If x- y and x- y are real numbers- and A - x2 - y2 - xy - x - y - 1- then

The correct option is A A > 0 A = 12 (2x^2 + 2y^2 2xy 2x 2y + 2) = 12 [(x y)^2 + (x 1)^2 + (y 1)^2] gt; 0 . ...

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If x≠ y and...

Question

If $x \neq y$ and $x, y$ are real numbers; and $A = x^{2} + y^{2} - x y - x - y + 1$ , then

A > 0

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A = 0

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A < 0

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0 < A < 1

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O

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If x,y, z are non-zero real numbers, then the inverse of matrix

$A = ⎡ ⎢ ⎣ \begin{matrix} x & 0 & 0 0 & y & 0 0 & 0 & z \end{matrix} ⎤ ⎥ ⎦$ is

a) $⎡ ⎢ ⎣ \begin{matrix} x^{- 1} & 0 & 0 0 & y^{- 1} & 0 0 & 0 & z^{- 1} \end{matrix} ⎤ ⎥ ⎦$

b) $x y z ⎡ ⎢ ⎣ \begin{matrix} x^{- 1} & 0 & 0 0 & y^{- 1} & 0 0 & 0 & z^{- 1} \end{matrix} ⎤ ⎥ ⎦$

c) $\frac{1}{x y z} ⎡ ⎢ ⎣ \begin{matrix} x & 0 & 0 0 & y & 0 0 & 0 & z \end{matrix} ⎤ ⎥ ⎦$

d) $\frac{1}{x y z} ⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 0 & 1 & 0 0 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦$

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