Proving determinant with variables

Question

I have a problem that asks:

Prove that det$\begin{pmatrix} 1 && 1 && 1 \\ a && b && c \\ a^2 && b^2 && c^2 \end{pmatrix} = (c-a)(c-b)(b-a)$

I was thinking of solving it like normal and see if I can end up with $(c-a)(c-b)(b-a)$

What I did:

det$\begin{pmatrix} b && c \\ b^2 && c^2 \end{pmatrix}$ - det$\begin{pmatrix} a && c \\ a^2 && c^2 \end{pmatrix}$ + det$\begin{pmatrix} a && b \\ a^2 && b^2 \end{pmatrix}$

which gives

$(bc^2 - cb^2)-(ac^2-ca^2)+(ab^2-ba^2)$

which simplifies to

$(b-a)c^2 + (a-c)b^2 + (c-b)a^2$

and this is where I got stuck. Am I even approaching this problem correctly? Any help would be appreciated.

Answer 1

which simplifies to $\;\; (b-a)c^2 + (a-c)b^2 + (c-b)a^2$

Hint: $\;(b-a)c^2 + (a-c)b^2 + (c\color{red}{-a+a}-b)a^2\,=\,\ldots$

Answer 2

It might be useful to look for helpful column/row operations before expanding.

Here, subtraction of the first column from the second and the third one simplify the determinant considerably:

$$\det\begin{pmatrix} 1 && 1 && 1 \\ a && b && c \\ a^2 && b^2 && c^2 \end{pmatrix} = \det \begin{pmatrix} 1 && 0 && 0 \\ a && b-a && c-a \\ a^2 && b^2-a^2 && c^2-a^2 \end{pmatrix} = \ldots $$ $$ \ldots =(b-a)(c-a)\det\begin{pmatrix} 1 && 1 \\ b+a && c+a \end{pmatrix} = (b-a)(c-a)(c-b)$$

Answer 3

If a=b, b=c, c=a then determinant value is 0 so
(a-b)(b-c)(c-a) are factor of the determinant now comparing the leading coefficient so the Value of the determinant is (a-b)(b-c)(c-a)