GATE-CS-2001

Singular Matrix: A square matrix is singular if and only if its determinant value is 0. S1 is True: The sum of two singular n × n matrices may be non-singular It can be seen be taking following example. The following two matrices are singular, but their sum is non-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

S2 is True: The sum of two n × n non-singular matrices may be singular

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

So basically, we have to tell whether these relations are equivalence or not.

1. 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.
2. R2(a,b)
   • Reflexive : No, because (a+a) is even.
   So R2 is not equivalence relation.
3. 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.
4. 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.

So option (b) is correct.. Source: http://www.cse.iitd.ac.in/~mittal/gate/gate_math_2001.html

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. https://en.wikipedia.org/wiki/Tautology_(logic) This solution is contributed by Anil Saikrishna Devarasetty.

We can easily build a DFA for S1. All we need to check is whether input string has even number of 0's. Therefore S1 is regular. We can't make a DFA for S2. For S2, we need a stack. Therefore S2 is not regular.

Refer http://en.wikipedia.org/wiki/Context-free_language#Closure_properties

Refer http://en.wikipedia.org/wiki/Powerset_construction

See question 3 of http://www.geeksforgeeks.org/operating-systems-set-2/

See question 4 of http://www.geeksforgeeks.org/operating-systems-set-2/

A low memory can be connected to 8085 by using READY signal, Communication is only possible when READY signal is set .So (D) is correct option

Stack pointer register hold the address of top of stack, which is the location of memory at which CPU should resume its execution after servicing some interrupt or subroutine call. So if SP register is not available then no subroutine call instructions are possible. So (A) is correct option.