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Given an $n \times n$ matrix $A$, where $x$ is any real number:

$A = \left[ \begin{array}{ c c c c c c c c } 1 & 1 & 1 & 1 & 1 & 1 & \cdots & 1 \\ 1 & x & x & x & x & x & \cdots & x \\ 1 & x & 2x & 2x & 2x & 2x & \cdots & 2x \\ 1 & x & 2x & 3x & 3x & 3x & \cdots & 3x \\ 1 & x & 2x & 3x & 4x & 4x & \cdots & 4x \\ 1 & x & 2x & 3x & 4x & 5x & \cdots & 5x \\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\ 1 & x & 2x & 3x & 4x & 5x & \cdots & (n-1)x \end{array} \right]$

Find the determinant.

By using $n=2,3,4,5,...$ and random $x=1,2,3,4...$ I have found that $det(A) = (x-1)(x)^{n-2}$ through observing a pattern.

However, I would like to be able to prove this through a proof, yet I have no idea where to start.

When I try to solve for the determinant using the abstract matrix A and using the property that the determinant of a square matrix is $(-1)^r * (\text{products of pivots})$, where r is the number of row interchanges, my answer is of the form $(x-1)(x)(x)(x)(x)...(n-?)x$ where "?" depends on how many rows I include in the abstract form of A. How do I show that $(x)(x)(x)...(n-?)x$ equals $(x)^{n-2}$?

Here is my work: http://i.stack.imgur.com/tinDw.jpg

Any hints? Thanks for the help!

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I recall a similar problem (involving a determinant containing powers of x) that was solved through mathematical induction. –  barf Jun 12 '11 at 0:40
5  
You're already done. The matrix has $n$ diagonal entries after you simplify it. One is equal to $1$, one is equal to $x - 1$, and the rest are equal to $x$ by induction, so there must be $n-2$ copies of $x$. –  Qiaochu Yuan Jun 12 '11 at 0:56
3  
You can use Gaussian elimination to get an upper triangular matrix with same determinant (your matrix is especially well suited for that technique). –  Joel Cohen Jun 12 '11 at 1:02
    
Qiaochu, thanks for that insight. –  Taylor P. Jun 12 '11 at 1:32

3 Answers 3

up vote 10 down vote accepted

I think what you did is almost perfectly correct. In any case, here is another way that uses induction. Let $A_n$ refer to the $n\times n$ matrix. The base case of the $2\times 2$ matrix is $\det A_2 =(x-1)$.

Now, for the inductive step lets look at $A_{n+1}$ and use the linearity of the determinant on the last entry of the last row. We have that $$\det\left[\begin{array}{ccccc} 1 & 1 & \cdots & 1 & 1\\ 1 & x & \cdots & x & x\\ \vdots & \vdots & \ddots & \vdots & \vdots\\ 1 & x & \cdots & (n-1)x & (n-1)x\\ 1 & x & \cdots & (n-1)x & nx\end{array}\right]=$$ $$\det\left[\begin{array}{ccccc} 1 & 1 & \cdots & 1 & 1\\ 1 & x & \cdots & x & x\\ \vdots & \vdots & \ddots & \vdots & \vdots\\ 1 & x & \cdots & (n-1)x & (n-1)x\\ 1 & x & \cdots & (n-1)x & (n-1)x\end{array}\right]+\det\left[\begin{array}{ccccc} 1 & 1 & \cdots & 1 & 1\\ 1 & x & \cdots & x & x\\ \vdots & \vdots & \ddots & \vdots & \vdots\\ 1 & x & \cdots & (n-1)x & (n-1)x\\ 0 & 0 & \cdots & 0 & x\end{array}\right]$$ The first determinant will be zero since two rows are the same. The second is $x\det A_n$.

Hope that helps,

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It didn't even occur to me to use induction using the linearity. Thank you very much. –  Taylor P. Jun 12 '11 at 1:45

Expand the determinant using the usual sum-product formula. Each term in this expansion results by picking some permutation $\pi$. If $\pi(1) = 1$ then the term will be of the form $Cx^{n-1}$, otherwise it will be of the form $Cx^{n-2}$ (since $\pi(1) \neq \pi^{-1}(1)$). Therefore the determinant is of the general form $$ Ax^{n-1} + Bx^{n-2} = x^{n-2} (Ax + B). $$ Substituting $x = 1$, we get that the first two rows are equal, and so the determinant is $0$. So the determinant has the general form $$Cx^{n-2}(x-1).$$ The coefficient of $x^{n-1}$ is clearly equal to the determinant of the $(1,1)$-minor, and if we substitute $x = 1$ we will get a matrix whose determinant is $C$.

As mentioned in another answer, it is very easy to see that the determinant of the matrix is $1$. First we subtract the first row from all other rows. This leaves a lone $1$ on the first row at the first column, so we can erase the first row and column. Continuing this way, we eventually reach the singleton $1$ matrix.

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+1, I like this answer. Although I wish there was a nicer way to get that $C$ is $1$. (I can't think of one) –  Eric Naslund Jun 12 '11 at 1:28

Let $L$ be the lower triangular matrix whose lower triangular part (including the main diagonal) is filled with ones, and let $e_1=(1,0,0,\ldots,0)^\top$. Then $$ A=\pmatrix{1&e_1^\top L^\top\\ Le_1&xLL^\top} =\pmatrix{1\\ &L} \underbrace{\pmatrix{1&e_1^\top\\ e_1&xI_{n-1}}}_B \pmatrix{1\\ &L^\top}. $$ Therefore $\det(A)=\det(B)$. Using Schur complement, if $x$ is an indeterminate, we have $\det(B)=\det(xI_{n-1})\det(1-\frac1x e_1^\top e_1)=x^{n-1}(1-\frac1x)$. Hence $\det(A)=x^{n-2}(x-1)$.

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