Neither of these strategies are dominated by other. So, we have to use mixed strategies to find the nash equilibrium with the help of removing dominated strategies.
But, I can't not continue to solve this. I mix-up again and again and also confused.
I see the answer of What is the mixed strategy Nash equilibrium in this game? but he didn't describe the way to find the nash equilibrium using the value of alpha and bita. He just showed the equation.
Show a method that can solve for any strategy values. Thank you for help.
For Red, it is not true that neither of his pure strategies dominates the other. It is true that neither of his pure strategies strictly dominates the other, but since $$\ R\left(\mbox{Up, Left}\right) =-1 \ge R\left(\mbox{Up, Right}\right)=-1\ ,\ \mbox{and}\\ R\left(\mbox{Down, Left}\right)=2 > R\left(\mbox{Down, Right}\right)=-1\ ,$$ where $\ R\ $ is Red's payoff matrix, then his first pure strategy, Left, clearly dominates his second, Right. And since $$ B\left(\mbox{Up, Left}\right) = 3 > B\left(\mbox{Down, Left}\right)=-1\ ,$$ where $\ B\ $ is Blue's payoff matrix, then $\ \mbox{(Up, Left)}\ $ is a pure strategy Nash equilibrum.
Response to comment: Mixed strategy Nash equilibria for general bimatrix games can be found with the Lemke-Howson algorithm.
However, for the simple game described here (with Red's payoff matrix modified as proposed in the comment), it's possible to find a mixed strategy Nash equilibrium simply by solving a couple of linear equations. Blue's payoff matrix is $$ B=\begin{pmatrix} 3&-1\\ -1&2 \end{pmatrix}\ , $$ so if Red plays the mixed strategy $\ \left(q, 1-q\right)\ $, Blue's expected payoff will be $$ 3q-(1-q)$$ when he chooses his first pure strategy (Up), and $$-q+2(1-q)$$ when he chooses his second pure strategy (Down). Thus, if Red chooses $\ q\ $ to make both these expected payoffs equal, $$ 3q-(1-q)=-q+2(1-q)\ ,$$ $\ \left(\ q=\frac{3}{7}\ \right)\ $, then Blue's expected payoff will be $\ 3\cdot \frac{3}{7}- \frac{4}{7}=\frac{5}{7}=-\frac{3}{7}+2\cdot\frac{4}{7}\ $ regardless of what strategy (whether pure or mixed) he chooses.
Likewise, Red's payoff matrix (with the amendment proposed in the comment) is $$ R=\begin{pmatrix} -2&-1\\ 2&-1 \end{pmatrix}\ , $$ so if Blue chooses a mixed strategy $\ \left(p, 1-p\right)\ $ such that $$ -2p+2\left(1-p\right)=-p-\left(1-p\right)\ ,$$ that is, with $\ p=\frac{3}{4}\ $, then Red's expected payoff will be $-2\cdot\frac{3}{4}+2\cdot\frac{1}{4}\ = -1 = -\frac{3}{4}-\frac{1}{4}\ $ regardless of what strategy (whether pure or mixed) he chooses.
Thus, the mixed strategies $\ \left(\frac{3}{4}, \frac{1}{4}\right)\ $ for Blue, and $\ \left(\frac{3}{7}, \frac{4}{7}\right)\ $ for Red constitute a Nash equilibrium for this gane.