If $n (A) = m, m > 0$ , then number of reflexive relations from $A$ to $A$ is

2^{m}

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2^{m^{2}}

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2^{m^{2} + m}

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2^{m^{2} - m}

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Solution

The correct option is D $2^{m^{2} - m}$
Let $A = {a_{1}, a_{2}, a_{3}, \dots ., a_{m}}$
$\Rightarrow A \times A = ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} (a_{1}, a_{1}), (a_{1}, a_{2}), \dots, (a_{1}, a_{m}) (a_{2}, a_{1}), (a_{2}, a_{2}), \dots, (a_{2}, a_{m}) . . . . (a_{m}, a_{1}), (a_{m}, a_{2}), \dots, (a_{m}, a_{m}) \end{matrix} ⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭$
$n (A \times A) = m^{2}$
Let us Consider elements in $A \times A$ as
$1)$ Diagonal elements $D_{1} = {(a_{1}, a_{1}), (a_{2}, a_{2}) . . . (a_{m}, a_{m})}$
$2)$ Non diagonal elements $D_{2} = {(a_{1}, a_{2}), (a_{2}, a_{1}), (a_{1}, a_{3}), (a_{3}, a_{1}) . . . (a_{m - 1}, a_{m}), (a_{m}, a_{m - 1})}$
$n (D_{1}) = m, n (D_{2}) = m^{2} - m$
Now for relation to be reflexive
$1)$ there is only one choice for diagonal elements i.e It has to be included to make relation reflexive
$2)$ for each of $m^{2} - m$ non diagonal elements, there are two choices either it can be included or excluded to make relation reflexive.
$∴$ from principle of counting, number of reflexive relations
$= 1 \times 1 \times . . . \times 1 (m times) \times 2 \times 2 \times . . . \times 2 (m^{2} - m times)$
$= 2^{m^{2} - m}$

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