Prove the given determinant

Question

Prove the given determinant: $$\left| \begin{matrix} a&b&ax+by \\ b&c&bx+cy \\ ax+by&bx+cy&0 \\ \end{matrix}\right|=(b^2-ac)(ax^2+2bxy+cy^2)$$

I didn't get any idea. Please help me solve this.

Answer 1

HINT:

Set $$C_3'=C_3-xC_1-yC_2$$ to find

$$\left| \begin{matrix} a&b&0 \\ b&c&0 \\ ax+by&bx+cy&-x(ax+by)-y(bx+cy) \\ \end{matrix}\right|=?$$

Answer 2

Develop the determinant along the last line. You get

$$D_{a,b,c}(x,y)=(ax+by)\begin{vmatrix}b&ax+by\\c&bx+cy\end{vmatrix}-(bx+cy)\begin{vmatrix}a&ax+by\\b&bx+cy\end{vmatrix}$$

Now compute the $2\times 2$ déterminants to get

$$D_{a,b,c}(x,y)=(ax+by)\left(b(bx+cy)-c(ax+by)\right)-(bx+cy)\left(a(bx+cy)-b(ax+by)\right)$$

It simplifies quite well into

$$\begin{align}D_{a,b,c}(x,y) =&(ax+by(b^2-ac)x+(bx+cy)(b^2-ac)y\\ =&(b^2-ac)(ax^2+2bxy+cy^2) \end{align}$$

Answer 3

A bit late in the day, here is the symmetric matrix version of this, writing $P^TAP = B,$ where $B$ is also symmetric with obvious determinant. We say that $A$ and $B$ are "congruent"

$$ \left( \begin{array}{ccc} 1&0&0 \\ 0&1&0 \\ -x&-y&0 \\ \end{array} \right) \left( \begin{array}{ccc} a&b&ax+by \\ b&c&bx+cy \\ ax+by&bx+cy&0 \\ \end{array} \right) \left( \begin{array}{ccc} 1&0&-x \\ 0&1&-y \\ 0&0&1 \\ \end{array} \right) = \left( \begin{array}{ccc} a&b&0 \\ b&c&0 \\ 0&0& - \left(a x^2 + 2bxy+cy^2 \right) \\ \end{array} \right) $$