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Let A=[ x y z...

Question

Let $A = ⎡ ⎢ ⎣ \begin{matrix} x & y & z y & z & x z & x & y \end{matrix} ⎤ ⎥ ⎦$ , where $x, y$ and $z$ are real numbers such that $x + y + z > 0$ a $x y z = 2$ .
If $A^{2} = I_{3}$ , then the value of $x^{3} + y^{3} + z^{3}$ is

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Solution

$A = ⎡ ⎢ ⎣ \begin{matrix} x & y & z y & z & x z & x & y \end{matrix} ⎤ ⎥ ⎦$ $∴ | A | = (x^{3} + y^{3} + z^{3} - 3 x y z)$
$A^{2} = I_{3}$
$| A^{2} | = 1$
$∴ (x^{3} + y^{3} + z^{3} - 3 x y z)^{2} = 1$
$\Rightarrow x^{3} + y^{3} + z^{3} - 3 x y z = 1$ as $(x + y + z > 0)$
$\Rightarrow x^{3} + y^{3} + z^{3} = 6 + 1 = 7$

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Similar questions

A = ⎡ ⎢ ⎣ \begin{matrix} x & y & z y & z & x z & x & y \end{matrix} ⎤ ⎥ ⎦

x, y

z

are real numbers such that

x + y + z > 0

x y z = 2

A^{2} = I_{3}

, then the value of

x^{3} + y^{3} + z^{3}

x, y, z

be non-zero real numbers such that

\frac{x}{y} + \frac{y}{z} + \frac{z}{x} = 7

\frac{y}{x} + \frac{z}{y} + \frac{x}{z} = 9

\frac{x^{3}}{y^{3}} + \frac{y^{3}}{z^{3}} + \frac{z^{3}}{x^{3}} - 3

x, y, z

be positive real numbers such that

x + y + z = 12

x^{3} y^{4} z^{5} = (0.1) {(600)}^{3}

x^{3} + y^{3} + z^{3}

x^{3} + y^{3} + z^{3} = 6 x y z

, where x, y, z are non-zero integers,

x + y + z \neq 0

x y z = x + y + z

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(x - y)^{2} + (y - z)^{2} + (z - x)^{2}

x + y + z = 0

\frac{x^{3} + y^{3}}{x y z} = \frac{y^{3} + z^{3}}{x y z} = \frac{x^{3} + z^{3}}{x y z} = a

, then which of the following can be a?

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