Determinant with variables What is the following determinant?

$$\begin{vmatrix}1+a &amp; b &amp; c &amp; d \\a &amp; 1+b &amp; c &amp; d \\a &amp; b &amp; 1+c &amp; d \\a &amp; b &amp; c &amp; 1+d \end{vmatrix}$$

I calculated it as $0$ but I do not think it is right. Thanks in advance.
Subtract row $2$ from row $1$, row $3$ from row $2$, and row $4$ from row $3$.  You get
$$
\begin{bmatrix}
1 &amp; -1 &amp; 0 &amp; 0 \\
0 &amp; 1 &amp; -1 &amp; 0 \\
0 &amp; 0 &amp; 1 &amp; -1 \\
a &amp; b &amp; c &amp; 1+d 
\end{bmatrix}
$$
If you expand along the fourth row, all the $3 \times 3$ determinants will be upper or lower triangular.  So, the determinants will be the product of the diagonals.
Let $\mathrm v := \begin{bmatrix} a &amp; b &amp; c &amp; d\end{bmatrix}^{\top}$. Using the Weinstein-Aronszajn determinant identity,
$$\det \begin{bmatrix}1+a &amp; b &amp; c &amp; d\\ a &amp; 1+b &amp; c &amp; d\\ a &amp; b &amp; 1+c &amp; d\\ a &amp; b &amp; c &amp; 1+d\end{bmatrix} = \det \left( \mathrm I_4 + 1_4 \mathrm v^{\top}\right) = 1 + \mathrm v^{\top} 1_4 = \color{blue}{1 + a + b + c + d}$$