Strategic form: Nash equilbrium I am currently working through a question where I have to find any Nash equilibrium not in pure strategies, together with the associated payoffs.

I have managed to identify the pure strategy Nash equilibria: (4,10) and (5,7).
As far as I can see, there are no pure strategies for either player which are strictly dominated.
In order to find any Nash equilibrium not in pure strategies, do I delete (4,10) and (5,7) from the matrix?
I.e. so the matrix becomes:

I'm not too sure what to do from this point onward, do I need to set up some sort of a linear system such as:


*

*3σ+1-σ-τ = 2(1-σ-τ)

*9π= 8π+6(1-π-ρ)
And substitute τ=1-σ and ρ=1-π into the system?
 A: Hint: No, do not delete (4,10), that would be a mistake. Observe instead that the mixed stategy $$x=(0,\epsilon, 1-\epsilon)$$ for $\epsilon>0$ (small enough) of player A (rows) dominates his first strategy (u). Thus you can delete (u) to obtain $$A=\pmatrix{0&4&0 \\5&0&2}\qquad \text{ vs } \qquad B=\pmatrix{0 &10& 0\\7 &0 &6}$$ Now similarly the mixed strategy $$y=(1-\epsilon,\epsilon, 0)$$ for $\epsilon>0$ (small enough) of player B (columns) dominates his strategy (r) and therefore you can delete it. This leaves you with $$A=\pmatrix{0&4 \\5&0}\qquad \text{ vs } \qquad B=\pmatrix{0 &10\\7 &0}$$ form which you can find all three Nash equilibria of the game (2 in pure strategies and one completely mixed)!
A: Here is my solution:



*

*The pure strategy nash equilibria are $(m,c)$ and $(d,l)$.


In order to proceed with finding mixed strategies, I am going to delete $\mathcal r$ and $\mathcal u$ as they seem to be weakly dominated.

After solving a few simple linear equations:
$\left[ \pi^{*},\rho^{*},\sigma^{*},\tau^{*} \right]$=$\left[ 0,\frac{7}{17},\frac{4}{9},\frac{5}{9} \right]$
