Determinants of $I+A$ and $I-A$

Question

Let $A$ be a $4\times4$ matrix with $\det(A) = 2$, find $\det(I+A)$ and $\det(I-A)$/

I read that we need the trace to find $\det(I+A)$, but I do not know how to find it when we only know the determinant of $A$. Help is much appreciated.

Answer 1

Assuming your matrix takes values in $\mathbb{R}$, then there is no unique answer without additional information.

To see this, let $$ A:= \begin{bmatrix} c & 0 & 0 & a \\ 0 & 2 & 0 & 0 \\ 0 & 0 & 2 & 0 \\ 1 & 0 & 0 & 2 \end{bmatrix} $$ where $a=\frac{4c-1}{2}$ and $c\in\mathbb{R}$. We have $\det A=2$, but $\det (I+A)= 9c+\frac{63}{2}$ and $\det(I-A)=-c-\frac{1}{2}$. Since $c$ is arbitrary, $\det(I+A)$ and $\det(I-A)$ can be arbitrary.

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Even if we fix the trace of $A$ to be a constant $s$, the answer is still not unique. Let $$ A:= \begin{bmatrix} 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & h \\ 0 & 0 & 1 & 0 \\ 0 & 2 & 0 & s-1 \end{bmatrix} $$ where $h\in\mathbb{R}$. It is easy to verify that $\det(A)=2$, but $\det(I+A)=4+2s-4h$, which is arbitrary.

Sidenote: In the 2D case, $\det(I+A)$ and $\det(I-A)$ are actually uniquely determined by $\det(A)$ and $\text{tr}(A)$.