SCARY BUT EASY

Algebra Level 5

$\Delta=\huge{\left| \begin{matrix} \frac { 1 }{ z } & \frac { 1 }{ z } & \frac { -(x+y) }{ { z }^{ 2 } } \\ \frac { -(y+z) }{ { x }^{ 2 } } & \frac { 1 }{ x } & \frac { 1 }{ x } \\ \frac { -y(y+z) }{ { x }^{ 2 }z } & \frac { (x+2y+z) }{ xz } & \frac { -y(x+y) }{ x{ z }^{ 2 } } \end{matrix} \right| }$

then $\Delta$ is

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dependent of

sin(cot^{-1}(tan(cos^{-1}(x)))),y,z

independent of

sin(cot^{-1}(tan(cos^{-1}(x)))),y,z

independent of

z

independent of

sin(cot^{-1}(tan(cos^{-1}(x)))),y

independent of

y,z

1 solution

Aastik Guru
Jun 11, 2020

Good problem @Tanishq Varshney The trick here is to multiply 1st row and 1st column by $x$ ,2nd row and 2nd column by $y$ and third row and third column by $z$ . Now simply add $C1+C2+C3$ to get the final answer as zero.

Hi there, Aastik Guru. That is a marvelous trick! I did it the long way and got $(xy^2+y^2z+yz^2-y^2-yz)/(x^3y^3)$ as the determinant. I guess I made a mistake. :(

James Wilson - 5 months ago

SCARY BUT EASY

1 solution

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