Game Theory 2x2 Static Game: Finding the Pure Strategy and Mixed Strategy Nash Equilibria with Weakly Dominant Strategies Payoff Matrix
I was trying to solve the payoff matrix above for the pure strategy and the mixed strategy Nash equilibria. It seems that both players have a weakly dominant strategy, by choosing U, player 1 would be indifferent or better off compared to choosing D, and while choosing R, player 2 would be indifferent or better off compared to choosing L. This brings me to the pure strategy Nash equilibrium UR or RU, with the payoff 1,1.
However, I'm mainly concerned with the result of the mixed strategy Nash equilibrium. When I derived the probabilities of player 1 choosing U (p), and player 2 choosing L (q), I got probabilities p=0 and q=1.
Does this imply that LD is a pure strategy Nash equilibrium or a mixed strategy Nash equilibrium? Also, why would player 1 and player 2 both opt for their weakly dominated strategies instead of their weakly dominant strategies?
Thank you in advance!

I think the solution shows UR is the only Nash Equilibrium. Using the convention that payoffs are (row player, column player) the mixed payoff for Row is $$p[q*1+(1-q)*1]+(1-p)[q*1-(1-q)*1]$$
$$p+2q-1-2pq+p=(2-2q)p+2q-1.$$
Taking the derivative with respect to $p$, we get
$$2-2q.$$
Now $q\leq 1$ so the derivative is positive except when $q=1$. So Row chooses Up with probability one.
For Column, a similar calculation shows
$$q[-p+(1-p)]+(1-q)[p+1-p]=1-2pq$$
Taking the derivative with respect to $q$, we get $-2p$ and that is maximized at zero, so Column plays L with probability zero, that is plays Right with probability one. And UR is the weakly dominant strategy as you point out.
One final pedantic note: if $q=1$, Row is indifferent, but Column won't be  choosing that if $p$ isn't 1, so it's not an equilibrium.
