How to find determinant of this matrix?

Question

Is there a manual method to find $\det\left(XY^{-1}\right)$ ?

Let $$X=\left[ {\begin{array}{cc} 1 & 2 & 2^2 & \cdots & 2^{2012} \\ 1 & 3 & 3^2 & \cdots & 3^{2012} \\ 1 & 4 & 4^2 & \cdots & 4^{2012} \\ \vdots & \vdots & \vdots & \cdots & \vdots \\ 1 & 2014 & 2014^2 & \cdots & 2014^{2012} \\ \end{array} } \right], $$

$$Y=\left[ {\begin{array}{cc}\frac{2^2}{4} & \frac{3^2}{5} & \dfrac{4^2}{6} & \cdots & \dfrac{2014^2}{2016} \\ 2 & 3 & 4 & \cdots & 2014 \\ 2^2 & 3^2 & 4^2 & \cdots & 2014^{2} \\ \vdots & \vdots & \vdots & \cdots & \vdots \\ 2^{2012} & 3^{2012} & 4^{2012} & \cdots & 2014^{2012} \\ \end{array} } \right] $$.

Thanks in advance.

Answer 1

Consider something a bit more general. Let $$X=\left[ {\begin{array}{cc} 1 & x_1 & x_1^2 & \cdots & x_1^{n-1} \\ 1 & x_2 & x_2^2 & \cdots & x_2^{n-1} \\ 1 & x_3 & x_3^2 & \cdots & x_3^{n-1} \\ \vdots & \vdots & \vdots & \cdots & \vdots \\ 1 & x_n & x_n^2 & \cdots & x_n^{n-1} \\ \end{array} } \right], $$ and $$Y=\left[ {\begin{array}{cc}\frac{x_1^2}{x_1+r} & \frac{x_2^2}{x_2+r} & \dfrac{x_3^2}{x_3+r} & \cdots & \dfrac{x_n^2}{x_n+r} \\ x_1 & x_2 & x_3 & \cdots & x_n \\ x_1^2 & x_2^2 & x_3^2 & \cdots & x_n^2 \\ \vdots & \vdots & \vdots & \cdots & \vdots \\ x_1^{n-1} & x_2^{n-1} & x_3^{n-1} & \cdots & x_n^{n-1} \\ \end{array} } \right]. $$

Your problem corresponds to $x_i=i+1,$ $n=2013,$ $r=2.$

Let $Y'$ be the matrix obtained by rescaling the columns of $Y$ by multiplying column $i$ by $(x_i+r)/x_i.$ So $$\det Y'=\det Y\prod_{i=1}^n\frac{x_i+r}{x_i} $$ and $$Y'=\left[ {\begin{array}{cc}x_1 & x_2 & x_3 & \cdots & x_n \\ x_1+r & x_2+r & x_3+r & \cdots & x_n+r \\ x_1(x_1+r) & x_2(x_2+r) & x_3(x_3+r) & \cdots & x_n(x_n+r) \\ \vdots & \vdots & \vdots & \cdots & \vdots \\ x_1^{n-2}(x_1+r) & x_2^{n-2}(x_2+r) & x_3^{n-2}(x_3+r) & \cdots & x_n^{n-2}(x_n+r) \\ \end{array} } \right]. $$

Now show that $$\det Y'=-r\det X,$$ from which your determinant can easily be evaluated. This is done by a series of row operations on $Y'.$ First subtract row $1$ from row $2.$ Then swap the first two rows. Then divide row $1$ by $r$. (These steps account for the factor $-r.$) We now have the matrix $$\left[ {\begin{array}{cc}1 & 1 & 1 & \cdots & 1 \\ x_1 & x_2 & x_3 & \cdots & x_n \\ x_1(x_1+r) & x_2(x_2+r) & x_3(x_3+r) & \cdots & x_n(x_n+r) \\ \vdots & \vdots & \vdots & \cdots & \vdots \\ x_1^{n-2}(x_1+r) & x_2^{n-2}(x_2+r) & x_3^{n-2}(x_3+r) & \cdots & x_n^{n-2}(x_n+r) \\ \end{array} } \right]. $$ Now subtract $r$ times row $2$ from row $3$. Then subtract $r$ times row $3$ from row $4.$ Continue in this way, finally subtracting $r$ times row $n-1$ from row $n.$ The resulting matrix will be $X^T.$

Answer 2

Not a solution, but a possible step in the right direction (too long for a comment):

Following Cramer's rule, we note that the solution $\vec y = (y_1,\dots,y_{2014})^T$ to $$X \,\vec y = \pmatrix{ \frac{2^2}{4} & \frac{3^2}{5} & \dfrac{4^2}{6} & \cdots & \dfrac{2014^2}{2016} }^T$$ Will satisfy $$ y_1 = \det(Y^T)/\det(X) = [\det(XY^{-1})]^{-1} $$ With that in mind: if there is another way to solve this equation, it will probably be easier than actually computing the determinant.

Note also that $X$ is a Vandermonde matrix.