Here's one sensible sequence of steps:
Step 1: Notice that T strictly dominates $B$, since $(3,1,4)$ is componentwise strictly greater than $(1,0,3)$. Remove $B$ and we are left with a $2 \times 3$ game.
Step 2: In this new game, with $B$ removed, $R$ dominates $C$, since $(2,3)$ is componentwise strictly greater than $(1,2)$. After removing $C$ we are left with a $2 \times 2$ game:
$$ \left(
\begin{array}{c|cc}
& L & R\\
\hline
T &3,0& 4,2\\
M &3,4& 2,3\\
\end{array}
\right) $$
Step 3: Having found two pure equilibria already, look for non-pure equilibria. Player 2 can be made indifferent between $L$ and $R$ as we see below. But, player 1 cannot be made indifferent between $T$ and $M$ because $T$ weakly dominates $M$: as soon as there is any positive probability on $R$, player 1 strictly prefers $T$. Thus player 2 cannot mix in equilibrium, and actually the pure equilibrium $(M, L)$ is actually only the endpoint of a range of equilibria:
$$ ((1-p, p), L)\ \text{where } p \in [2/3, 1] $$
The threshold of $p=2/3$ is the point at which player II is indifferent between $L$ and $R$ against $(1-p,p)$. When $p=2/3$ both L and R give expected payoff $1/3 \cdot 0 + 2/3 \cdot 4 = 1/3 \cdot 2 + 2/3 \cdot 3 = 8/3$.
A range of equilibria like this is only possible is a degenerate game. A $2 \times 2$ game is necessarily degenerate. More generally, a game is degenerate if there exists a mixed strategy $x$ with support size $k$ (i.e. , $|\{x_i \ | \ x_i >0\}| = k$) and more than $k$ best responses against $x$. In this example, the strategy $x$ is the pure strategy $L$, which has support size 1 but two best responses against it, $TM$ and $M$.
You can check the equilibria at:
The output from the former for this game is:
2 x 2 Payoff matrix A:
3 4
3 2
2 x 2 Payoff matrix B:
0 2
4 3
EE = Extreme Equilibrium, EP = Expected Payoff
Decimal Output
EE 1 P1: (1) 0.333333333333 0.666666666667 EP= 3.0 P2: (1) 1.0 0.0 EP= 2.66666666667
EE 2 P1: (2) 1.0 0.0 EP= 4.0 P2: (2) 0.0 1.0 EP= 2.0
EE 3 P1: (3) 0.0 1.0 EP= 3.0 P2: (1) 1.0 0.0 EP= 4.0
Rational Output
EE 1 P1: (1) 1/3 2/3 EP= 3 P2: (1) 1 0 EP= 8/3
EE 2 P1: (2) 1 0 EP= 4 P2: (2) 0 1 EP= 2
EE 3 P1: (3) 0 1 EP= 3 P2: (1) 1 0 EP= 4
Connected component 1:
{1, 3} x {1}
Connected component 2:
{2} x {2}
Here one of the two equilibrium components (the range of equilibria described above in terms of $p$) is denoted by
Connected component 1:
{1, 3} x {1}
The other equilibrium component is the pure strategy equilibrium $(T,R)$.
Yes, there are infinitely many, with player 2 mixing.
Assume that player 2 mixes: Here it may be possible because the payoff for player 2 in $(s,s,n)$ and in $(s,n,n)$ is equal. You only need to check if $(s,(y,1-y),n)$ is an equilibrium for $0<y<1$. Given that $3$ plays $n$, $s$ is a best response for player $1$ for any $y$ of player $2$ such that $$(-1)y+(1)(1-y)\ge (1)y+0(1-y)\implies y\le \frac13$$ Now, given that $1$ plays $s$, $n$ is a best response for player $3$ for any $y$ of player $2$ such that $$(-2)y+0(1-y)\ge(1)y+(-1)(1-y)\implies y\le \frac14$$ Hence, any strategy profile of the form $$(s,(y,1-y),n)\quad \text{with } 0<y\le \frac14$$ is a Nash equilibrium where only player 2 uses a mixed strategy.
To see that there are no other such equilibria:
Assume that player $3$ plays the mixed strategy $(z,1-z)$ where $0<z<1$ is the probability of playing $s$. If players $1$ and $2$ play the pure strategy profile $(s,s)$ then player $3$ has an incentive to choose $z=1$, hence this is not an equilibrium. If players $1$ and $2$ play any other pure strategy profile, then player $3$ has an incentive to choose $z=0$, hence there is no equilibrium where $3$ uses a mixed strategy and $1$ and $2$ use pure strategies.
Repeat for player 1: Assume that player $1$ plays the mixed strategy $(x,1-x)$ where $0<x<1$ is the probability of playing $s$. If players $2$ and $3$ play the pure strategy profile $(s,s)$ then player $1$ has an incentive to choose $x=1$, hence this is not an equilibrium. Similarly, if players $2$ and $3$ play any other pure strategy profile, then player $1$ has an incentive to choose either $z=1$ or $x=0$, hence there is no equilibrium where $1$ uses a mixed strategy and $2$ and $3$ use pure strategies.
Best Answer
Indeed, finding a Nash Equilibrium for larger games than $2\times 2$ is usually not a fun process, as there are many conditions to verify and (simple) equations to solve. The general process would involve checking all possibilities: only pure equilibria, equilibria where only two actions are mixed, and equilibria where all actions are played (for each player!). Luckily, when these kind of questions are given to be solved by hand, usually there are tricks that make the solution more simple, such as symmetry or dominance.
Indeed, lets start with dominance. Note that all the payoffs for the row player in the T row are higher than in the M, so T dominates M and can be deleted from the matrix. It makes the work simpler.
More tricky but also quite visible is that $C$ is dominated, this time by a mixed action. If column player mixes $L$ and $R$ with equal probabilities, his payoff is $2.5$ against $T$ and $1.5$ against $B$, which are higher than his payoffs when playing $C$.
Now you have only a $2\times 2$ games where it is easy to see that there are two pure equilibria ($(T,L)$ and $(B,R)$) and one mixed (each player plays each action with probability half).