$\det (A+B)=\det (A-B)$, prove that $B^{-1}$ exists iff $b_{11}\neq b_{21}$

Question

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?

Answer 1

$$det(A+B) = det(B) + det \begin{pmatrix} b_{11} & 1 & b_{13}\\ b_{21} & 1 & b_{23}\\ b_{31} & 1 & b_{33} \end{pmatrix} + det \begin{pmatrix} b_{11} & b_{12} & 2\\ b_{21} & b_{22} & 2\\ b_{31} & b_{32} & 3 \end{pmatrix} + det \begin{pmatrix} b_{11} & 1 & 2\\ b_{21} & 1 & 2\\ b_{31} & 2 & 3 \end{pmatrix} $$ $$det(A-B) = -det(B) + det \begin{pmatrix} b_{11} & 1 & b_{13}\\ b_{21} & 1 & b_{23}\\ b_{31} & 1 & b_{33} \end{pmatrix} + det \begin{pmatrix} b_{11} & b_{12} & 2\\ b_{21} & b_{22} & 2\\ b_{31} & b_{32} & 3 \end{pmatrix} - det \begin{pmatrix} b_{11} & 1 & 2\\ b_{21} & 1 & 2\\ b_{31} & 2 & 3 \end{pmatrix} $$ 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 $b_{11}=b_{21}$.