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$\left|\begin{array}{cccc}1&a&b&c+d\\1&b&c&a+d\\1&c&d&a+b\\1&d&a&b+c\end{array}\right|= \left|\begin{array}{cccc}1&a&b&c\\1&b&c&a\\1&c&d&a\\1&d&a&b\end{array}\right|+ \left|\begin{array}{cccc}1&a&b&d\\1&b&c&d\\1&c&d&b\\1&d&a&c\end{array}\right|$

I tried to calculate the determinant but I couldn't do it after separating the determinant by the property. How should I calculate it?

Senex Ægypti Parvi
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maki
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    notice that the sum in each line is the same (a+b+c+d+1). – hamam_Abdallah Oct 16 '16 at 21:29
  • Perhaps you could show what you've done? That way, it might be easier to help-also try to learn maths jax-http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference – John_dydx Oct 16 '16 at 21:29
  • Oh, I didn't notice that the sum of each line is the same. I know if the determinant has the same row or column, it's 0. Is there some property about the sum of a line? – maki Oct 16 '16 at 21:36
  • Instead of $a+d$, write $d+a$ and you will see that the two determinants are identical to a permutation of the rows and columns. –  Oct 16 '16 at 21:55

4 Answers4

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$${\begin{vmatrix}1&a & b &c+d\\1 &b &c &d+a \\1 &c &d &a+b\\1&d &a &b+c &\end{vmatrix}} \space c_2+c_3+c_4 \to c_4 \\ {\begin{vmatrix}1&a & b &a+b+c+d\\1 &b &c &a+b+c+d \\1 &c &d &a+b+c+d\\1&d &a &a+b+c+d &\end{vmatrix}} \space factor \space (a+b+c+d)=\\(a+b+c+d) {\begin{vmatrix}1&a & b &1\\1 &b &c &1 \\1 &c &d &1\\1&d &a &1 &\end{vmatrix}}$$now $c_1-c_4 \to c_1$ $${\begin{vmatrix}0&a & b &1\\0 &b &c &1 \\0 &c &d &1\\0&d &a &1 &\end{vmatrix}} $$

Khosrotash
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    I was under the wrong impression that I could add only a row to another row for determinant. Thank you Khosrotash :). – maki Oct 16 '16 at 21:59
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Looks like homework to me, so I will suggest ideas instead of just the answer.

Take the first row of the original matrix and add it to the third row. Then add the second row to the fourth row. How do the new third and fourth rows compare? What does that imply about the determinant?

Oscar Lanzi
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  • Honestly, it's not my homework. I know the answer which is 0. This is the practice to understand determinant profoundly. Anyway thank you for your idea. – maki Oct 16 '16 at 21:46
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Compute $$ \begin{bmatrix} 1 & a & b & c+d \\ 1 & b & c & d+a \\ 1 & c & d & a+b \\ 1 & d & a & b+c \end{bmatrix} \begin{bmatrix} a+b+c+d \\ -1 \\ -1 \\ -1 \end{bmatrix} $$

egreg
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$${\begin{vmatrix}1&a & b &c\\1 &b &c &d \\1 &c &d &a\\1&d &a &b &\end{vmatrix}}=-{\begin{vmatrix}1&a & b &d\\1 &b &c &a \\1 &c &d &b\\1&d &a &c &\end{vmatrix}}$$

because the second determinant is the same as the first by cyclically permuting the columns $2,3$ and $4$, an cyclically permuting the four rows.

  • Does a cyclic permutation represent an even or an odd prrmutarion? Does that depend on how many items are cyclically permuted? I would indicate that the column permutation us even and does not change the sign of rhe seterminabt, while the column permutation is odd and does change the sign. – Oscar Lanzi Oct 16 '16 at 22:08