One pure Nash equilibrium, no mixed equilibrium?

I’m given a 4×4 payoff matrix

begin{bmatrix}(0,7)&(2,5)&(7,0)&(0,1)(5,2)&(3,3)&(5,2)&(0,1)(7,0)&(2,5)&(0,7)&(0,1)(0,0)&(0,-2)&(0,0)&(10,-1)end{bmatrix}

There isn’t any strategy that is dominated for both players.
As a combination of best responses, I see that (3,3) is a pure strategy Nash equilibrium. I tried to compute mixed strategy Nash equilibrium by setting probabilities of row player’s actions as p,q,r,s and equating the payoffs for the column player, which yields p= -1/3

Does this mean there is no mixed strategy Nash equilibrium? Thus only one Nash equilibrium exists?