What change does a determinant undergo if to each column (beginning with
the second) we add the preceding column and at the same time we add the
last to the first?
Solution
Let us write $a_1,\ldots,a_n$ for the column vectors of the original matrix $A$. Then the new matrix is $B = (a_1+a_2|a_2+a_3|\cdots|a_n+a_1)$ and by linearity we have
$$ \det(B) = \sum_{i_1,\ldots,i_n \in\{0,1\}} \det(a_{1+i_1}|\cdots |a_{n+i_n})$$ where addition in indices is done modulo $n$. If $i_1=\cdots=i_n = 0$ we have $\det(a_{1+i_1}|\cdots |a_{n+i_n}) = \det(A).$ If $i_1=\cdots=i_n = 1$ we have $\det(a_{1+i_1}|\cdots |a_{n+i_n}) = \det(a_2|a_3|\cdots|a_n|a_1) = (-1)^{n-1}.$ Finally if $i_s = 0$ but $i_t=1$ for some $s,t$ then there is an $r$ with $i_r=1,i_{r+1}=0$. But then $(a_{1+i_1}|\cdots |a_{n+i_n})$ has two equal columns and so its determinant is equal to $0$. Therefore $$\det(B) = (1 + (-1)^{n-1})\det(A)$$
The interested reader can see one more solution using fundamental elements in mathimatikoi.org .
Solution
Let us write $a_1,\ldots,a_n$ for the column vectors of the original matrix $A$. Then the new matrix is $B = (a_1+a_2|a_2+a_3|\cdots|a_n+a_1)$ and by linearity we have
$$ \det(B) = \sum_{i_1,\ldots,i_n \in\{0,1\}} \det(a_{1+i_1}|\cdots |a_{n+i_n})$$ where addition in indices is done modulo $n$. If $i_1=\cdots=i_n = 0$ we have $\det(a_{1+i_1}|\cdots |a_{n+i_n}) = \det(A).$ If $i_1=\cdots=i_n = 1$ we have $\det(a_{1+i_1}|\cdots |a_{n+i_n}) = \det(a_2|a_3|\cdots|a_n|a_1) = (-1)^{n-1}.$ Finally if $i_s = 0$ but $i_t=1$ for some $s,t$ then there is an $r$ with $i_r=1,i_{r+1}=0$. But then $(a_{1+i_1}|\cdots |a_{n+i_n})$ has two equal columns and so its determinant is equal to $0$. Therefore $$\det(B) = (1 + (-1)^{n-1})\det(A)$$
The interested reader can see one more solution using fundamental elements in mathimatikoi.org .
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