
Let we have the sets as \[A = \left\{ {1,2,3} \right\},B = \left\{ {1,3,5} \right\}\]. If relation R from A to B is given by \[R = \left\{ {\left( {1,3} \right),\left( {2,5} \right),\left( {3,3} \right)} \right\}\]. Then \[{R^{ - 1}}\] is
1. \[\left\{ {\left( {3,3} \right),\left( {3,1} \right),\left( {5,2} \right)} \right\}\]
2. \[\left\{ {\left( {1,3} \right),\left( {2,5} \right),\left( {3,3} \right)} \right\}\]
3. \[\left\{ {\left( {1,3} \right),\left( {5,2} \right)} \right\}\]
4. None of these
Answer
497.4k+ views
Hint: We are given a relation and we are supposed to find its inverse. An inverse relation can be obtained by interchanging the values in each ordered pair in the relation. In the inverse of a relation, the x and y values are interchanged. If the values of both x and y are the same in an ordered pair, then the same appears in the inverse of the relation.
Complete step-by-step solution:
Suppose, x and y are two sets of ordered pairs. And set x has relation with set y, then the values of set x are called domain whereas the values of set y are called range.
An inverse relation is a set of ordered pairs obtained by interchanging the first and the second elements of each pair in the original function. If a function contains a point \[\left( {a,b} \right)\] then the inverse relation of this function contains the point\[\left( {b,a} \right)\].
Generally, the method of calculating an inverse is swapping of coordinates x and y. This newly created inverse is a relation but not necessarily a function.
The original function has to be a one-to-one function to assure that its inverse will also be a function. A function is said to be a one to one function only if every second element corresponds to the first value (values of x and y are used only once).
We are given \[A = \left\{ {1,2,3} \right\}\]
\[B = \left\{ {1,3,5} \right\}\]
\[R = \left\{ {\left( {1,3} \right),\left( {2,5} \right),\left( {3,3} \right)} \right\}\]
Therefore we get \[{R^{ - 1}} = \left\{ {\left( {3,3} \right),\left( {3,1} \right),\left( {5,2} \right)} \right\}\]
Therefore option (1) is the correct answer.
Note: A relation is the relationship between two or more sets of values. If your function is defined as a list of ordered pairs, simply swap the x and y values to find the inverse of relation. Remember, the inverse relation will be a function only if the original function is one-to-one and onto. If the graph of a function contains a point (a, b), then the graph of the inverse relation of this function contains the point (b, a).
Complete step-by-step solution:
Suppose, x and y are two sets of ordered pairs. And set x has relation with set y, then the values of set x are called domain whereas the values of set y are called range.
An inverse relation is a set of ordered pairs obtained by interchanging the first and the second elements of each pair in the original function. If a function contains a point \[\left( {a,b} \right)\] then the inverse relation of this function contains the point\[\left( {b,a} \right)\].
Generally, the method of calculating an inverse is swapping of coordinates x and y. This newly created inverse is a relation but not necessarily a function.
The original function has to be a one-to-one function to assure that its inverse will also be a function. A function is said to be a one to one function only if every second element corresponds to the first value (values of x and y are used only once).
We are given \[A = \left\{ {1,2,3} \right\}\]
\[B = \left\{ {1,3,5} \right\}\]
\[R = \left\{ {\left( {1,3} \right),\left( {2,5} \right),\left( {3,3} \right)} \right\}\]
Therefore we get \[{R^{ - 1}} = \left\{ {\left( {3,3} \right),\left( {3,1} \right),\left( {5,2} \right)} \right\}\]
Therefore option (1) is the correct answer.
Note: A relation is the relationship between two or more sets of values. If your function is defined as a list of ordered pairs, simply swap the x and y values to find the inverse of relation. Remember, the inverse relation will be a function only if the original function is one-to-one and onto. If the graph of a function contains a point (a, b), then the graph of the inverse relation of this function contains the point (b, a).
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