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Are there any $n \times n$ $(n \ne 1)$ matrices satisfying $$Det(A+B)=Det(A)+Det(B)$$

Trivially the above equation is True for either $A=O$ or $B=O$ or $ A=B=O$

For $2 \times 2$ matrix i have tried as follows:

Let $$A=\begin{bmatrix} a &b \\ c & d \end{bmatrix}$$ and $$B=\begin{bmatrix} p &q \\ r &s \end{bmatrix}$$ Then we have

$$Det(A+B)=(a+p)(d+s)-(b+q)(c+r)$$

$$Det(A)+Det(B)=ad-bc+ps-rq$$ Equating both we get

$$as+pd=br+qc$$ which is satisfied by infinite values of $a,b,c,d,p,q,r,s$

One such pair is

$$A=\begin{bmatrix} 2 &10 \\ 2 & 3 \end{bmatrix} $$ and

$$B=\begin{bmatrix} 4 &1 \\ 2 &5 \end{bmatrix}$$

is there any general way to find the matrices of higher order?

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    $\begingroup$ It's not too hard to come up with examples of singular matrices $A$ and $B$ such that $A+B$ is also singular. This leads to lots of examples where $\det(A)+\det(B)=\det(A+B)$. $\endgroup$ – carmichael561 May 24 '17 at 5:40
  • $\begingroup$ by example a diagonal matrix and a matrix where $a_{1,n}\neq 0$ and all other entries are zero. $\endgroup$ – Masacroso May 24 '17 at 5:41
  • $\begingroup$ exmple $$A=\begin{bmatrix}2 & 0 & 0 \\0 & 3&0 \\0 &0&5\end{bmatrix}\\B=\begin{bmatrix}0 & m & n \\0 & 0&p \\0 &0&0\end{bmatrix}\\det(A+B)=30+0,det(A)+det(B)=30+0$$ $\endgroup$ – Khosrotash May 24 '17 at 5:43
  • $\begingroup$ I though the question is focused not onto the trivial case when both (or one of them!) is zero. $\endgroup$ – Brethlosze May 24 '17 at 5:46
  • $\begingroup$ Check the PolyaPal answer in the indicated question. Addressing a way in how to build matrix in such way, just permutating rows and change sign of one rows. But not higher order. $\endgroup$ – Brethlosze May 24 '17 at 5:50
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Consider $f(A,B) = \det(A+B) - \det(A) - \det(B)$ as a continuous function (in fact a multivariate polynomial) on pairs of $n\times n$ real matrices $A, B$. There are some $A, B$ where $f(A,B) > 0$ and others where $f(A,B) < 0$ (in fact $f(AC, BC) = \det(C) f(A,B)$ for any $n \times n$ matrix $C$, so we can get one case from the other). On any continuous curve $(A_t, B_t)$ with $f(A_0,B_0) > 0 > f(A_1, B_1)$, by the Intermediate Value Theorem we have $f(A_t, B_t) = 0$ for some $t$ with $0 < t < 1$.

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