GATE-CS-2001 - GeeksforGeeks

22 Nov, 2021
S1: The sum of two singular n × n matrices may be non-singular
S2: The sum of two n × n non-singular matrices may be singular. 
M1 and M2 are singular
M1 =  1  1
      1  1
       
M2 =   1  -1
      -1   1

But M1+M2 is non-singular  
M1+M2 =  2  0
         0  2

M1 and M2 are non-singular
M1 =  1  0
      0  1
       
M2 =   -1  0
        0  -1

But M1+M2 is singular  
M1+M2 =  0  0
         0  0

R1(a,b) iff (a+b) is even over the set of integers
R2(a,b) iff (a+b) is odd over the set of integers
R3(a,b) iff a.b > 0 over the set of non-zero rational numbers
R4(a,b) iff |a - b| <= 2 over the set of natural numbers
R1(a,b)Reflexive : Yes, because (a+a) is even.Symmetrix : Yes, (a+b) is even ⟹ (b+a) is even.Transitive : Yes, because (a+b) is even and (b+c) is even ⟹ (a+c) is even.So R1 is equivalence relation.R2(a,b)Reflexive : No, because (a+a) is even.So R2 is not equivalence relation.R3(a,b)Reflexive : Yes, because a.a > 0.Symmetrix : Yes, a.b > 0 ⟹ b.a > 0.Transitive : Yes, because a.b > 0 and b.c > 0 ⟹ a.c > 0.So R3 is equivalence relation.R4(a,b)Reflexive : Yes, because |a-a| ≤ 2.Symmetrix : Yes, |a-b| ≤ 2 ⟹ |b-a| ≤ 2.Transitive : No, because |a-b| ≤ 2 and |b-c| ≤ 2 ⇏ (a-c) is even.So R4 is not equivalence relation.
R1(a,b)Reflexive : Yes, because (a+a) is even.Symmetrix : Yes, (a+b) is even ⟹ (b+a) is even.Transitive : Yes, because (a+b) is even and (b+c) is even ⟹ (a+c) is even.So R1 is equivalence relation.
Reflexive : Yes, because (a+a) is even.
Symmetrix : Yes, (a+b) is even ⟹ (b+a) is even.
Transitive : Yes, because (a+b) is even and (b+c) is even ⟹ (a+c) is even.
R2(a,b)Reflexive : No, because (a+a) is even.So R2 is not equivalence relation.
Reflexive : No, because (a+a) is even.
R3(a,b)Reflexive : Yes, because a.a > 0.Symmetrix : Yes, a.b > 0 ⟹ b.a > 0.Transitive : Yes, because a.b > 0 and b.c > 0 ⟹ a.c > 0.So R3 is equivalence relation.
Reflexive : Yes, because a.a > 0.
Symmetrix : Yes, a.b > 0 ⟹ b.a > 0.
Transitive : Yes, because a.b > 0 and b.c > 0 ⟹ a.c > 0.
R4(a,b)Reflexive : Yes, because |a-a| ≤ 2.Symmetrix : Yes, |a-b| ≤ 2 ⟹ |b-a| ≤ 2.Transitive : No, because |a-b| ≤ 2 and |b-c| ≤ 2 ⇏ (a-c) is even.So R4 is not equivalence relation.
Reflexive : Yes, because |a-a| ≤ 2.
Symmetrix : Yes, |a-b| ≤ 2 ⟹ |b-a| ≤ 2.
Transitive : No, because |a-b| ≤ 2 and |b-c| ≤ 2 ⇏ (a-c) is even.
Consider two well-formed formulas in prepositional logic. F1 : P ⇒ ¬P
F2 : ( P⇒¬P)∨(¬P⇒P)
Which of the following statements is correct?     
F1 and F2 are both satisfiable 
F1 unsatisfiable, F2 is satisfiable 
F1 is unsatisfiable, F2 is valid  
F1 is satisfiable, F2 is valid 
The concept behind this solution is: a) Satisfiable If there is an assignment of truth values which makes that expression true. b) UnSatisfiable If there is no such assignment which makes the expression true c) Valid If the expression is Tautology Here, P => Q is nothing but –P v Q F1: P => -P = -P v –P = -P F1 will be true if P is false and F1 will be false when P is true so F1 is Satisfiable F2: (P => -P) v (-P => P) which is equals to (-P v-P) v (-(-P) v P) = (-P) v (P) = Tautology So, F1 is Satisfiable and F2 is valid Option (a) is correct.  
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