Let $t\in \mathbb R$ be a parameter, and $$|A(t)|= \begin{vmatrix} a_{11}+t &a_{12}+t &\cdots &a_{1n}+t\\ a_{21}+t &a_{22}+t &\cdots &a_{2n}+t\\ \vdots &\vdots & &\vdots\\ a_{n1}+t &a_{n2}+t &\cdots &a_{nn}+t \end{vmatrix} $$ Prove:
$$|A(t)|=|A(0)|+\sum_{i,j=1}^{n}A_{ij}$$ where $A_{ij}$ denotes the cofactor of $a_{ij}$ within $|A(0)|$.
My attempt:
I have tried the two following ways, but both don't seem to work:
1).Induction. The hypothesis is that it holds for $A_k(t)$ of the size $k\times k$ and I am to show it also holds for $A_{k+1}(t)$. I tried expanding $A_{k+1}(t)$ by the first row or column, but the result was too complicated to deal with, although I did obtain a $|A_{k+1}(0)|$ from it.
2).Another thread was to use
$$|A_{k+1}(t)|=
\begin{vmatrix}
1 &t &t &\cdots&t \\
0 &a_{11}+t &a_{12}+t &\cdots &a_{1,k+1}+t\\
0 &a_{21}+t &a_{22}+t &\cdots &a_{2,k+1}+t\\
\vdots &\vdots &\vdots & &\vdots\\
0 &a_{k+1,1}+t &a_{k+1,2}+t &\cdots &a_{k+1,k+1}+t
\end{vmatrix}
$$
and then substracted the first row from out all the others, I obtained
$$|A_{k+1}(t)|=
\begin{vmatrix}
1 &t &t &\cdots&t \\
-1 &a_{11} &a_{12} &\cdots &a_{1,k+1}\\
-1 &a_{21} &a_{22} &\cdots &a_{2,k+1}\\
\vdots &\vdots &\vdots & &\vdots\\
-1 &a_{k+1,1} &a_{k+1,2} &\cdots &a_{k+1,k+1}
\end{vmatrix}$$
Then I thought I should proceed to expanding the first row or column, in both cases I also obtained a $|A_{k+1}(0)|$, but the rest was just a terrible mess.
I expect that maybe I need some extra tricks to deal with the second term of RHS, i.e., the sum of all the cofactors of $a_{ij}$. I know the Laplace expansion, but it doesn't seem to suit this case. Could you help me? Best regards.
Let $v$ be the vector with $1$ on each coordinate. Let $A_i$ be column vectors of $A$. Then :
$$det(A)=det(A_1+tv,A_2+tv,...,A_n+tv)$$
Let us set $B_k^0:=A_k$ and $B_k^1:=v$ then expanding $det(A)$ by multilinearity :
$$det(A)=\sum_{i=0}^n\sum_{E\in\mathcal{P}_i(\{1,...,n\})}t^k.det((B_k^{\chi_E(k)})_{1\leq k\leq n}) $$
But if $i\geq 2$ and $E\in \mathcal{P}_i(\{1,...,n\})$ then for at least two $k$ we have that $B_k^{\chi_E(k)}=v$ then in that case :
$$det((B_k^{\chi_E(k)})_{1\leq k\leq n})=0$$
This means that :
$$det(A)=t^0det(A_1,...,A_n)+t\sum_{r=1}^ndet(A_1,...,A_{r-1},v,A_{r+1},...,A_n)$$
Now (expanding the determinant with respect to the $r$-th column) :
$$det(A_1,...,A_{r-1},v,A_{r+1},...,A_n)=\sum_{j=1}^nA_{j,r}$$
$$det(A)=det(A(0))+t\sum_{r=1}^n\sum_{j=1}^nA_{j,r}$$