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Calculate the determinant of the following matrix:

$$ \begin{pmatrix} a+x & b & c \\ a & b+x & c \\ a & b & c+x \\ \end{pmatrix} $$

without using the rule of sarrus or minors

I got the answer using this property multiple times: $$ \begin{pmatrix} a+x & b & c \\ a & b+x & c \\ a & b & c+x \\ \end{pmatrix} = \begin{pmatrix} x & b & c \\ 0 & b+x & c \\ 0 & b & c+x \\ \end{pmatrix} + \begin{pmatrix} a & b & c \\ a & b+x & c \\ a & b & c+x \\ \end{pmatrix} $$

but that got very tedious and long for higher dimensions

The answer is: $$ x^2(a+b+c+x) $$

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    $\begingroup$ What you show is "a general rule": to use the multilinearity of the determinant. Why not better to write "to develop this determinant without using 1,2,3,4,5,6 ", or better: to develop it using this or that only ? Foggy instructions bring foggy answers, misunderstandings, etc. $\endgroup$
    – DonAntonio
    Oct 22, 2016 at 19:02
  • $\begingroup$ @DonAntonio By a general rule, I mean a method for calculating any kind of determinant. Using the rule I show above I can solve this particular determinant, but not all. $\endgroup$
    – Msegade
    Oct 22, 2016 at 19:16
  • $\begingroup$ @Msegade For the general approach lookup the characteristic polynomial of a matrix. $\endgroup$
    – dxiv
    Oct 22, 2016 at 19:19
  • $\begingroup$ @Mse That's false: the determinant function is always multilinear, no matter what the order of the matrix, how many zero entries or over what field the matrix is defined. Again, that is as a "General rule" as one can expect, so again: you better specify in a very detailed way what you want, or else people can begin calculating in different, wild ways. $\endgroup$
    – DonAntonio
    Oct 22, 2016 at 19:19
  • $\begingroup$ @DonAntonio I'm not saying the rule does not apply to all determinants. I'm saying that aplying that rule doesn't always leads you to the answer ( the numeric value of the determinant) in a general matrix. I'm fine with wild ways of solving this problem :) $\endgroup$
    – Msegade
    Oct 22, 2016 at 19:23

3 Answers 3

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The determinant is quite obviously a $3^{rd}$ degree monic polynomial $P(x)$. By inspection $x=0$ must be a double root since all the rows become equal and the matrix degenerates to rank $1$. The remaining root can be guessed as $x=-a-b-c$ by noticing that the sum of all columns becomes $0$. Therefore $P(x) = x^2(x+a+b+c)$.

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The determinant you seek is the characteristic polynomial of the matrix $-A$, where $$ A = \left( \begin{array}{ccc} a & b & c \\ a & b & c \\ a & b & c \end{array}\right) $$

This matrix has at most one linearly independent row, so at least two of the three eigenvalues are $0$ from which we conclude that the characteristic polynomial of $-A$ is of the form $x^2 (x-r)$ where $r$ is the remaining eigenvalue of $-A$.

Since all but one of the eigenvalues is zero, it is also correct to say that $r$ is the sum of the eigenvalues of $-A$. The sum of the eigenvalues of a square matrix is the trace of the matrix, and it's a standard result that the trace is equal to the sum of the diagonal elements.

So the trace of $-A$ is $-(a+b+c)$ and the characteristic polynomial of $-A$ is therefore $x^2 (x+a+b+c)$.

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$$det \begin{pmatrix} a+x & b & c \\ a & b+x & c \\ a & b & c+x \\ \end{pmatrix} $$ row1=row 1-row3 $$ det\begin{pmatrix} x & 0 & -x \\ a & b+x & c \\ a & b & c+x \\ \end{pmatrix} $$ row2=row2-row3 $$det \begin{pmatrix} x & 0 & -x \\ 0 & x & -x \\ a & b & c+x \\ \end{pmatrix} $$ colum3=colum1+colum2+colum3 $$det \begin{pmatrix} x & 0 & 0 \\ 0 & x & 0\\ a & b & a+b+c+x \\ \end{pmatrix} $$ finally we have $$(a+b+c+x)det \begin{pmatrix} x & 0 \\ 0 & x \\ \end{pmatrix} $$ given the result wanted. $$x^2(a+b+c+x) $$

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  • $\begingroup$ Careful: this may enter under the definition of "general rule" the OP meant...or not, of course. $\endgroup$
    – DonAntonio
    Oct 22, 2016 at 19:30
  • $\begingroup$ sorry, i miis that detail $\endgroup$
    – user245074
    Oct 22, 2016 at 19:32
  • $\begingroup$ @juliogodoy Thats a good answer $\endgroup$
    – Msegade
    Oct 22, 2016 at 19:34

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