Let \[A\] and \[B\] be two sets such that \[n\left( A \right) = 3\] and\[n\left( B \right) = 2\]. If \[\left( {x,1} \right),\left( {y,2} \right),\left( {z,1} \right)\] are in \[A \times B\], write \[A\] and \[B\].
Answer
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Hint: In the given question, we have been given that there are two sets with some given number of elements. Then, we have been given some elements of the product of the two sets. And with that given information of the elements of the product of two sets, we have to deduce the elements of both the sets. A thing to note is that when two sets are multiplied, then the elements in the product form an ordered pair whose first element belongs to the elements of first set in the multiplication and the second element belongs to the elements of second set in the multiplication.
Formula Used:
Let there be two sets \[A\] and \[B\] with some elements in them. Let one element of \[A\] be \[x\] and one element of \[B\] be \[y\]. Then, \[A \times B\] will contain one ordered pair which is \[\left( {x,y} \right)\].
Complete step-by-step answer:
In the question, we have been given two sets \[A\] and \[B\] with \[3\] and \[2\] elements respectively.
In the product of them, there are going to be \[3 \times 2 = 6\] elements. Out of the \[6\] elements, we have been given \[3\] elements which are \[\left( {x,1} \right),\left( {y,2} \right),\left( {z,1} \right)\].
Now, in the \[3\] elements, each of the elements which is on the left of the bracket belongs to set \[A\] and the other to \[B\].
Clearly, the \[3\] elements of the product have \[3\] elements of set \[A\] as all the left elements of the \[3\] brackets are unique. Hence, the \[3\] elements which are all of \[A\] are \[A = \left\{ {x,y,z} \right\}\].
Similarly for \[B\], the \[2\] elements which are all of \[B\] are \[B = \left\{ {1,2} \right\}\].
Note: So, for solving questions of such type, we first write what has been given to us. Then we write down what we have to find. Then we think about the formulae which contain the known and the unknown and pick the one which is the most suitable and the most effective for finding the answer of the given question. Then we put in the knowns into the formula, evaluate the answer and find the unknown. It is really important to follow all the steps of the formula to solve the given expression very carefully and in the correct order, because even a slightest error is going to make the whole expression awry and is going to give us an incorrect answer.
Formula Used:
Let there be two sets \[A\] and \[B\] with some elements in them. Let one element of \[A\] be \[x\] and one element of \[B\] be \[y\]. Then, \[A \times B\] will contain one ordered pair which is \[\left( {x,y} \right)\].
Complete step-by-step answer:
In the question, we have been given two sets \[A\] and \[B\] with \[3\] and \[2\] elements respectively.
In the product of them, there are going to be \[3 \times 2 = 6\] elements. Out of the \[6\] elements, we have been given \[3\] elements which are \[\left( {x,1} \right),\left( {y,2} \right),\left( {z,1} \right)\].
Now, in the \[3\] elements, each of the elements which is on the left of the bracket belongs to set \[A\] and the other to \[B\].
Clearly, the \[3\] elements of the product have \[3\] elements of set \[A\] as all the left elements of the \[3\] brackets are unique. Hence, the \[3\] elements which are all of \[A\] are \[A = \left\{ {x,y,z} \right\}\].
Similarly for \[B\], the \[2\] elements which are all of \[B\] are \[B = \left\{ {1,2} \right\}\].
Note: So, for solving questions of such type, we first write what has been given to us. Then we write down what we have to find. Then we think about the formulae which contain the known and the unknown and pick the one which is the most suitable and the most effective for finding the answer of the given question. Then we put in the knowns into the formula, evaluate the answer and find the unknown. It is really important to follow all the steps of the formula to solve the given expression very carefully and in the correct order, because even a slightest error is going to make the whole expression awry and is going to give us an incorrect answer.
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