116841
Define a relation \(R\) over a class of \(n \times n\) real matrices \(A\) and \(B\) as "ARB, if there exists a non -singular matrix \(P\) such that \(\mathbf{P A P}^{-1}=B^{\prime \prime}\). Then which of the following is true?
1 \(\mathrm{R}\) is symmetric, transitive but not reflexive.
2 \(\mathrm{R}\) is reflexive, symmetric but not transitive.
3 \(\mathrm{R}\) is an equivalence relation.
4 \(\mathrm{R}\) is reflexive, transitive but not symmetric.
Explanation:
C A and B are matrices of \(n \times n\) order and ARB if there exists a non-singular matrix \(\mathrm{P}(\operatorname{det}(\mathrm{P}) \neq 0\) ) Such that \(\mathrm{PAP}^{-1}=\mathrm{B}\) For reflexive - \(\mathrm{ARA} \Rightarrow \mathrm{PAP}^{-1}=\mathrm{A}\) (i) must be true For \(\quad P=I\), Equation (i) is true so ' \(R\) ' is reflexive For symmetric - \(\mathrm{ARB} \Leftrightarrow \mathrm{PAP}^{-1}=\mathrm{B}\) For \(B R A\) if \(\mathrm{PBP}^{-1}=\mathrm{A}\) (ii) must be true \(\because \mathrm{PAP}^{-1}=\mathrm{B}\) \(\mathrm{P}^{-1} \mathrm{PAP}^{-1}=\mathrm{P}^{-1} \mathrm{~B}\) \(\mathrm{IAP}^{-1} \mathrm{P}=\mathrm{P}^{-1} \mathrm{BP}\) \(\mathrm{A}=\mathrm{P}^{-1} \mathrm{BP}\) From equation (ii) and (iii) \(\mathrm{PBP}^{-1}=\mathrm{P}^{-1} \mathrm{BP}\) can be true some \(\mathrm{P}=\mathrm{P}^{-1}\) \(\Rightarrow \quad \mathrm{P}^2=\mathrm{I}\) \((\because \operatorname{det}(\mathrm{P}) \neq 0)\) So, \(\quad \mathrm{R}\) is symmetric. For transitive - \(\mathrm{ARB} \Leftrightarrow \mathrm{PAP}^{-1}=\mathrm{B}\) \(\mathrm{BRC} \Leftrightarrow \mathrm{PBP}^{-1}=\mathrm{C}\) .is true Now, is true \(\mathrm{P} \mathrm{PAP}^{-1} \mathrm{P}^{-1}=\mathrm{C}\) \(\mathrm{P}^2 \mathrm{~A}\left(\mathrm{P}^2\right)^{-1}=\mathrm{C}\) \(\Rightarrow \quad\) ARC So, ' \(R\) ' is transitive relation \(\Rightarrow\) Hence, \(\mathrm{R}\) is equivalence.
Shift-II
Sets, Relation and Function
116842
If \(A=\{2,3,4,5\}, B=\{36,45,49,60,77,90\}\) and let \(R\) be the relation 'is factor of ' from \(A\) to \(B\). Then the range of \(R\) is the set
1 \(\{60\}\)
2 \(\{36,45,60,90\}\)
3 \(\{49,77\}\)
4 \(\{49,60,77\}\)
5 \(\{36,45,49,6077,90\}\)
Explanation:
B We have, A \(\{2,3,4,5\}\) \(\mathrm{B}=\{36,45,49,60,77,90\}\) \(R\) : number form \(B\) with factor from \(A\) {l|l} |2 | 36| |---|---| |2 | 18| |2 | 9| |3 | 3| | 1 {c|c} |3 | 45 | | |---|---|---| |3 | 15 | | 5 | | {c|c} |7 | 49| |---|---| |7 | 7| | 1| | Factor of \(49=7\) \(\therefore\) factor of \(36=2,3\) factor of \(45=3,5\) {l|l} |2 | 60| |---|---| |2 | 30| |3 | 15| |5 | 5\(\quad\){l|l} |2 | 77| |---|---| |11 | 11| | 1| | Factor of \(60=2,3,5\) \(\therefore\) factor of \(77=7\) and 11 {l|l} |2 | 90| |---|---| |3 | 45| |3 | 15| |5 | 5| | 1 factor of \(90=2,3,5\) \(R=\{36,45,60,90\}\)
Kerala CEE-2020
Sets, Relation and Function
116843
On the set \(\mathbf{N}\) of all natural numbers define the relation \(R\) by a \(R b\) if and only of the GCD of a and \(b\) is 2 , then \(R\) is
1 reflexive but not symmetric
2 symmetric only
3 reflexive and transitive
4 reflexive symmetric and transitive
5 not reflexive not symmetric and not transitive
Explanation:
B The relation \(\mathrm{R}\) is defined by \(\mathrm{aRb}\), if and only if the GCD of \(a\) and \(b\) is 2 . \(a R b=G C D\) of \(a\) and \(b\) is 2 (i) \(a R b\), then GCD of \(a\) and \(a\) is \(a\). \(\therefore \mathrm{R}\) is not reflexive. (ii) \(\mathrm{aRb} \Rightarrow \mathrm{bRa}\) if GCD of \(a\) and \(b\) is 2 , then GCD of \(b\) and \(a\) is 2 . \(\therefore \mathrm{R}\) is symmetiric. (iii) \(\mathrm{aRb}, \mathrm{bRc} \Rightarrow \mathrm{cRa}\) if GCD of \(a\) and \(b\) is 2 and GCD of \(b\) and \(c\) is 2 . Then it is need net to be GCD of \(c\) and is 2 . \(\therefore \mathrm{R}\) is transitive.
Kerala CEE-2007
Sets, Relation and Function
116844
If \(n(A)=2\) and total number of possible relations from set \(A\) to set \(B\) is 1024 , then \(n(B)\) is
1 20
2 10
3 5
4 512
Explanation:
C Given, \(\mathrm{n}(\mathrm{A})=2\) And, total number of possible relations from set A to set \(\mathrm{B}\) is 1024 . Then, find \(\mathrm{n}(\mathrm{B})=\) ? \(2^{\{\mathrm{n}(\mathrm{A}) \cdot \mathrm{n}(\mathrm{B})\}}=\) Total number of possible relations \(2^{\{n(A) \cdot n(B)\}}= \text { from set } A \text { to set } B\) \(1024\) \(2^{\{\mathrm{n}(\mathrm{A}) \cdot \mathrm{n}(\mathrm{B})\}}=2^{10}\) \(\mathrm{n}(\mathrm{A}) \cdot \mathrm{n}(\mathrm{B})=10\) \(2 \mathrm{n}(\mathrm{B})=10\) \(\mathrm{n}(\mathrm{B})=5\)
Karnataka CET 2020
Sets, Relation and Function
116845
If a relation \(R\) on the set \(\{1,2,3\}\) be defined by \(R=\{(1, \mathbf{1})\}\), then \(R\) is
1 Reflective and transitive
2 Symmetric and transitive
3 Only symmetric
4 Reflexive and symmetric
Explanation:
B Given, A set \(\{1,2,3\}\) be defined by relation \(\mathrm{R}=\{(1,1)\}\) Then, check relation are - (a) Reflexive :- In this relation, \((1,1) \in \mathrm{R}\) but \((2,2)(3,3) \in \mathrm{R}\) So, it is not reflexive relation. (b) Symmetric : - \({ }_1 R_2 \Rightarrow{ }_2 R_1\) Means \(-(\mathrm{a}, \mathrm{b})=(1,1) \Rightarrow \in \mathrm{R}\) Then, \((\mathrm{b}, \mathrm{a})=(1,1) \in \mathrm{R}\) So, it is symmetric relation (c) Transitive - \(\text { If }(a, b)=(1,1) \in R\) \((b, c)=(1,1) \in R\) Then, \((\mathrm{a}, \mathrm{c})=(1,1) \in \mathrm{R}\) So, it is transitive relation. Hence, \(\mathrm{R}\) is symmetric and transitive but not reflexive relation.
116841
Define a relation \(R\) over a class of \(n \times n\) real matrices \(A\) and \(B\) as "ARB, if there exists a non -singular matrix \(P\) such that \(\mathbf{P A P}^{-1}=B^{\prime \prime}\). Then which of the following is true?
1 \(\mathrm{R}\) is symmetric, transitive but not reflexive.
2 \(\mathrm{R}\) is reflexive, symmetric but not transitive.
3 \(\mathrm{R}\) is an equivalence relation.
4 \(\mathrm{R}\) is reflexive, transitive but not symmetric.
Explanation:
C A and B are matrices of \(n \times n\) order and ARB if there exists a non-singular matrix \(\mathrm{P}(\operatorname{det}(\mathrm{P}) \neq 0\) ) Such that \(\mathrm{PAP}^{-1}=\mathrm{B}\) For reflexive - \(\mathrm{ARA} \Rightarrow \mathrm{PAP}^{-1}=\mathrm{A}\) (i) must be true For \(\quad P=I\), Equation (i) is true so ' \(R\) ' is reflexive For symmetric - \(\mathrm{ARB} \Leftrightarrow \mathrm{PAP}^{-1}=\mathrm{B}\) For \(B R A\) if \(\mathrm{PBP}^{-1}=\mathrm{A}\) (ii) must be true \(\because \mathrm{PAP}^{-1}=\mathrm{B}\) \(\mathrm{P}^{-1} \mathrm{PAP}^{-1}=\mathrm{P}^{-1} \mathrm{~B}\) \(\mathrm{IAP}^{-1} \mathrm{P}=\mathrm{P}^{-1} \mathrm{BP}\) \(\mathrm{A}=\mathrm{P}^{-1} \mathrm{BP}\) From equation (ii) and (iii) \(\mathrm{PBP}^{-1}=\mathrm{P}^{-1} \mathrm{BP}\) can be true some \(\mathrm{P}=\mathrm{P}^{-1}\) \(\Rightarrow \quad \mathrm{P}^2=\mathrm{I}\) \((\because \operatorname{det}(\mathrm{P}) \neq 0)\) So, \(\quad \mathrm{R}\) is symmetric. For transitive - \(\mathrm{ARB} \Leftrightarrow \mathrm{PAP}^{-1}=\mathrm{B}\) \(\mathrm{BRC} \Leftrightarrow \mathrm{PBP}^{-1}=\mathrm{C}\) .is true Now, is true \(\mathrm{P} \mathrm{PAP}^{-1} \mathrm{P}^{-1}=\mathrm{C}\) \(\mathrm{P}^2 \mathrm{~A}\left(\mathrm{P}^2\right)^{-1}=\mathrm{C}\) \(\Rightarrow \quad\) ARC So, ' \(R\) ' is transitive relation \(\Rightarrow\) Hence, \(\mathrm{R}\) is equivalence.
Shift-II
Sets, Relation and Function
116842
If \(A=\{2,3,4,5\}, B=\{36,45,49,60,77,90\}\) and let \(R\) be the relation 'is factor of ' from \(A\) to \(B\). Then the range of \(R\) is the set
1 \(\{60\}\)
2 \(\{36,45,60,90\}\)
3 \(\{49,77\}\)
4 \(\{49,60,77\}\)
5 \(\{36,45,49,6077,90\}\)
Explanation:
B We have, A \(\{2,3,4,5\}\) \(\mathrm{B}=\{36,45,49,60,77,90\}\) \(R\) : number form \(B\) with factor from \(A\) {l|l} |2 | 36| |---|---| |2 | 18| |2 | 9| |3 | 3| | 1 {c|c} |3 | 45 | | |---|---|---| |3 | 15 | | 5 | | {c|c} |7 | 49| |---|---| |7 | 7| | 1| | Factor of \(49=7\) \(\therefore\) factor of \(36=2,3\) factor of \(45=3,5\) {l|l} |2 | 60| |---|---| |2 | 30| |3 | 15| |5 | 5\(\quad\){l|l} |2 | 77| |---|---| |11 | 11| | 1| | Factor of \(60=2,3,5\) \(\therefore\) factor of \(77=7\) and 11 {l|l} |2 | 90| |---|---| |3 | 45| |3 | 15| |5 | 5| | 1 factor of \(90=2,3,5\) \(R=\{36,45,60,90\}\)
Kerala CEE-2020
Sets, Relation and Function
116843
On the set \(\mathbf{N}\) of all natural numbers define the relation \(R\) by a \(R b\) if and only of the GCD of a and \(b\) is 2 , then \(R\) is
1 reflexive but not symmetric
2 symmetric only
3 reflexive and transitive
4 reflexive symmetric and transitive
5 not reflexive not symmetric and not transitive
Explanation:
B The relation \(\mathrm{R}\) is defined by \(\mathrm{aRb}\), if and only if the GCD of \(a\) and \(b\) is 2 . \(a R b=G C D\) of \(a\) and \(b\) is 2 (i) \(a R b\), then GCD of \(a\) and \(a\) is \(a\). \(\therefore \mathrm{R}\) is not reflexive. (ii) \(\mathrm{aRb} \Rightarrow \mathrm{bRa}\) if GCD of \(a\) and \(b\) is 2 , then GCD of \(b\) and \(a\) is 2 . \(\therefore \mathrm{R}\) is symmetiric. (iii) \(\mathrm{aRb}, \mathrm{bRc} \Rightarrow \mathrm{cRa}\) if GCD of \(a\) and \(b\) is 2 and GCD of \(b\) and \(c\) is 2 . Then it is need net to be GCD of \(c\) and is 2 . \(\therefore \mathrm{R}\) is transitive.
Kerala CEE-2007
Sets, Relation and Function
116844
If \(n(A)=2\) and total number of possible relations from set \(A\) to set \(B\) is 1024 , then \(n(B)\) is
1 20
2 10
3 5
4 512
Explanation:
C Given, \(\mathrm{n}(\mathrm{A})=2\) And, total number of possible relations from set A to set \(\mathrm{B}\) is 1024 . Then, find \(\mathrm{n}(\mathrm{B})=\) ? \(2^{\{\mathrm{n}(\mathrm{A}) \cdot \mathrm{n}(\mathrm{B})\}}=\) Total number of possible relations \(2^{\{n(A) \cdot n(B)\}}= \text { from set } A \text { to set } B\) \(1024\) \(2^{\{\mathrm{n}(\mathrm{A}) \cdot \mathrm{n}(\mathrm{B})\}}=2^{10}\) \(\mathrm{n}(\mathrm{A}) \cdot \mathrm{n}(\mathrm{B})=10\) \(2 \mathrm{n}(\mathrm{B})=10\) \(\mathrm{n}(\mathrm{B})=5\)
Karnataka CET 2020
Sets, Relation and Function
116845
If a relation \(R\) on the set \(\{1,2,3\}\) be defined by \(R=\{(1, \mathbf{1})\}\), then \(R\) is
1 Reflective and transitive
2 Symmetric and transitive
3 Only symmetric
4 Reflexive and symmetric
Explanation:
B Given, A set \(\{1,2,3\}\) be defined by relation \(\mathrm{R}=\{(1,1)\}\) Then, check relation are - (a) Reflexive :- In this relation, \((1,1) \in \mathrm{R}\) but \((2,2)(3,3) \in \mathrm{R}\) So, it is not reflexive relation. (b) Symmetric : - \({ }_1 R_2 \Rightarrow{ }_2 R_1\) Means \(-(\mathrm{a}, \mathrm{b})=(1,1) \Rightarrow \in \mathrm{R}\) Then, \((\mathrm{b}, \mathrm{a})=(1,1) \in \mathrm{R}\) So, it is symmetric relation (c) Transitive - \(\text { If }(a, b)=(1,1) \in R\) \((b, c)=(1,1) \in R\) Then, \((\mathrm{a}, \mathrm{c})=(1,1) \in \mathrm{R}\) So, it is transitive relation. Hence, \(\mathrm{R}\) is symmetric and transitive but not reflexive relation.
NEET Test Series from KOTA - 10 Papers In MS WORD
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Sets, Relation and Function
116841
Define a relation \(R\) over a class of \(n \times n\) real matrices \(A\) and \(B\) as "ARB, if there exists a non -singular matrix \(P\) such that \(\mathbf{P A P}^{-1}=B^{\prime \prime}\). Then which of the following is true?
1 \(\mathrm{R}\) is symmetric, transitive but not reflexive.
2 \(\mathrm{R}\) is reflexive, symmetric but not transitive.
3 \(\mathrm{R}\) is an equivalence relation.
4 \(\mathrm{R}\) is reflexive, transitive but not symmetric.
Explanation:
C A and B are matrices of \(n \times n\) order and ARB if there exists a non-singular matrix \(\mathrm{P}(\operatorname{det}(\mathrm{P}) \neq 0\) ) Such that \(\mathrm{PAP}^{-1}=\mathrm{B}\) For reflexive - \(\mathrm{ARA} \Rightarrow \mathrm{PAP}^{-1}=\mathrm{A}\) (i) must be true For \(\quad P=I\), Equation (i) is true so ' \(R\) ' is reflexive For symmetric - \(\mathrm{ARB} \Leftrightarrow \mathrm{PAP}^{-1}=\mathrm{B}\) For \(B R A\) if \(\mathrm{PBP}^{-1}=\mathrm{A}\) (ii) must be true \(\because \mathrm{PAP}^{-1}=\mathrm{B}\) \(\mathrm{P}^{-1} \mathrm{PAP}^{-1}=\mathrm{P}^{-1} \mathrm{~B}\) \(\mathrm{IAP}^{-1} \mathrm{P}=\mathrm{P}^{-1} \mathrm{BP}\) \(\mathrm{A}=\mathrm{P}^{-1} \mathrm{BP}\) From equation (ii) and (iii) \(\mathrm{PBP}^{-1}=\mathrm{P}^{-1} \mathrm{BP}\) can be true some \(\mathrm{P}=\mathrm{P}^{-1}\) \(\Rightarrow \quad \mathrm{P}^2=\mathrm{I}\) \((\because \operatorname{det}(\mathrm{P}) \neq 0)\) So, \(\quad \mathrm{R}\) is symmetric. For transitive - \(\mathrm{ARB} \Leftrightarrow \mathrm{PAP}^{-1}=\mathrm{B}\) \(\mathrm{BRC} \Leftrightarrow \mathrm{PBP}^{-1}=\mathrm{C}\) .is true Now, is true \(\mathrm{P} \mathrm{PAP}^{-1} \mathrm{P}^{-1}=\mathrm{C}\) \(\mathrm{P}^2 \mathrm{~A}\left(\mathrm{P}^2\right)^{-1}=\mathrm{C}\) \(\Rightarrow \quad\) ARC So, ' \(R\) ' is transitive relation \(\Rightarrow\) Hence, \(\mathrm{R}\) is equivalence.
Shift-II
Sets, Relation and Function
116842
If \(A=\{2,3,4,5\}, B=\{36,45,49,60,77,90\}\) and let \(R\) be the relation 'is factor of ' from \(A\) to \(B\). Then the range of \(R\) is the set
1 \(\{60\}\)
2 \(\{36,45,60,90\}\)
3 \(\{49,77\}\)
4 \(\{49,60,77\}\)
5 \(\{36,45,49,6077,90\}\)
Explanation:
B We have, A \(\{2,3,4,5\}\) \(\mathrm{B}=\{36,45,49,60,77,90\}\) \(R\) : number form \(B\) with factor from \(A\) {l|l} |2 | 36| |---|---| |2 | 18| |2 | 9| |3 | 3| | 1 {c|c} |3 | 45 | | |---|---|---| |3 | 15 | | 5 | | {c|c} |7 | 49| |---|---| |7 | 7| | 1| | Factor of \(49=7\) \(\therefore\) factor of \(36=2,3\) factor of \(45=3,5\) {l|l} |2 | 60| |---|---| |2 | 30| |3 | 15| |5 | 5\(\quad\){l|l} |2 | 77| |---|---| |11 | 11| | 1| | Factor of \(60=2,3,5\) \(\therefore\) factor of \(77=7\) and 11 {l|l} |2 | 90| |---|---| |3 | 45| |3 | 15| |5 | 5| | 1 factor of \(90=2,3,5\) \(R=\{36,45,60,90\}\)
Kerala CEE-2020
Sets, Relation and Function
116843
On the set \(\mathbf{N}\) of all natural numbers define the relation \(R\) by a \(R b\) if and only of the GCD of a and \(b\) is 2 , then \(R\) is
1 reflexive but not symmetric
2 symmetric only
3 reflexive and transitive
4 reflexive symmetric and transitive
5 not reflexive not symmetric and not transitive
Explanation:
B The relation \(\mathrm{R}\) is defined by \(\mathrm{aRb}\), if and only if the GCD of \(a\) and \(b\) is 2 . \(a R b=G C D\) of \(a\) and \(b\) is 2 (i) \(a R b\), then GCD of \(a\) and \(a\) is \(a\). \(\therefore \mathrm{R}\) is not reflexive. (ii) \(\mathrm{aRb} \Rightarrow \mathrm{bRa}\) if GCD of \(a\) and \(b\) is 2 , then GCD of \(b\) and \(a\) is 2 . \(\therefore \mathrm{R}\) is symmetiric. (iii) \(\mathrm{aRb}, \mathrm{bRc} \Rightarrow \mathrm{cRa}\) if GCD of \(a\) and \(b\) is 2 and GCD of \(b\) and \(c\) is 2 . Then it is need net to be GCD of \(c\) and is 2 . \(\therefore \mathrm{R}\) is transitive.
Kerala CEE-2007
Sets, Relation and Function
116844
If \(n(A)=2\) and total number of possible relations from set \(A\) to set \(B\) is 1024 , then \(n(B)\) is
1 20
2 10
3 5
4 512
Explanation:
C Given, \(\mathrm{n}(\mathrm{A})=2\) And, total number of possible relations from set A to set \(\mathrm{B}\) is 1024 . Then, find \(\mathrm{n}(\mathrm{B})=\) ? \(2^{\{\mathrm{n}(\mathrm{A}) \cdot \mathrm{n}(\mathrm{B})\}}=\) Total number of possible relations \(2^{\{n(A) \cdot n(B)\}}= \text { from set } A \text { to set } B\) \(1024\) \(2^{\{\mathrm{n}(\mathrm{A}) \cdot \mathrm{n}(\mathrm{B})\}}=2^{10}\) \(\mathrm{n}(\mathrm{A}) \cdot \mathrm{n}(\mathrm{B})=10\) \(2 \mathrm{n}(\mathrm{B})=10\) \(\mathrm{n}(\mathrm{B})=5\)
Karnataka CET 2020
Sets, Relation and Function
116845
If a relation \(R\) on the set \(\{1,2,3\}\) be defined by \(R=\{(1, \mathbf{1})\}\), then \(R\) is
1 Reflective and transitive
2 Symmetric and transitive
3 Only symmetric
4 Reflexive and symmetric
Explanation:
B Given, A set \(\{1,2,3\}\) be defined by relation \(\mathrm{R}=\{(1,1)\}\) Then, check relation are - (a) Reflexive :- In this relation, \((1,1) \in \mathrm{R}\) but \((2,2)(3,3) \in \mathrm{R}\) So, it is not reflexive relation. (b) Symmetric : - \({ }_1 R_2 \Rightarrow{ }_2 R_1\) Means \(-(\mathrm{a}, \mathrm{b})=(1,1) \Rightarrow \in \mathrm{R}\) Then, \((\mathrm{b}, \mathrm{a})=(1,1) \in \mathrm{R}\) So, it is symmetric relation (c) Transitive - \(\text { If }(a, b)=(1,1) \in R\) \((b, c)=(1,1) \in R\) Then, \((\mathrm{a}, \mathrm{c})=(1,1) \in \mathrm{R}\) So, it is transitive relation. Hence, \(\mathrm{R}\) is symmetric and transitive but not reflexive relation.
116841
Define a relation \(R\) over a class of \(n \times n\) real matrices \(A\) and \(B\) as "ARB, if there exists a non -singular matrix \(P\) such that \(\mathbf{P A P}^{-1}=B^{\prime \prime}\). Then which of the following is true?
1 \(\mathrm{R}\) is symmetric, transitive but not reflexive.
2 \(\mathrm{R}\) is reflexive, symmetric but not transitive.
3 \(\mathrm{R}\) is an equivalence relation.
4 \(\mathrm{R}\) is reflexive, transitive but not symmetric.
Explanation:
C A and B are matrices of \(n \times n\) order and ARB if there exists a non-singular matrix \(\mathrm{P}(\operatorname{det}(\mathrm{P}) \neq 0\) ) Such that \(\mathrm{PAP}^{-1}=\mathrm{B}\) For reflexive - \(\mathrm{ARA} \Rightarrow \mathrm{PAP}^{-1}=\mathrm{A}\) (i) must be true For \(\quad P=I\), Equation (i) is true so ' \(R\) ' is reflexive For symmetric - \(\mathrm{ARB} \Leftrightarrow \mathrm{PAP}^{-1}=\mathrm{B}\) For \(B R A\) if \(\mathrm{PBP}^{-1}=\mathrm{A}\) (ii) must be true \(\because \mathrm{PAP}^{-1}=\mathrm{B}\) \(\mathrm{P}^{-1} \mathrm{PAP}^{-1}=\mathrm{P}^{-1} \mathrm{~B}\) \(\mathrm{IAP}^{-1} \mathrm{P}=\mathrm{P}^{-1} \mathrm{BP}\) \(\mathrm{A}=\mathrm{P}^{-1} \mathrm{BP}\) From equation (ii) and (iii) \(\mathrm{PBP}^{-1}=\mathrm{P}^{-1} \mathrm{BP}\) can be true some \(\mathrm{P}=\mathrm{P}^{-1}\) \(\Rightarrow \quad \mathrm{P}^2=\mathrm{I}\) \((\because \operatorname{det}(\mathrm{P}) \neq 0)\) So, \(\quad \mathrm{R}\) is symmetric. For transitive - \(\mathrm{ARB} \Leftrightarrow \mathrm{PAP}^{-1}=\mathrm{B}\) \(\mathrm{BRC} \Leftrightarrow \mathrm{PBP}^{-1}=\mathrm{C}\) .is true Now, is true \(\mathrm{P} \mathrm{PAP}^{-1} \mathrm{P}^{-1}=\mathrm{C}\) \(\mathrm{P}^2 \mathrm{~A}\left(\mathrm{P}^2\right)^{-1}=\mathrm{C}\) \(\Rightarrow \quad\) ARC So, ' \(R\) ' is transitive relation \(\Rightarrow\) Hence, \(\mathrm{R}\) is equivalence.
Shift-II
Sets, Relation and Function
116842
If \(A=\{2,3,4,5\}, B=\{36,45,49,60,77,90\}\) and let \(R\) be the relation 'is factor of ' from \(A\) to \(B\). Then the range of \(R\) is the set
1 \(\{60\}\)
2 \(\{36,45,60,90\}\)
3 \(\{49,77\}\)
4 \(\{49,60,77\}\)
5 \(\{36,45,49,6077,90\}\)
Explanation:
B We have, A \(\{2,3,4,5\}\) \(\mathrm{B}=\{36,45,49,60,77,90\}\) \(R\) : number form \(B\) with factor from \(A\) {l|l} |2 | 36| |---|---| |2 | 18| |2 | 9| |3 | 3| | 1 {c|c} |3 | 45 | | |---|---|---| |3 | 15 | | 5 | | {c|c} |7 | 49| |---|---| |7 | 7| | 1| | Factor of \(49=7\) \(\therefore\) factor of \(36=2,3\) factor of \(45=3,5\) {l|l} |2 | 60| |---|---| |2 | 30| |3 | 15| |5 | 5\(\quad\){l|l} |2 | 77| |---|---| |11 | 11| | 1| | Factor of \(60=2,3,5\) \(\therefore\) factor of \(77=7\) and 11 {l|l} |2 | 90| |---|---| |3 | 45| |3 | 15| |5 | 5| | 1 factor of \(90=2,3,5\) \(R=\{36,45,60,90\}\)
Kerala CEE-2020
Sets, Relation and Function
116843
On the set \(\mathbf{N}\) of all natural numbers define the relation \(R\) by a \(R b\) if and only of the GCD of a and \(b\) is 2 , then \(R\) is
1 reflexive but not symmetric
2 symmetric only
3 reflexive and transitive
4 reflexive symmetric and transitive
5 not reflexive not symmetric and not transitive
Explanation:
B The relation \(\mathrm{R}\) is defined by \(\mathrm{aRb}\), if and only if the GCD of \(a\) and \(b\) is 2 . \(a R b=G C D\) of \(a\) and \(b\) is 2 (i) \(a R b\), then GCD of \(a\) and \(a\) is \(a\). \(\therefore \mathrm{R}\) is not reflexive. (ii) \(\mathrm{aRb} \Rightarrow \mathrm{bRa}\) if GCD of \(a\) and \(b\) is 2 , then GCD of \(b\) and \(a\) is 2 . \(\therefore \mathrm{R}\) is symmetiric. (iii) \(\mathrm{aRb}, \mathrm{bRc} \Rightarrow \mathrm{cRa}\) if GCD of \(a\) and \(b\) is 2 and GCD of \(b\) and \(c\) is 2 . Then it is need net to be GCD of \(c\) and is 2 . \(\therefore \mathrm{R}\) is transitive.
Kerala CEE-2007
Sets, Relation and Function
116844
If \(n(A)=2\) and total number of possible relations from set \(A\) to set \(B\) is 1024 , then \(n(B)\) is
1 20
2 10
3 5
4 512
Explanation:
C Given, \(\mathrm{n}(\mathrm{A})=2\) And, total number of possible relations from set A to set \(\mathrm{B}\) is 1024 . Then, find \(\mathrm{n}(\mathrm{B})=\) ? \(2^{\{\mathrm{n}(\mathrm{A}) \cdot \mathrm{n}(\mathrm{B})\}}=\) Total number of possible relations \(2^{\{n(A) \cdot n(B)\}}= \text { from set } A \text { to set } B\) \(1024\) \(2^{\{\mathrm{n}(\mathrm{A}) \cdot \mathrm{n}(\mathrm{B})\}}=2^{10}\) \(\mathrm{n}(\mathrm{A}) \cdot \mathrm{n}(\mathrm{B})=10\) \(2 \mathrm{n}(\mathrm{B})=10\) \(\mathrm{n}(\mathrm{B})=5\)
Karnataka CET 2020
Sets, Relation and Function
116845
If a relation \(R\) on the set \(\{1,2,3\}\) be defined by \(R=\{(1, \mathbf{1})\}\), then \(R\) is
1 Reflective and transitive
2 Symmetric and transitive
3 Only symmetric
4 Reflexive and symmetric
Explanation:
B Given, A set \(\{1,2,3\}\) be defined by relation \(\mathrm{R}=\{(1,1)\}\) Then, check relation are - (a) Reflexive :- In this relation, \((1,1) \in \mathrm{R}\) but \((2,2)(3,3) \in \mathrm{R}\) So, it is not reflexive relation. (b) Symmetric : - \({ }_1 R_2 \Rightarrow{ }_2 R_1\) Means \(-(\mathrm{a}, \mathrm{b})=(1,1) \Rightarrow \in \mathrm{R}\) Then, \((\mathrm{b}, \mathrm{a})=(1,1) \in \mathrm{R}\) So, it is symmetric relation (c) Transitive - \(\text { If }(a, b)=(1,1) \in R\) \((b, c)=(1,1) \in R\) Then, \((\mathrm{a}, \mathrm{c})=(1,1) \in \mathrm{R}\) So, it is transitive relation. Hence, \(\mathrm{R}\) is symmetric and transitive but not reflexive relation.
116841
Define a relation \(R\) over a class of \(n \times n\) real matrices \(A\) and \(B\) as "ARB, if there exists a non -singular matrix \(P\) such that \(\mathbf{P A P}^{-1}=B^{\prime \prime}\). Then which of the following is true?
1 \(\mathrm{R}\) is symmetric, transitive but not reflexive.
2 \(\mathrm{R}\) is reflexive, symmetric but not transitive.
3 \(\mathrm{R}\) is an equivalence relation.
4 \(\mathrm{R}\) is reflexive, transitive but not symmetric.
Explanation:
C A and B are matrices of \(n \times n\) order and ARB if there exists a non-singular matrix \(\mathrm{P}(\operatorname{det}(\mathrm{P}) \neq 0\) ) Such that \(\mathrm{PAP}^{-1}=\mathrm{B}\) For reflexive - \(\mathrm{ARA} \Rightarrow \mathrm{PAP}^{-1}=\mathrm{A}\) (i) must be true For \(\quad P=I\), Equation (i) is true so ' \(R\) ' is reflexive For symmetric - \(\mathrm{ARB} \Leftrightarrow \mathrm{PAP}^{-1}=\mathrm{B}\) For \(B R A\) if \(\mathrm{PBP}^{-1}=\mathrm{A}\) (ii) must be true \(\because \mathrm{PAP}^{-1}=\mathrm{B}\) \(\mathrm{P}^{-1} \mathrm{PAP}^{-1}=\mathrm{P}^{-1} \mathrm{~B}\) \(\mathrm{IAP}^{-1} \mathrm{P}=\mathrm{P}^{-1} \mathrm{BP}\) \(\mathrm{A}=\mathrm{P}^{-1} \mathrm{BP}\) From equation (ii) and (iii) \(\mathrm{PBP}^{-1}=\mathrm{P}^{-1} \mathrm{BP}\) can be true some \(\mathrm{P}=\mathrm{P}^{-1}\) \(\Rightarrow \quad \mathrm{P}^2=\mathrm{I}\) \((\because \operatorname{det}(\mathrm{P}) \neq 0)\) So, \(\quad \mathrm{R}\) is symmetric. For transitive - \(\mathrm{ARB} \Leftrightarrow \mathrm{PAP}^{-1}=\mathrm{B}\) \(\mathrm{BRC} \Leftrightarrow \mathrm{PBP}^{-1}=\mathrm{C}\) .is true Now, is true \(\mathrm{P} \mathrm{PAP}^{-1} \mathrm{P}^{-1}=\mathrm{C}\) \(\mathrm{P}^2 \mathrm{~A}\left(\mathrm{P}^2\right)^{-1}=\mathrm{C}\) \(\Rightarrow \quad\) ARC So, ' \(R\) ' is transitive relation \(\Rightarrow\) Hence, \(\mathrm{R}\) is equivalence.
Shift-II
Sets, Relation and Function
116842
If \(A=\{2,3,4,5\}, B=\{36,45,49,60,77,90\}\) and let \(R\) be the relation 'is factor of ' from \(A\) to \(B\). Then the range of \(R\) is the set
1 \(\{60\}\)
2 \(\{36,45,60,90\}\)
3 \(\{49,77\}\)
4 \(\{49,60,77\}\)
5 \(\{36,45,49,6077,90\}\)
Explanation:
B We have, A \(\{2,3,4,5\}\) \(\mathrm{B}=\{36,45,49,60,77,90\}\) \(R\) : number form \(B\) with factor from \(A\) {l|l} |2 | 36| |---|---| |2 | 18| |2 | 9| |3 | 3| | 1 {c|c} |3 | 45 | | |---|---|---| |3 | 15 | | 5 | | {c|c} |7 | 49| |---|---| |7 | 7| | 1| | Factor of \(49=7\) \(\therefore\) factor of \(36=2,3\) factor of \(45=3,5\) {l|l} |2 | 60| |---|---| |2 | 30| |3 | 15| |5 | 5\(\quad\){l|l} |2 | 77| |---|---| |11 | 11| | 1| | Factor of \(60=2,3,5\) \(\therefore\) factor of \(77=7\) and 11 {l|l} |2 | 90| |---|---| |3 | 45| |3 | 15| |5 | 5| | 1 factor of \(90=2,3,5\) \(R=\{36,45,60,90\}\)
Kerala CEE-2020
Sets, Relation and Function
116843
On the set \(\mathbf{N}\) of all natural numbers define the relation \(R\) by a \(R b\) if and only of the GCD of a and \(b\) is 2 , then \(R\) is
1 reflexive but not symmetric
2 symmetric only
3 reflexive and transitive
4 reflexive symmetric and transitive
5 not reflexive not symmetric and not transitive
Explanation:
B The relation \(\mathrm{R}\) is defined by \(\mathrm{aRb}\), if and only if the GCD of \(a\) and \(b\) is 2 . \(a R b=G C D\) of \(a\) and \(b\) is 2 (i) \(a R b\), then GCD of \(a\) and \(a\) is \(a\). \(\therefore \mathrm{R}\) is not reflexive. (ii) \(\mathrm{aRb} \Rightarrow \mathrm{bRa}\) if GCD of \(a\) and \(b\) is 2 , then GCD of \(b\) and \(a\) is 2 . \(\therefore \mathrm{R}\) is symmetiric. (iii) \(\mathrm{aRb}, \mathrm{bRc} \Rightarrow \mathrm{cRa}\) if GCD of \(a\) and \(b\) is 2 and GCD of \(b\) and \(c\) is 2 . Then it is need net to be GCD of \(c\) and is 2 . \(\therefore \mathrm{R}\) is transitive.
Kerala CEE-2007
Sets, Relation and Function
116844
If \(n(A)=2\) and total number of possible relations from set \(A\) to set \(B\) is 1024 , then \(n(B)\) is
1 20
2 10
3 5
4 512
Explanation:
C Given, \(\mathrm{n}(\mathrm{A})=2\) And, total number of possible relations from set A to set \(\mathrm{B}\) is 1024 . Then, find \(\mathrm{n}(\mathrm{B})=\) ? \(2^{\{\mathrm{n}(\mathrm{A}) \cdot \mathrm{n}(\mathrm{B})\}}=\) Total number of possible relations \(2^{\{n(A) \cdot n(B)\}}= \text { from set } A \text { to set } B\) \(1024\) \(2^{\{\mathrm{n}(\mathrm{A}) \cdot \mathrm{n}(\mathrm{B})\}}=2^{10}\) \(\mathrm{n}(\mathrm{A}) \cdot \mathrm{n}(\mathrm{B})=10\) \(2 \mathrm{n}(\mathrm{B})=10\) \(\mathrm{n}(\mathrm{B})=5\)
Karnataka CET 2020
Sets, Relation and Function
116845
If a relation \(R\) on the set \(\{1,2,3\}\) be defined by \(R=\{(1, \mathbf{1})\}\), then \(R\) is
1 Reflective and transitive
2 Symmetric and transitive
3 Only symmetric
4 Reflexive and symmetric
Explanation:
B Given, A set \(\{1,2,3\}\) be defined by relation \(\mathrm{R}=\{(1,1)\}\) Then, check relation are - (a) Reflexive :- In this relation, \((1,1) \in \mathrm{R}\) but \((2,2)(3,3) \in \mathrm{R}\) So, it is not reflexive relation. (b) Symmetric : - \({ }_1 R_2 \Rightarrow{ }_2 R_1\) Means \(-(\mathrm{a}, \mathrm{b})=(1,1) \Rightarrow \in \mathrm{R}\) Then, \((\mathrm{b}, \mathrm{a})=(1,1) \in \mathrm{R}\) So, it is symmetric relation (c) Transitive - \(\text { If }(a, b)=(1,1) \in R\) \((b, c)=(1,1) \in R\) Then, \((\mathrm{a}, \mathrm{c})=(1,1) \in \mathrm{R}\) So, it is transitive relation. Hence, \(\mathrm{R}\) is symmetric and transitive but not reflexive relation.