A question about intersection of two cones in $\mathbb{R}^{n\times n}$

Question

I've been frustrated with this problem for several days:

In $\mathbb{R}^n$, let $K\subset \lbrace 1,\dots,n\rbrace$, and $A\in \mathbb{R}^{n\times n}$ matrix with $a_{ii}=1$, and $a_{ij}<0 \ \forall i\neq j$. We define the transform $T_K:\mathbb{R}^{n\times n} \rightarrow \mathbb{R}^{n\times n}$, $T_K(A)=B$ as follows:

1. If $i,j\in K, i=j$, then $b_{ij}=1$,
2. If $i,j\notin K, i=j$, then $b_{ij}=\pm 1$,
3. For each $j\in K$, $b_{ij}=\lambda_j a_{ij}, \lambda_j\in \lbrace 0,1\rbrace, \ \forall i\in K, i\neq j$,
4. Otherwise, $b_{ij}=0$.

I have two questions about $A$ and his transform $T_K(A)$:

1. If I define $C(A)$ as the interior of the cone generated by A, that's it, \begin{equation} C(A)=\{Ax:x>0\} \end{equation} and define $C(T_K(A))$ as the interior of the cone generated by $T_K(A)$, It is possible assert that \begin{equation} C(A)\cap C(T_K(A))\neq \emptyset ? \end{equation}
2. If I consider $K,K'\subset \lbrace 1,\dots,n\rbrace, K\neq K'$, It is possible assert that \begin{equation} C(T_K(A))\cap C(T_{K'}(A))\neq \emptyset ? \end{equation}

Thanks a lot if anyone bring me a light.

EDIT: I forgot to mention that in the condition

2. If $i,j\notin K, i=j$, then $b_{ij}=\pm 1$

There is always at least one $b_{ij}$ with $i,j\notin K, i=j$ and $b_{ij}=1$.