<s>[INST] $\det (A+B)=\det (A-B)$, prove that $B^{-1}$ exists iff $b_{11}\neq b_{21}$ 
Let $A=\begin{pmatrix}
0 & 1 & 2 \\
0 & 1 & 2 \\
0 & 2 & 3 
\end{pmatrix}\in \mathbb{R}^{3\times 3}$ and let $B=\left [b_{ij}\right ]\in \mathbb{R}^{3\times 3}$ such that $\det \left (A+B\right )=\det \left (A-B\right )$. Prove that $B$ is invertible if and only if $b_{11}\neq b_{21}$.

After simplifying by using properties of the determinant, the condition $\det \left (A+B\right )=\det \left (A-B\right )$ becomes $\det \begin{pmatrix}
b_{11} & b_{12} & 2 \\
b_{21} & b_{22} & 2 \\
b_{31} & b_{23} & 3 
\end{pmatrix}=-\det \begin{pmatrix}
b_{11} & 1 & b_{13} \\
b_{21} & 1 & b_{23} \\
b_{31} & 2 & b_{33} 
\end{pmatrix}$, providing that I did not commit any mistake. This does not seem concluding and I don't think it's the right approach to the solution. Any help?
 [/INST] $$det(A+B) = det(B) +
det
\begin{pmatrix}
b_{11} & 1 & b_{13}\\
b_{21} & 1 & b_{23}\\
b_{31} & 1 & b_{33}
\end{pmatrix}
+
det
\begin{pmatrix}
b_{11} & b_{12} & 2\\
b_{21} &  b_{22} & 2\\
b_{31} &  b_{32} & 3
\end{pmatrix}
+
det
\begin{pmatrix}
b_{11} & 1 & 2\\
b_{21} &  1 & 2\\
b_{31} &  2 & 3
\end{pmatrix}
$$
$$det(A-B) = -det(B) +
det
\begin{pmatrix}
b_{11} & 1 & b_{13}\\
b_{21} & 1 & b_{23}\\
b_{31} & 1 & b_{33}
\end{pmatrix}
+
det
\begin{pmatrix}
b_{11} & b_{12} & 2\\
b_{21} &  b_{22} & 2\\
b_{31} &  b_{32} & 3
\end{pmatrix}
-
det
\begin{pmatrix}
b_{11} & 1 & 2\\
b_{21} &  1 & 2\\
b_{31} &  2 & 3
\end{pmatrix}
$$
so you get
$$
det(B) = -det
\begin{pmatrix}
b_{11} & 1 & 2\\
b_{21} &  1 & 2\\
b_{31} &  2 & 3
\end{pmatrix} = b_{11} - b_{21}
$$
and it is zero if and only if $b_{11}=b_{21}$.
</s>