Cofactor expansion method for finding the determinant of a matrix

Question

Use the determinant properties to simplify the given matrix and show that $\det(A) = (x - y)(x - z)(x - w)(y - z)(y - w)(z - w)$ for

$$A = \begin{pmatrix} 1 & x & x^2 & x^3 \\ 1 & y & y^2 & y^3 \\ 1 & z & z^2 & z^3\\ 1 & w & w^2 & w^3 \\ \end{pmatrix} $$

For this problem, I noticed that the area of interest is the second column of matrix $A$. Will using the cofactor expansion method of finding the determinant yield the desired result if I choose the second column? The second column has $x$,$y$,$z$,$w$ and what I am looking for is $\det(A) = (x-y)(x-z)(x-w)(y-z)(y-w)(z-w)$. Thanks for any help!

Answer 1

Here is a non-clever approach that is still better than brute force. Subtract the bottom row from each of the preceding rows. This does not change the determinant. The result is $$\begin{pmatrix} 0 & x-w & x^2-w^2 & x^3-w^3 \\ 0 & y-w & y^2-w^2 & y^3-w^3 \\ 0 & z-w & z^2-w^2 & z^3-w^3 \\ 1 & w & w^2 & w^3 \end{pmatrix}$$ Expanding along the first column, find that the determinant is $$-\det \begin{pmatrix} x-w & x^2-w^2 & x^3-w^3 \\ y-w & y^2-w^2 & y^3-w^3 \\ z-w & z^2-w^2 & z^3-w^3 \end{pmatrix}$$ Factor out $(x-w)(y-w)(z-w)$: $$-(x-w)(y-w)(z-w) \det \begin{pmatrix} 1 & x+w & x^2+xw+w^2 \\ 1 & y+w & y^2+yw+w^2 \\ 1 & z+w & z^2+zw+w^2 \end{pmatrix}$$ Again subtract the bottom row from the first two. $$-(x-w)(y-w)(z-w) \det \begin{pmatrix} 0 & x-z & x^2-z^2+(x-z)w \\ 0 & y-z & y^2-z^2+(y-z)w \\ 1 & z+w & z^2+zw+w^2 \end{pmatrix}$$ This becomes a $2\times 2$ determinant, from which you can factor $(x-z)(y-z)$: $$-(x-w)(y-w)(z-w)(x-z)(y-z) \det \begin{pmatrix} 1 & x+z+1 \\ 1 & y+z+1 \end{pmatrix}$$ and conclude.