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I am trying to solve the determinant of a matrix of the following form:

$$ A_n = \begin{bmatrix} x & b & b & \cdots & b \\ a & x & b & \cdots & b \\ a & a & x & \cdots & b \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ a & a & a & \cdots & x \end{bmatrix} $$

I attempted to use recursion to find a general formula for the determinant $D_n$ of the matrix $A_n$. My idea was to subtract successive columns to simplify the matrix and find a general expression.

After some calculations with the first and the second column, I arrived at the following formula for the determinant:

$$ D_n(x) = (x - b) D_{n-1}(x) + (x - a)^{n-1} \cdot b $$

This formula seems to work well for $(n = 2, 3)$, but I am not entirely sure it is correct. My professor' ’s question on this problem is:

Demonstrate that $ D_n(x)' = n D_{n-1}(x)'$ (where $D_n(x)'$ is the derivative with respect to $x$),

which does not seem to work with my formula.

Can anyone help me verify the formula or suggest a simpler approach to solving this determinant?

Rócherz
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MiguelCG
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  • I feel like there's something wrong with the property you're being asked, it clearly doesn't hold for $n = 2$ or $n=3$. Did you mean $D_n(x)' = n D_{n-1}(x)$, without the derivative on the right-hand side? – Carlos Santi Toledo May 13 '25 at 10:03
  • That’s a special case of https://math.stackexchange.com/q/1801627/42969. – Martin R May 13 '25 at 10:04
  • @CarlosSantiToledo I have checked the statement again and it is with both derivatives. But as I said, it may be an error from my teacher. – MiguelCG May 13 '25 at 10:11
  • @MiguelCG you can easily check that $D_3(x) = x^3 - 3abx+ab^2+a^2b$, while $D_2(x) = x^2 -ab$. You have that $D_3(x)' = 3(x^2-ab) = 3D_2(x)$, while $D_3(x)' \neq 6x = 3D_2(x)'$, so I think your teacher got it mixed up. – Carlos Santi Toledo May 13 '25 at 10:15
  • From https://math.stackexchange.com/q/1801627/42969 one gets $D_n(x) = (b(x-a)^n - a(x-b)^n)/(b-a)$, so that $D_n(x)' = n D_{n-1}(x)$, without the derivative on the right-hand side, is the correct formula. – Martin R May 13 '25 at 10:16
  • @MartinR Very nice proof in the other answer, thanks for the link! – Carlos Santi Toledo May 13 '25 at 10:21
  • @MartinR Thanks for the related question, now I am trying to go from my recurrence to that formula. – MiguelCG May 13 '25 at 10:41
  • Your recurrence formulat looks correct, but I am not sure if it is suited to derive the other formula $D_n(x)' = n D_{n-1}(x)$. Differentiating the determinant $D_n$ (by differentiating every row with respect to $x$) seems to be the easier approach. – Martin R May 13 '25 at 10:43

2 Answers2

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The determinant of a matrix with entries which are functions of $x$ can be differentiated with respect to $x$ by differentiating every row of the matrix and adding the results, see for example Derivative of a determinant whose entries are functions.

Here we get $$ D_n(x)' = \begin{vmatrix} 1 & 0 & 0 & \cdots & 0 \\ a & x & b & \cdots & b \\ a & a & x & \cdots & b \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ a & a & a & \cdots & x \end{vmatrix} + \begin{vmatrix} x & b & b & \cdots & b \\ 0 & 1 & 0 & \cdots & 0 \\ a & a & x & \cdots & b \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ a & a & a & \cdots & x \end{vmatrix} \\ + \begin{vmatrix} x & b & b & \cdots & b \\ a & x & b & \cdots & b \\ 0 & 0 & 1 & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ a & a & a & \cdots & x \end{vmatrix} + \cdots + \begin{vmatrix} x & b & b & \cdots & b \\ a & x & b & \cdots & b \\ a & a & x & \cdots & b \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & 1 \end{vmatrix} \, . $$ Developing each determinant on the right along the $n$-th row shows that they are all equal to $D_{n-1}(x)$, so that $$ D_n(x)' = n D_{n-1}(x) \, . $$

Martin R
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There have been some good answers and annotations from the comments. However, I think I would not realize that development in an exam, so sticking with my recurrence, I will prove by induction that $D_{n}(x)' = n D_{n-1}(x)$.

Case $n = 3$: $$ D_3(x)' = 3x^2 - 3ab = 3(x^2 - ab) = nD_2(x) $$

Case $n$: Suppose that it is true for $n - 1$ and compute: $$ D_n(x)' = D_{n-1}(x) + D_{n-1}(x)'(x - b) + (n - 1)(x - a)^{n - 2}b = D_{n-1}(x) + (n - 1) D_{n-2}(x) (x - b) + (n - 1) (x - a)^{n - 2}b = D_{n-1}(x) + (n - 1) D_{n-1}(x) = n D_{n-1}(x). $$

MiguelCG
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