We have two $2\times 2$ matrices. The first matrix is $A=\begin{bmatrix} a & b \\ c & d \end{bmatrix}$, and the second matrix is obtained from the first: $B=\begin{bmatrix} 0 & 1 \\ -\det(A) & \operatorname{tr}(A) \end{bmatrix}$, or in other words $B=\begin{bmatrix} 0 & 1 \\ bc-ad & a+d \end{bmatrix}$.

The characteristic polynomial of $A$ is $\lambda^2 - \operatorname{tr}(A) \lambda + \det(A)$.  

The characteristic polynomial of $B$ is the same, i.e. $\det(A-\lambda E) = \det(B-\lambda E)$.

Since $\det(A-\lambda E) = \det(B-\lambda E)$, then $A$ is similar to $B$, if $A$ isn't equal to $\lambda E$. Prove why, if $\det(A-\lambda E) = \det(B-\lambda E)$, then $A$ is similar to $B$..?

Thank you.