ∣$ is equal to

1

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- 16

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{log}_{z} x y z

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None of these

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Solution

The correct option is D None of these
$∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} {log}_{a} (\frac{x}{y}) & {log}_{a} (\frac{y}{z}) & {log}_{a} (\frac{z}{x}) {log}_{b} (\frac{y}{z}) & {log}_{b} (\frac{z}{x}) & {log}_{b} (\frac{x}{y}) {log}_{c} (\frac{z}{x}) & {log}_{c} (\frac{x}{y}) & {log}_{c} (\frac{y}{z}) \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣$

Apply $C_{1} = C_{1} + C_{2} + C_{3}$

$∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} 0 & l o g_{a} (\frac{y}{z}) & l o g_{a} (\frac{z}{x}) 0 & l o g_{b} (\frac{z}{x}) & l o g_{b} (\frac{x}{y}) 0 & l o g_{c} (\frac{x}{y}) & l o g_{c} (\frac{y}{z}) \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣$

because $C_{1} + C_{2} + C_{3}$ will give three terms
$l o g x, l o g y, l o g z$ and three terms $= l o g x, - l o g y, - l o g z$
which will get cancelled
D is correct.

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Similar questions

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the value of the Determinant ∣∣ ∣ ∣ ∣ ∣∣loga(xy)loga(yz)loga(zx)logb(yz)logb(zx)logb(xy)logc(zx)logc(xy)logc(yz)∣∣ ∣ ∣ ∣ ∣∣ is equal to