How should I calculate the determinant?

Question

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

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

Answer 1

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

Answer 2

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?

Answer 3

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

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.

Answer 4

Compute $$ \begin{bmatrix} 1 & a & b & c+d \\ 1 & b & c & d+a \\ 1 & c & d & a+b \\ 1 & d & a & b+c \end{bmatrix} \begin{bmatrix} a+b+c+d \\ -1 \\ -1 \\ -1 \end{bmatrix} $$