CBSE Class 11 Mathematics Relations and Functions Assignment Set 08

A-Cross - Product

Question 1. If \(\left(\frac{x}{3} - 1, y + \frac{2}{3}\right) = \left(\frac{5}{3}, \frac{1}{3}\right)\), find \(x, y\).
Answer: Equating the corresponding coordinates of the given ordered pairs:
\(\frac{x}{3} - 1 = \frac{5}{3}\)
\(\frac{x}{3} = \frac{5}{3} + 1 = \frac{8}{3}\)
\(x = 8\)
And for the second coordinate:
\(y + \frac{2}{3} = \frac{1}{3}\)
\(y = \frac{1}{3} - \frac{2}{3} = -\frac{1}{3}\)
Therefore, \((x, y) = \left(8, -\frac{1}{3}\right)\).
In simple words: Set the x-parts equal to each other and the y-parts equal to each other to solve for x and y.

Exam Tip: Two ordered pairs are equal if and only if their corresponding first and second elements are equal.

Question 2. Let \(A = \{1, 3, 5\}\), \(B = \{2, 4, 6\}\), find \(A \times B\) also \(B \times A\). Is \(A \times B = B \times A\)? Also find \(n(A \times B)\).
Answer:
\(A \times B = \{(1, 2), (1, 4), (1, 6), (3, 2), (3, 4), (3, 6), (5, 2), (5, 4), (5, 6)\}\)
\(B \times A = \{(2, 1), (2, 3), (2, 5), (4, 1), (4, 3), (4, 5), (6, 1), (6, 3), (6, 5)\}\)
Since the ordered pairs in \(A \times B\) and \(B \times A\) are not identical, \(A \times B \neq B \times A\).
The number of elements is \(n(A \times B) = 3 \times 3 = 9\).
In simple words: The Cartesian product pairs every element of the first set with every element of the second set. Order matters in ordered pairs.

Exam Tip: Remember that Cartesian product is not commutative, meaning \(A \times B \neq B \times A\) for non-equal sets.

Question 3. If \(n(A) = 10\), \(n(B) = 5\), find \(n(A \times B)\), \(n(B \times A)\). Is \(n(A \times B) = n(B \times A)\)?
Answer:
\(n(A \times B) = n(A) \times n(B) = 10 \times 5 = 50\)
\(n(B \times A) = n(B) \times n(A) = 5 \times 10 = 50\)
Yes, \(n(A \times B) = n(B \times A) = 50\).
In simple words: The number of elements in a Cartesian product is the product of the cardinalities of the two sets.

Exam Tip: While the sets of ordered pairs \(A \times B\) and \(B \times A\) are different, their cardinalities (number of elements) are always equal.

Question 4. (i) If set \(A = \phi\), set \(B = \phi\), find \(A \times B\).
(ii) If \(A \times B \neq \phi\) then discuss about set \(A, B\).
(iii) If \(A\) and \(B\) are non-empty sets and any one of the two sets is infinite then what you can say about \(A \times B\)?
Answer:
(i) If \(A = \phi\) and \(B = \phi\), then \(A \times B = \phi \times \phi = \phi\).
(ii) If \(A \times B \neq \phi\), then neither set \(A\) can be empty nor set \(B\) can be empty. Therefore, \(A \neq \phi\) and \(B \neq \phi\).
(iii) If \(A\) and \(B\) are non-empty sets and at least one of them is infinite, then their Cartesian product \(A \times B\) is an infinite set.
In simple words: The cross product of empty sets is empty. If the product has elements, both sets must have elements. Infinite sets make infinite products.

Exam Tip: \(A \times B = \phi\) if and only if \(A = \phi\) or \(B = \phi\).

Question 5. If \(A \times B = \{(a, b), (a, c), (d, b), (d, c)\}\), find sets \(A\) and \(B\).
Answer:
Set \(A\) consists of all first coordinates of the ordered pairs in \(A \times B\):
\(A = \{a, d\}\)
Set \(B\) consists of all second coordinates of the ordered pairs in \(A \times B\):
\(B = \{b, c\}\)
In simple words: Extract all the first elements to get set A, and all the second elements to get set B.

Exam Tip: When listing elements of sets \(A\) and \(B\), ensure duplicates are removed since sets contain distinct elements.

Question 6. Let \(A = \{a, b\}\), find \(A \times A \times A\).
Answer:
First find \(A \times A = \{(a, a), (a, b), (b, a), (b, b)\}\).
Then find \(A \times A \times A = (A \times A) \times A\):
\(A \times A \times A = \{(a, a, a), (a, a, b), (a, b, a), (a, b, b), (b, a, a), (b, a, b), (b, b, a), (b, b, b)\}\)
In simple words: For a triple Cartesian product, form all possible ordered triplets by taking one element from A for each position.

Exam Tip: If \(n(A) = k\), then the number of elements in the Cartesian product of \(m\) sets is \(k^m\). Here \(2^3 = 8\) triplets.

Question 7. Let \(A = \{1, 2, 3\}\), \(B = \{1, 2, 3, 4\}\), \(C = \{4, 5\}\). Verify:
(i) \(A \times (B \cap C) = (A \times B) \cap (A \times C)\)
(ii) \(A \times (B \cup C) = (A \times B) \cup (A \times C)\)
(iii) \(A \times (B - C) = (A \times B) - (A \times C)\)
Answer:
Given sets: \(A = \{1, 2, 3\}\), \(B = \{1, 2, 3, 4\}\), \(C = \{4, 5\}\).

(i) Verification of \(A \times (B \cap C) = (A \times B) \cap (A \times C)\):
\(B \cap C = \{4\}\)
LHS: \(A \times (B \cap C) = \{1, 2, 3\} \times \{4\} = \{(1, 4), (2, 4), (3, 4)\}\)
RHS: \((A \times B) \cap (A \times C)\)
\(A \times B = \{(1,1), (1,2), (1,3), (1,4), (2,1), (2,2), (2,3), (2,4), (3,1), (3,2), (3,3), (3,4)\}\)
\(A \times C = \{(1,4), (1,5), (2,4), (2,5), (3,4), (3,5)\}\)
Intersection gives: \{(1, 4), (2, 4), (3, 4)\}
Thus, LHS = RHS.

(ii) Verification of \(A \times (B \cup C) = (A \times B) \cup (A \times C)\):
\(B \cup C = \{1, 2, 3, 4, 5\}\)
LHS: \(A \times (B \cup C) = \{1, 2, 3\} \times \{1, 2, 3, 4, 5\}\)
\(= \{(1,1), (1,2), (1,3), (1,4), (1,5), (2,1), (2,2), (2,3), (2,4), (2,5), (3,1), (3,2), (3,3), (3,4), (3,5)\}\)
RHS: \((A \times B) \cup (A \times C)\) yields the exact same set of 15 ordered pairs.
Thus, LHS = RHS.

(iii) Verification of \(A \times (B - C) = (A \times B) - (A \times C)\):
\(B - C = \{1, 2, 3\}\)
LHS: \(A \times (B - C) = \{1, 2, 3\} \times \{1, 2, 3\} = \{(1,1), (1,2), (1,3), (2,1), (2,2), (2,3), (3,1), (3,2), (3,3)\}\)
RHS: Removing elements of \(A \times C\) from \(A \times B\) leaves precisely those pairs where the second coordinate is not in \(C\), which gives the same set.
Thus, LHS = RHS.
In simple words: Cartesian product distributes over set intersection, union, and set difference.

Exam Tip: These distributive properties of Cartesian product over set operations are fundamental identities in set theory.

Question 8. The Cartesian product \(A \times A\) has 9 elements among which are found \((-1, 0)\) and \((0, 1)\). Find the set \(A\) and the remaining elements of \(A \times A\).
Answer:
Since \(n(A \times A) = 9\), we have \(n(A) = \sqrt{9} = 3\).
The ordered pairs \((-1, 0)\) and \((0, 1)\) belong to \(A \times A\), which implies that \(-1, 0, 1 \in A\).
Since \(n(A) = 3\) and we have found 3 distinct elements \(\{-1, 0, 1\}\), the set must be:
\(A = \{-1, 0, 1\}\)
Now, \(A \times A = \{-1, 0, 1\} \times \{-1, 0, 1\}\)
\(A \times A = \{(-1,-1), (-1,0), (-1,1), (0,-1), (0,0), (0,1), (1,-1), (1,0), (1,1)\}\)
The remaining elements of \(A \times A\) (excluding the two given ones) are:
\(\{(-1,-1), (-1,1), (0,-1), (0,0), (1,-1), (1,0), (1,1)\}\)
In simple words: Find all unique numbers appearing in the given coordinates to form set A, then generate all 9 pairs of A x A.

Exam Tip: Use the cardinality formula \(n(A \times A) = [n(A)]^2\) to determine the size of set A first.

Question 9. Let \(A\) and \(B\) be the two sets such that \(n(A) = 3\) and \(n(B) = 2\). If \((a, 1), (b, 2), (c, 1)\) are in \(A \times B\), find \(A\) and \(B\) where \(a, b, c\) are distinct elements.
Answer:
Set \(A\) consists of the first coordinates of the elements in \(A \times B\):
\(A = \{a, b, c\}\)
Set \(B\) consists of the second coordinates of the elements in \(A \times B\):
\(B = \{1, 2\}\)
In simple words: Collect all first elements for set A and all second elements for set B.

Exam Tip: Verify that the cardinalities match: \(n(A) = 3\) and \(n(B) = 2\), which is consistent with the given information.

Question 10. Let \(A = \{1, 2, 3\}\), \(B = \{5, 6\}\). Write \(A \times B\) also write its subsets. Subsets of \(A \times B = \phi, \{(1, 5)\}, \{(1, 6)\}, \{(2, 5)\}, \{(2, 6)\}, \{(3, 5)\}, \{(3, 6)\}, \dots\)
Answer:
\(A \times B = \{(1, 5), (1, 6), (2, 5), (2, 6), (3, 5), (3, 6)\}\)
Here, \(n(A \times B) = 3 \times 2 = 6\).
The total number of subsets of \(A \times B\) is \(2^6 = 64\).
Some of the subsets include:
\(\phi, \{(1, 5)\}, \{(1, 6)\}, \{(2, 5)\}, \{(2, 6)\}, \{(3, 5)\}, \{(3, 6)\},\)
\(\{(1, 5), (1, 6)\}, \{(1, 5), (2, 5)\}, \dots, A \times B\).
In simple words: A set with 6 elements has 64 subsets. List them starting from the empty set up to the set itself.

Exam Tip: The power set of a Cartesian product contains all possible relations from set A to set B.