Answer to Question #102915 in Linear Algebra for Jake

Question

Let a 4x4 matrix V =

1 x x^2 x^3 //

1 y y^2 y^3 //

1 z z^2 z^3 //

1 w w^2 w^3

Prove that det(V)=(y-x)(z-x)(w-x)(z-y)(w-y)(w-z)

Answer 1

$Det(V)=\begin{vmatrix} 1 & x & x^2 & x^3\\ 1 & y & y^2 & y^3\\ 1 & z & z^2 & z^3\\ 1 & w & w^2 & w^3\\ \end{vmatrix}$

$=\begin{vmatrix} 1 & x & x^2 & x^3\\ 0 & y-x & y^2-x^2 & y^3-x^3\\ 0 & z-x & z^2-x^2 & z^3-x^3\\ 0 & w-x & w^2-x^2 & w^3-x^3\\ \end{vmatrix}$

Row 2 = Row 2 - Row 1; Row 3 = Row 3 - Row 1; Row 4 = Row 4 - Row 1

$=\begin{vmatrix} 1 & x & x^2 & x^3\\ 0 & y-x & (y-x)(y+x) & (y-x)(y^2+yx+x^2)\\ 0 & z-x & (z-x)(z+x) & (z-x)(z^2+zx+x^2)\\ 0 & w-x & (w-x)(w+x) & (w-x)(w^2+wx+x^2)\\ \end{vmatrix}$

$=(y-x)(z-x)(w-x) \begin{vmatrix} 1 & x & x^2 & x^3\\ 0 & 1 & (y+x) & (y^2+yx+x^2)\\ 0 & 1 & (z+x) & (z^2+zx+x^2)\\ 0 & 1 & (w+x) & (w^2+wx+x^2)\\ \end{vmatrix}$

By taking out $(y-x)$ from Row 2, $(z-x)$ from Row 3, $(w-x)$ from Row 4

$=(y-x)(z-x)(w-x) \begin{vmatrix} 1 & x & x^2 & x^3\\ 0 & 1 & y+x & (y^2+yx+x^2)\\ 0 & 0 & z-y & (z^2-y^2+zx-yx)\\ 0 & 0 & w-y & (w^2-y^2+wx-yx)\\ \end{vmatrix}$

Row 3 = Row 3 - Row 2; Row 4 = Row 4 - Row 2

$=(y-x)(z-x)(w-x) \begin{vmatrix} 1 & x & x^2 & x^3\\ 0 & 1 & y+x & (y^2+yx+x^2)\\ 0 & 0 & z-y & ((z-y)(z+y)+(z-y)x)\\ 0 & 0 & w-y & ((w-y)(w+y)+(w-y)x)\\ \end{vmatrix}​\\ =(y-x)(z-x)(w-x) \begin{vmatrix} 1 & x & x^2 & x^3\\ 0 & 1 & y+x & (y^2+yx+x^2)\\ 0 & 0 & z-y & (z-y)(z+y+x)\\ 0 & 0 & w-y & (w-y)(w+y+x)\\ \end{vmatrix}$

$=(y-x)(z-x)(w-x)(z-y)(w-y) \begin{vmatrix} 1 & x & x^2 & x^3\\ 0 & 1 & y+x & (y^2+yx+x^2)\\ 0 & 0 & 1 & (z+y+x)\\ 0 & 0 & 1 & (w+y+x)\\ \end{vmatrix}$

By taking out $(z-y)$ from Row 3, $(w-y)$ from Row 4

$=(y-x)(z-x)(w-x)(z-y)(w-y) \begin{vmatrix} 1 & x & x^2 & x^3\\ 0 & 1 & y+x & (y^2+yx+x^2)\\ 0 & 0 & 1 & (z+y+x)\\ 0 & 0 & 0 & (w-z)\\ \end{vmatrix}$

Row 4 = Row 4 - Row 3

$=(y-x)(z-x)(w-x)(z-y)(w-y)(w-z) \begin{vmatrix} 1 & x & x^2 & x^3\\ 0 & 1 & y+x & (y^2+yx+x^2)\\ 0 & 0 & 1 & (z+y+x)\\ 0 & 0 & 0 & 1\\ \end{vmatrix}$

By taking out $(w-z)$ from Row 4

$=(y-x)(z-x)(w-x)(z-y)(w-y)(w-z).$

Since $\begin{vmatrix} 1 & x & x^2 & x^3\\ 0 & 1 & y+x & (y^2+yx+x^2)\\ 0 & 0 & 1 & (z+y+x)\\ 0 & 0 & 0 & 1\\ \end{vmatrix} =1$

Hence proved.