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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?

Let \(\displaystyle{A=\left ( a_{ij} \right )}\) an \(\displaystyle{n\times n}\) matrix.

By applying the changes referred to it's determinant we are taking the following determinant

\(\left | B \right |=\begin{vmatrix}
a_{11}+a_{1n} &a_{12}+a_{11} &\cdots &a_{1n}+a_{1\left ( n-1 \right )} \\
a_{21}+a_{2n} &a_{22}+a_{21} & &a_{2n}+a_{2\left ( n-1 \right )} \\
\vdots &\ddots & &\vdots \\
a_{n1}+a_{nn} &a_{n2}+a_{n1} &\cdots &a_{nn}+a_{n\left ( n-1 \right )}
\end{vmatrix}\)

\(=\begin{vmatrix}
a_{11} &a_{12}+a_{11} & &a_{1n}+a_{1\left (n-1 \right )} \\
a_{21} &a_{22}+a_{21} & &a_{2n}+a_{2\left ( n-1 \right )} \\
\vdots &\vdots &\ddots & \\
a_{n1} &a_{n2}+a_{n1} & &a_{nn}+a_{n\left ( n-1 \right )}
\end{vmatrix}+\begin{vmatrix}
a_{1n} &a_{12}+a_{11} & &a_{1n}+a_{1\left (n-1 \right )} \\
a_{2n} &a_{22}+a_{21} & &a_{2n}+a_{2\left ( n-1 \right )} \\
\vdots &\vdots &\ddots & \\
a_{nn} &a_{n2}+a_{n1} & &a_{nn}+a_{n\left ( n-1 \right )}
\end{vmatrix}\)

but,

\(\left | K \right |=\begin{vmatrix}
a_{11} &a_{12}+a_{11} & &a_{1n}+a_{1\left (n-1 \right )} \\
a_{21} &a_{22}+a_{21} & &a_{2n}+a_{2\left ( n-1 \right )} \\
\vdots &\vdots &\ddots & \\
a_{n1} &a_{n2}+a_{n1} & &a_{nn}+a_{n\left ( n-1 \right )}
\end{vmatrix}\xrightarrow[]{C_{2}\rightarrow C_{2}-C_{1}}\begin{vmatrix}
a_{11} &a_{12} & &a_{1n}+a_{1\left (n-1 \right )} \\
a_{21} &a_{22} & &a_{2n}+a_{2\left ( n-1 \right )} \\
\vdots &\vdots &\ddots & \\
a_{n1} &a_{n2} & &a_{nn}+a_{n\left ( n-1 \right )}
\end{vmatrix}\)

\(\xrightarrow[]{C_{3}\rightarrow C_{3}-C_{2}}\cdots \xrightarrow[]{C_{n}\rightarrow C_{n}-C_{n-1}}\begin{vmatrix}
a_{11} &a_{12} & &a_{1n} \\
a_{21} &a_{22} & &a_{2n} \\
\vdots &\vdots &\ddots & \\
a_{n1} &a_{n2} & &a_{nn}
\end{vmatrix}=\left | A \right |\)

and

\(\begin{vmatrix}
a_{1n} &a_{12}+a_{11} & &a_{1n}+a_{1\left (n-1 \right )} \\
a_{2n} &a_{22}+a_{21} & &a_{2n}+a_{2\left ( n-1 \right )} \\
\vdots &\vdots &\ddots & \\
a_{nn} &a_{n2}+a_{n1} & &a_{nn}+a_{n\left ( n-1 \right )}
\end{vmatrix}\xrightarrow[]{C_{1}\rightarrow C_{1}+\left ( -1 \right )^{1}C_{n}}\begin{vmatrix}
-a_{1\left ( n-1 \right )} &a_{12}+a_{11} & &a_{1n}+a_{1\left (n-1 \right )} \\
-a_{2\left ( n-1 \right )} &a_{22}+a_{21} & &a_{2n}+a_{2\left ( n-1 \right )} \\
\vdots &\vdots &\ddots & \\
-a_{n\left ( n-1 \right )} &a_{n2}+a_{n1} & &a_{nn}+a_{n\left ( n-1 \right )}
\end{vmatrix}\)

\(\xrightarrow[]{C_{1}\rightarrow C_{1}+\left ( -1 \right )^{2}C_{n-1}}\begin{vmatrix}
a_{1\left ( n-2 \right )} &a_{12}+a_{11} & &a_{1n}+a_{1\left (n-1 \right )} \\
a_{2\left ( n-2 \right )} &a_{22}+a_{21} & &a_{2n}+a_{2\left ( n-1 \right )} \\
\vdots &\vdots &\ddots & \\
a_{n\left ( n-2 \right )} &a_{n2}+a_{n1} & &a_{nn}+a_{n\left ( n-1 \right )}
\end{vmatrix}\xrightarrow[]{C_{1}\rightarrow C_{1}+\left ( -1 \right )^{3}C_{n-2}}\cdots \)

\(\xrightarrow[]{C_{1}\rightarrow C_{1}+\left ( -1 \right )^{n-1}C_{2=n-\left ( n-2 \right )}}\begin{vmatrix}
\left ( -1 \right )^{n-1}a_{11} &a_{12}+a_{11} & &a_{1n}+a_{1\left (n-1 \right )} \\
\left ( -1 \right )^{n-1}a_{21} &a_{22}+a_{21} & &a_{2n}+a_{2\left ( n-1 \right )} \\
\vdots &\vdots &\ddots & \\
\left ( -1 \right )^{n-1}a_{n1} &a_{n2}+a_{n1} & &a_{nn}+a_{n\left ( n-1 \right )}
\end{vmatrix}=\left | K \right |=\left | A \right |\) if \(\displaystyle{n}\) is not even

and if \(\displaystyle{n}\) is even then

\(\begin{vmatrix}
\left ( -1 \right )^{n-1}a_{11} &a_{12}+a_{11} & &a_{1n}+a_{1\left (n-1 \right )} \\
\left ( -1 \right )^{n-1}a_{21} &a_{22}+a_{21} & &a_{2n}+a_{2\left ( n-1 \right )} \\
\vdots &\vdots &\ddots & \\
\left ( -1 \right )^{n-1}a_{n1} &a_{n2}+a_{n1} & &a_{nn}+a_{n\left ( n-1 \right )}
\end{vmatrix}=\begin{vmatrix}
-a_{11} &a_{12}+a_{11} & &a_{1n}+a_{1\left (n-1 \right )} \\
-a_{21} &a_{22}+a_{21} & &a_{2n}+a_{2\left ( n-1 \right )} \\
\vdots &\vdots &\ddots & \\
-a_{n1} &a_{n2}+a_{n1} & &a_{nn}+a_{n\left ( n-1 \right )}
\end{vmatrix}\)

\(=-\left | K \right |=-\left | A \right |\)

so \(\displaystyle{\left | B \right |=\left | K \right |+\left | K \right |=2\left | K \right |=2\left | A \right |}\)

if n is not even or

\(\displaystyle{\left | B \right |=\left | K \right |-\left | K \right |=0}\)

if n is even.

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 \(\det(B) = (1 + (-1)^{n-1})\det(A).\)

Great Demetres.