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.


2 Answers 2


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
    Commented Feb 14, 2016 at 9:47
  • $\begingroup$ @user26977 Thank you, I edited answer. I was told before not to give complete answers, though. $\endgroup$
    – DonAntonio
    Commented Feb 14, 2016 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.) $\endgroup$ Commented Feb 14, 2016 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.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .