Answer:
Follows are the solution to the given point:
Step-by-step explanation:
In point a:
¬∃y∃xP (x, y)
∀x∀y(>P(x,y))
In point b:
¬∀x∃yP (x, y)
∃x∀y ¬P(x,y)
In point c:
¬∃y(Q(y) ∧ ∀x¬R(x, y)) ∀y(> Q(y) V ∀ ¬ (¬R(x,y)))
∀y(¬Q(Y)) V ∃xR(x,y) )
In point d:
¬∃y(∃xR(x, y) ∨ ∀xS(x, y))
∀y(∀x>R(x,y)) ∃x>s(x,y))
In point e:
¬∃y(∀x∃zT (x, y, z) ∨ ∃x∀zU (x, y, z))
∀y(∃x ∀z)>T(x,y,z) ∀x ∃z> V (x,y,z))
Answer:
If k = −1 then the system has no solutions.
If k = 2 then the system has infinitely many solutions.
The system cannot have unique solution.
Step-by-step explanation:
We have the following system of equations
The augmented matrix is
The reduction of this matrix to row-echelon form is outlined below.
The last row determines, if there are solutions or not. To be consistent, we must have k such that
Case k = −1:
If k = −1 then the last equation becomes 0 = −2 which is impossible.Therefore, the system has no solutions.
Case k = 2:
This gives the infinite many solution.