Determinant with variables What is the following determinant?
$$\begin{vmatrix}1+a & b & c & d \\a & 1+b & c & d \\a & b & 1+c & d \\a & b & c & 1+d \end{vmatrix}$$
I calculated it as $0$ but I do not think it is right. Thanks in advance.

If you are not familiar with the concepts used in Thomas Andrews' (elegant) answer; you can try to compute the determinant by using properties to create zeroes.
For example; subtract the second row from the other rows and then add the first column to the second to get:
$$\begin{vmatrix}1+a & b & c & d \\a & 1+b & c & d \\a & b & 1+c & d \\a & b & c & 1+d \end{vmatrix} = 
\begin{vmatrix}
1 & -1 & 0 & 0 \\
a & 1+b & c & d \\
0 & -1 & 1 & 0 \\
0 & -1 & 0 & 1 \end{vmatrix}
= 
\begin{vmatrix}
1 & 0 & 0 & 0 \\
a & 1+a+b & c & d \\
0 & -1 & 1 & 0 \\
0 & -1 & 0 & 1 \end{vmatrix}$$
Now expand the determinant to the first row, reducing it to a $3 \times 3$-determinant. Subtracting $c$ times the second row from the first and  $d$ times the third row from the first, you get:
$$\begin{vmatrix}
 1+a+b & c & d \\
 -1 & 1 & 0 \\
 -1 & 0 & 1 \end{vmatrix}=
\begin{vmatrix}
 1+a+b+c+d & 0 & 0 \\
 -1 & 1 & 0 \\
 -1 & 0 & 1 \end{vmatrix}$$
Expanding to the first row again agrees with Thomas Andrews' answer: $1+a+b+c+d$.
