$P(X,Y)=$\begin{bmatrix} P(x_1,y_1) & P(x_1,y_2)\\ P(x_2,y_1) & P(x_2,y_2) \end{bmatrix} =
\begin{bmatrix} 0.54 & 0.06\\ 0.06 & 0.34 \end{bmatrix}
And i calculate the $H(x)=-1log_21=0$ and $H(y)=-1log_21=0$,and $H(X|Y)=H(Y|X)=-2log_22=-2$ ,so $I(X,Y)=H(X)-H(X|Y)=0-(-2)=2$,but according the explanation of mutual information in wiki https://en.wikipedia.org/wiki/Conditional_probability ,the picture is as below,it seems that it is impossible that the mutual information equal to $2$ when H(X) and H(Y)=$0$ ,is my calculation wrong?or it does possible that mutual information equal to $2$ when H(X) and H(Y)=$0$?
I would interpret the $P$ matrix in the statement of the problem as telling me that the random variable $x$ has the value $x_1$ with probability $0.54+0.06=0.6$ and has the value $x_2$ with probability $0.6+0.34=0.4$. So $$ H(x)=-(0.6)\log(0.6)-(0.4)\log(0.4), $$ and similarly for $H(y)$. Neither of these is zero.
Your calculation of $H(x|y)$ and $H(y|x)$ also looks wrong (nor do I understand why the variables $x$ and $y$ became $X$ and $Y$ half way through the question).Unless I've messed up the arithmetic, $$ H(x|y)=H(y|x)=-(0.54)\log(0.9)-(0.06)\log(0.1)-(0.06)\log(0.15)-(0.34)\log(0.85). $$