Matrix Matching
If $X > Y$ and $X+Y$ is odd, then

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$- 1$

neither positive nor negative

${(- 1)}^{X} + {(- 1)}^{Y}$

${(- 1)}^{X - Y}$

${(- 1)}^{X Y}$

${(- 1)}^{X + Y}$

answer is C, B, A.

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Detailed Solution

$X > Y$ and $X+Y$ is odd

Let $X = 2, Y = 1$

$\begin{array}{l} O p t i o n (A) : {(- 1)}^{X} + {(- 1)}^{Y} \\ = {(- 1)}^{2} + {(- 1)}^{3} \\ = 1 + (- 1) \\ = 1 - 1 = 0 \end{array}$

0 is also called neither positive nor negative

$\begin{array}{l} O p t i o n (B) : {(- 1)}^{X - Y} \\ = {(- 1)}^{2 - 1} \\ = - 1 \end{array}$

$\begin{array}{l} O p t i o n (C) : {(- 1)}^{X Y} \\ = {(- 1)}^{2 \times 1} \\ = {(- 1)}^{2} \\ = 1 \end{array}$

$\begin{array}{l} O p t i o n (D) : {(- 1)}^{X + Y} \\ = {(- 1)}^{2 + 1} \\ = {(- 1)}^{3} \\ = - 1 \end{array}$

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