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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 $B = A - I$?

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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 $$\det (A+B)=\det(A)+\det(B)+\sum_{i=1}^{n-1}\Gamma_n^i\det(A/B^i)$$ Where $\Gamma_n^i\det(A/B^i)$ 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

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    $\begingroup$ Note: This formula is just a consequence of applying the multilinearity of the determinant once to each row (consecutively, so you end up with a sum of $2^n $ addends). $\endgroup$ – darij grinberg Mar 12 '18 at 8:20
  • $\begingroup$ Do you have an answer to this?: math.stackexchange.com/questions/3184171/… $\endgroup$ – Susan_Math123 Apr 11 at 19:53
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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.

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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)

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  • $\begingroup$ You don't need diagonalizable in your arguments. You can always work with the Jordan form, or with the Schur decomposition, and reason exactly like you did. $\endgroup$ – Martin Argerami Dec 12 '12 at 13:07
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The characteristic polynomial $P_A(\lambda)$ of a matrix $A$ is defined as $$ P_A(\lambda)=\det(A-I\lambda) $$ Therefore, $\det(A)=P_A(0)$, while $\det(A+I)=P_A(-1)$ and $\det(A-I)=P_A(1)$.

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  • $\begingroup$ Merry Christmas. $\endgroup$ – user1551 Dec 12 '12 at 9:43
  • $\begingroup$ @user1551: and Happy New Year! $\endgroup$ – robjohn Dec 12 '12 at 10:02
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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 $

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  • $\begingroup$ P.S Someone please help with formatting this to correct transposes. Didn't manage. $\endgroup$ – mexmex Dec 4 '18 at 9:54
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For I and A $N\times N$ matrices we have:

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

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