Find the value of the following $n \times n$ determinantes

Question

Find the value of the following $n \times n$ determinantes

1. $$\begin{vmatrix} a_1+x & x & x & \ldots & x \\ x & a_2+x & x & \ldots & x \\ x & x & a_3+x & \ldots & x \\ \vdots & \vdots& &\ddots& \vdots\\ x & x & x & \ldots & a_n+x \\ \end{vmatrix}$$
2. $$\begin{vmatrix} a_1+x & a_2 & a_3 & \ldots & a_n \\ a_1 & a_2+x & a_3 & \ldots & a_n \\ a_1 & a_2 & a_3+x & \ldots & a_n \\ \vdots & \vdots& &\ddots& \vdots\\ a_1 & a_2 & a_3 & \ldots & a_n+x \\ \end{vmatrix}$$

Both seem to be equally complicated to solve, I reckon that it's needed to subtract the $x$ from the diagonal in each term. I tried by subtracting the $(k-1)$-th row from the $k$-th row, however that doesn't really lead me to anything more comfortable whatsoever. So help is greatly appreciated, also perhaps a link to some methods on solving these kind of problems, would also be helpful.

Answer 1

For 2) Let $$A=\begin{bmatrix} a_1 & a_2 & .. &a_n \\ a_1 & a_2 & .. &a_n \\ a_1 & a_2 & .. &a_n \\ ...&...&...&...\\ a_1 & a_2 & .. &a_n \\ \end{bmatrix}$$

Then $A$ has rank $1$ and hence $\lambda=0$ is an eigenvalue with geometric multiplicity $n-1$, and hence has algebraic multiplicity at least $n-1$.

As the trace is the sum of eigenvalues, the last eigenvalue is $a_1+a_2+..+a_n$.

Thus $$ \det(\lambda I-A) =\lambda^n -(a_1+...+a_n) \lambda^{n-1} $$

Now replace $\lambda$ by $-x$.

For 1) Subtract the last row from each of the previous $n-1$. You get $$\begin{vmatrix} a_1+x & x & x & \ldots & x \\ x & a_2+x & x & \ldots & x \\ x & x & a_3+x & \ldots & x \\ \vdots & \vdots& &\ddots& \vdots\\ x & x & x & \ldots & a_n+x \\ \end{vmatrix}=\begin{vmatrix} a_1 & 0 & 0 & \ldots & -a_n \\ 0 & a_2 & 0 & \ldots & -a_n \\ 0 & 0 & a_3 & \ldots & -a_n \\ \vdots & \vdots& &\ddots& \vdots\\ x & x & x & \ldots & a_n+x \\ \end{vmatrix}$$

This determinant is a linear function in $x$. Therefore $$\begin{vmatrix} a_1+x & x & x & \ldots & x \\ x & a_2+x & x & \ldots & x \\ x & x & a_3+x & \ldots & x \\ \vdots & \vdots& &\ddots& \vdots\\ x & x & x & \ldots & a_n+x \\ \end{vmatrix}=ax+b$$

Now when $x=0$ we get $$b=a_1 ... a_n$$

All you have to do next is do row expansion by the last row in $$\begin{vmatrix} a_1 & 0 & 0 & \ldots & -a_n \\ 0 & a_2 & 0 & \ldots & -a_n \\ 0 & 0 & a_3 & \ldots & -a_n \\ \vdots & \vdots& &\ddots& \vdots\\ x & x & x & \ldots & a_n+x \\ \end{vmatrix}$$ in order to find the coefficient of $x$.

For 2), without eigenvalues: Transpose the matrix, and then add all rows to the first. We get:

$$\begin{vmatrix} a_1+x & a_2 & a_3 & \ldots & a_n \\ a_1 & a_2+x & a_3 & \ldots & a_n \\ a_1 & a_2 & a_3+x & \ldots & a_n \\ \vdots & \vdots& &\ddots& \vdots\\ a_1 & a_2 & a_3 & \ldots & a_n+x \\ \end{vmatrix}=\begin{vmatrix} a_1+x & a_1 & a_1 & \ldots & a_1 \\ a_2 & a_2+x & a_2 & \ldots & a_2 \\ a_3 & a_3 & a_3+x & \ldots & a_3 \\ \vdots & \vdots& &\ddots& \vdots\\ a_n & a_n & a_n & \ldots & a_n+x \\ \end{vmatrix}=\begin{vmatrix} a_1+a_2+..+a_n+x & a_1+a_2+..+a_n+x & a_1+a_2+..+a_n+x & \ldots &a_1+a_2+..+a_n+x\\ a_2 & a_2+x & a_2 & \ldots & a_2 \\ a_3 & a_3 & a_3+x & \ldots & a_3 \\ \vdots & \vdots& &\ddots& \vdots\\ a_n & a_n & a_n & \ldots & a_n+x \\ \end{vmatrix}=(a_1+a_2+..+a_n+x)\begin{vmatrix} 1 & 1 & 1 & \ldots &1\\ a_2 & a_2+x & a_2 & \ldots & a_2 \\ a_3 & a_3 & a_3+x & \ldots & a_3 \\ \vdots & \vdots& &\ddots& \vdots\\ a_n & a_n & a_n & \ldots & a_n+x \\ \end{vmatrix}$$

Now if you do each row minus $a_i$ row one you get $$=(a_1+a_2+..+a_n+x)\begin{vmatrix} 1 & 1 & 1 & \ldots &1\\ 0 & x & 0 & \ldots & 0 \\ 0 & 0 & x & \ldots & 0 \\ \vdots & \vdots& &\ddots& \vdots\\ 0& 0 & 0 & \ldots & x \\ \end{vmatrix}$$

Answer 2

I don't know if you are familiar with induction, but you can try it for both forms you mentioned in your question.

For the first one the determinant is $$D_1=\prod_{j=1}^n a_j + \sum_{1\leq i_1<i_2<\cdots<i_{n-1}\leq n} a_{i_1} a_{i_2} \cdots a_{i_{n-1}} x$$

For the second one $$ D_2= x^n +\sum_{i=1}^n a_i x^{n-1}$$

I get these formulae by trying the dterminants for $n=1,2,3,4$, you can check this and this for example, then easily you can extract such equalities for $D_1$ and $D_2$, then by induction you prove the result for every $n$. If you got stuck in any step of the induction, just drop a comment.

Hope this helps you.

Answer 3

Let

\begin{align} T_n &= \begin{vmatrix} a_1+x & x & x & \cdots & x \\ x & a_2+x & x & \cdots & x \\ x & x & a_3+x & \cdots & x \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ x & x & x & \cdots & a_n+x \end{vmatrix} \\ &= \begin{vmatrix} a_1+x & x & x & \cdots & x \\ -a_1 & a_2 & 0 & \cdots & 0 \\ -a_1 & 0 & a_3 & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ -a_1 & 0 & 0 & \cdots & a_n \end{vmatrix} \end{align}

Expand $T_n$ with the last column, and we can get

$$ T_n = x a_1 a_2 \cdots a_{n-1} + a_n T_{n-1} $$

It's obvious that

$$ T_1 = a_1 + x $$

By induction, we can get

$$ T_n = x \sum^n_{j=1} \prod^n_{\scriptstyle i=1 \atop \scriptstyle i\ne j} a_i + \prod^n_{i=1} a_i $$