My problem is on the specific determinant.

$$\det \begin{pmatrix} na_1+b_1 & na_2+b_2 & na_3+b_3 \\ nb_1+c_1 & nb_2+c_2 & nb_3+c_3 \\ nc_1+a_1 & nc_2+a_2 & nc_3+a_3 \end{pmatrix}= (n+1)(n^2-n+1) \det\begin{pmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{pmatrix}$$

All I can do is prove the factor $(n+1)$ and I think that we have to work only on one column and the do the exact same thing to the others.


Use multilinearity of determinant:

$$\begin{vmatrix}na_1+b_1&na_2+b_2&na_3+b_3\\ nb_1+c_1&nb_2+c_2&nb_3+c_3\\ nc_1+a_1&nc_2+a_2&nc_3+a_3\end{vmatrix}=\begin{vmatrix}na_1&na_2&na_3\\ nb_1&nb_2&nb_3\\ nc_1&nc_2&nc_3\end{vmatrix}+\begin{vmatrix}na_1&na_2&b_3\\ nb_1&nb_2&c_3\\ nc_1&nc_2&a_3\end{vmatrix}+$$

$$+\begin{vmatrix}na_1&b_2&na_3\\ nb_1&c_2&nb_3\\ nc_1&a_2&nc_3\end{vmatrix}+\begin{vmatrix}na_1&b_2&b_3\\ nb_1&c_2&c_3\\ nc_1&a_2&a_3\end{vmatrix}+\begin{vmatrix}b_1&na_2&na_3\\ c_1&nb_2&nb_3\\ a_1&nc_2&nc_3\end{vmatrix}+\ldots$$

Observe that if we put

$$\Delta=\begin{vmatrix}a_1&a_2&a_3\\ b_1&b_2&b_3\\ c_1&c_2&c_3\end{vmatrix}$$

then we have that the four first determinants above equal (factor out constants from rows/columns):


Well, develop the other three determinants left and sum up all.

  • $\begingroup$ you are right, up to point where you introduce $\Delta$. All $n^2$ terms should cancel because each of them has two identical columns, the same holds for linear terms. The last term contains no $n$ so the final result should be $(n^3+1)\Delta$ which is factored as OP suggests. I think you should add all 8 terms in the expansion $\endgroup$ – user26977 Feb 14 '16 at 9:47
  • $\begingroup$ @user26977 Thank you, I edited answer. I was told before not to give complete answers, though. $\endgroup$ – DonAntonio Feb 14 '16 at 10:11
  • 1
    $\begingroup$ This site has people with conflicting opinions. Some think you shouldn't give a complete answer to certain questions because the OP somehow doesn't deserve it. Others think you should always give a complete answer. The truth is you can write whatever you want within the bare minimum of acceptable guidelines (not offensive, doesn't leak personal information, doesn't threaten to commit a crime, etc.) $\endgroup$ – Matt Samuel Feb 14 '16 at 21:26

If you notice that $$ \begin{pmatrix} na_1+b_1 & na_2+b_2 & na_3+b_3 \\ nb_1+c_1 & nb_2+c_2 & nb_3+c_3 \\ nc_1+a_1 & nc_2+a_2 & nc_3+a_3 \end{pmatrix}= \begin{pmatrix} n & 1 & 0 \\ 0 & n & 1 \\ 1 & 0 & n \\ \end{pmatrix} \begin{pmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{pmatrix}$$ and if you know that $\det(AB)=\det(A)\det(B)$, then it only remains to calculate determinant of the matrix $\begin{pmatrix} n & 1 & 0 \\ 0 & n & 1 \\ 1 & 0 & n \\ \end{pmatrix}$.

It is not very difficult to notice the above identity, if you know, that multiplication (from the left) means simply making linear combinations of the rows. See, for example, this answer.


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