Determinant with variables

Question

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.

Answer 1

The matrix is $I-A$ where $A$ is a matrix with a triple eigenvalue $0$ and a fourth eigenvalue $-(a+b+c+d)$. So the characteristic polynomial $\det(I\lambda - A)=\lambda^3(\lambda +(a+b+c+d))$. Setting $\lambda=1$ you get $\det(I-A)=1+a+b+c+d$.

The left eigenvector for $A$ are any vectors $(x,y,z,w)$ with $x+y+z+w=0$ (these have eigenvalues $0$) and vector $(a,b,c,d)$ has eigenvalue $-(a+b+c+d)$. The vectors with eigenvalue $0$ are generated by vectors $(1,-1,0,0), (1,0,-1,0),$ and $(1,0,0,-1)$.

Right eigenvectors are a little messier. If $(x,y,z,w)^T$ satisfies $ax+by+cz+dw=0$, then it is an eigenvector with eigenvalue $0$. And $(1,1,1,1)$ has eigenvalue $-(a+b+c+d)$.

Answer 2

Subtract row $2$ from row $1$, row $3$ from row $2$, and row $4$ from row $3$. You get $$ \begin{bmatrix} 1 & -1 & 0 & 0 \\ 0 & 1 & -1 & 0 \\ 0 & 0 & 1 & -1 \\ a & b & c & 1+d \end{bmatrix} $$ 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.

Answer 3

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: $1+a+b+c+d$.

Answer 4

$$\begin{vmatrix}1+a& b & c & d \\a & 1+b & c & d \\a & b & 1+c & d \\a & b & c & 1+d \end{vmatrix}=\begin{vmatrix}1 & b & c & d \\0 & 1+b & c & d \\0 & b & 1+c & d \\0 & b & c & 1+d \end{vmatrix}+\begin{vmatrix}a & b & c & d \\a & 1+b & c & d \\a & b & 1+c & d \\a & b & c & 1+d \end{vmatrix}\\ =\begin{vmatrix}1+b & c & d \\b & 1+c & d \\ b & c & 1+d \end{vmatrix}+\begin{vmatrix}a & b & c & d \\0 & 1 & 0 & 0 \\0 &0 & 1 & 0 \\0&0 & 0 & 1\end{vmatrix}\\ =\begin{vmatrix}1 & c & d \\0 & 1+c & d \\ 0 & c & 1+d \end{vmatrix}+\begin{vmatrix}b & c & d \\b & 1+c & d \\ b & c & 1+d \end{vmatrix}+a\\ =\begin{vmatrix}1+c & d \\ c & 1+d \end{vmatrix}+\begin{vmatrix}b & c & d \\0& 1 & d \\ 0 & 0& 1 \end{vmatrix}+a\\ =\begin{vmatrix}1 & d \\ 0 & 1+d \end{vmatrix}+\begin{vmatrix}c & d \\ c & 1+d \end{vmatrix}+b+a\\ =\begin{vmatrix}1+d \end{vmatrix}+\begin{vmatrix}c & d \\ 0& 1 \end{vmatrix}+b+a\\ =1+d+c+b+a.$$

Answer 5

Let $\mathrm v := \begin{bmatrix} a & b & c & d\end{bmatrix}^{\top}$. Using the Weinstein-Aronszajn determinant identity,

$$\det \begin{bmatrix}1+a & b & c & d\\ a & 1+b & c & d\\ a & b & 1+c & d\\ a & b & c & 1+d\end{bmatrix} = \det \left( \mathrm I_4 + 1_4 \mathrm v^{\top}\right) = 1 + \mathrm v^{\top} 1_4 = \color{blue}{1 + a + b + c + d}$$