Krdytkyu

Suppose there is $n times n$ matrix $A$. If we form matrix $B = A+I$ where $I$ is $n times n$ identity matrix, how does $|B|$ - determinant of $B$ - change compared to $|A|$? And what about the case where $B = A - I$?

In case you are interested, there is a result which expresses $det(A+B)$ in terms of $det(A)$ and $det (B)$, it is given by following $$det (A+B)=det(A)+det(B)+sum_{i=1}^{n-1}Gamma_n^idet(A/B^i)$$
Where $Gamma_n^idet(A/B^i)$ is defined as a sum of the combination
of determinants, in which the $i$ rows of $A$ are substituted
by the corresponding rows of $B$. You can find the proof in this IEEE article: http://ieeexplore.ieee.org/stamp/stamp.jsp?tp=&arnumber=262036&userType=inst

There is no simple answer. $P(lambda) = det(A-lambda I)$ is the characteristic polynomial of $A$. This is a polynomial of degree $n$: the coefficient of $lambda^n$ is $(-1)^n$ and the coefficient of $lambda^0$ is $det(A)$. The coefficients of other powers of $lambda$ are various functions of the entries of $A$. $det(A+I)$ and $det(A-I)$ are $P(-1)$ and $P(1)$; that's about all there is to say about them.

As others already have pointed out, there is no simple relation. Here is one answer more for the intuition.
Consider the (restricting) codition, that $A_{n times n}$ is diagonalizable, then $$det(A) = lambda_0 cdot lambda_1 cdot lambda_2 cdot cdots lambda _{n-1} $$
Now consider you add the identity matrix. The determinant changes to
$$det(B) = (lambda_0+1) cdot (lambda_1+1) cdot (lambda_2+1) cdot cdots (lambda _{n-1} +1)$$
I think it is obvious how irregular the result depends on the given eigenvalues of A. If some $lambda_k=0$ then $det(A)=0$ but that zero-factor changes to $(lambda_k+1)$ and det(B) need not be zero. Other way round - if some factor $lambda_k=-1$ then the addition by I makes that factor $lambda_k+1=0$ and the determinant $det(B)$ becomes zero. If some $0 gt lambda_k gt -1$ then the determinant may change its sign...

So there is no hope to make one single statement about the behave of B related to A - except that where @pritam linked to, or except you would accept a statement like $$det(A)=e_n(Lambda_n) to det(B)= sum_{j=0}^n e_j(Lambda) $$ where $ Lambda = {lambda_k}_{k=0..n-1} $ and $e_k(Lambda)$ denotes k'th elementary symmetric polynomial over $Lambda$... (And this is only for diagonalizable matrices)

The characteristic polynomial $P_A(lambda)$ of a matrix $A$ is defined as
$$
P_A(lambda)=det(A-Ilambda)
$$
Therefore, $det(A)=P_A(0)$, while $det(A+I)=P_A(-1)$ and $det(A-I)=P_A(1)$.

For I and A $Ntimes N$ matrices we have:

$$det(I-A)=1-sum_{j}^{N}A_{jj}+sum_{ngeq 2}^{N}(-1)^{n}sum_{1leq j_{1}<...<j_{n}leq N}det((A_{j_{a},j_{b}})^{n}).$$

I would like to contribute the following special case which I encountered when I was looking for an answer and stumbled upon this thread.

If A=uv^intercal, then det(I+A) = 1+u^Tv