As the title says, I would like to write a formula using index notation for the determinant of the matrix $I + A$, where $A$ is a $4$ x $4$ matrix and $I$ is the indentity matrix, using the trace of the matrix.
The problem is then: $$ \text{det}\ (I + A) .$$
My attempt: Why I have is that $ \text{det}\ (I + A) = \text{det}\ (I) + \text{det}\ (A) = 1 + \text{det} (A).$ So the problem is reduced to calculating the determinant of the matrix $A$. Now, using index notation I have $$ \text{det} (A)= \epsilon_{abcd} \epsilon_{ijkl} A_{ai} A_{bj} A_{ck} A_{dl} ,$$ then, I rewrote the product $\epsilon_{abcd} \epsilon_{ijkl}$ in terms of the Kronecker delta as follows: $$ \epsilon_{abcd} \epsilon_{ijkl} = \delta_{ai} (\epsilon_{jkl} \epsilon_{bcd} \delta_{bj} \delta_{ck} \delta_{dl}) - \\ \delta_{aj} (\epsilon_{ikl} \epsilon_{bcd} \delta_{bi} \delta_{ck} \delta_{dl}) + \delta_{ak} (\epsilon_{ijl} \epsilon_{bcd} \delta_{bi} \delta_{cj} \delta_{dl}) - \delta_{al} (\epsilon_{ijk} \epsilon_{bcd} \delta_{bi} \delta_{cj} \delta_{dk}).$$
Plugging in this result in the original expression for the determinant I found: $$ \text{det} (I + A) = 1 + \epsilon_{jkl} \epsilon_{bcd} (A_{ai} \delta_{ai}) (A_{bj} \delta_{bj}) (A_{ck} \delta_{ck}) (A_{dl} \delta_{dl}) \\ - \epsilon_{ikl} \epsilon_{bcd} A_{ai} \delta_{aj} A_{bj} \delta_{bi} A_{ck} \delta_{ck} A_{dl} \delta_{dl} + \epsilon_{ijl} \epsilon_{bcd} A_{ai} \delta_{ak} A_{bj} \delta_{bi} A_{ck} \delta_{cj} A_{dl} \delta_{dl} \\ - \epsilon_{ijk} \epsilon_{bcd} A_{ai} \delta_{al} A_{bj} \delta_{bi} A_{ck} \delta_{cj} A_{dl} \delta_{dk}.$$ I recognize that the terms in parentheses are the definition of the trace of A, but for some reason I think this is a very messy approach because the Levi-Civita symbol has now three indices and I don't know how I could find an expression for them, in order to clean up the result.
Firstly, is my the result correct?
Is there a property or expression that could help me simplify the problem?