1. ## determinant

prove that

|a b ax+by |
|b c bx+cy |
|ax+by bx+cy 0 |

=(b^2-ac)(ax^2+2bxy+cy^2)

2. Originally Posted by minaxijgupta
prove that

|a b ax+by |
|b c bx+cy |
|ax+by bx+cy 0 |

=(b^2-ac)(ax^2+2bxy+cy^2)
Compute this determinant by using the 3rd collossus.
Thus, we have,
$(-1)^{1+3}(ax+by)\left| \begin{array}{cc} b&c\\ ax+by& bx+cy \end{array} \right|$ $+(-1)^{2+3}(bx+cy)\left| \begin{array}{cc} a&b\\ ax+by&bx+cy \end{array} \right| +(-1)^{3+3}0 \left| \begin{array}{cc} a&b\\b&c \end{array} \right|$
Thus,
$(ax+by)(b^2x+bcy-acx-bcy)$ $-(bx+cy)(abx+acy-abx-b^2y)$
Thus,
$(ax+by)(b^2x-acx)-(bx+cy)(acy-b^2y)$
Now, just open this expression up and also open up the expression you are trying to prove to see that they match!

3. Originally Posted by ThePerfectHacker
...
Thus,
$(ax+by)(b^2x+bcy-acx-bcy)$ $-(bx+cy)(abx+acy-abx-b^2y)$
Thus,
$(ax+by)(b^2x-acx)-(bx+cy)(acy-b^2y)$
Now, just open this expression up and also open up the expression you are trying to prove to see that they match!
Hello,

with this result (which I never would have found) I suggest factorization:

$(ax+by)(b^2x-acx)-(bx+cy)(acy-b^2y)$

$(ax+by)\cdot x\cdot (b^2-ac)-(bx+cy)\cdot (-y)\cdot (b^2-ac)$

$(b^2-ac)\cdot \left[(ax+by)\cdot x+(bx+cy)\cdot (+y) \right]$

And now, minaxijgupta, the next steps are for you.

Best wishes

EB

4. Hello, minaxijgupta!

I think I've come up with a good approach . . .

Prove that: . $\begin{vmatrix}a & b & ax+by \\ b & c & bx+cy \\ ax+by & bx+cy & 0\end{vmatrix}\;=\;(b^2-ac)(ax^2+2bxy+cy^2)$

Perform some row operations to simplify the determinant . . .

$\begin{array}{ccc}x\cdot R_1 \\ y\cdot R_2 \\. \end{array}\;\begin{vmatrix}ax & bx & ax^2+bxy \\ by & cy & bxy + cy^2 \\ ax+by & bx + cy & 0\end{vmatrix}$

$\begin{array}{ccc}R_1+R_2\\.\\.\end{array}\;\begin {vmatrix}ax+by & bc + cy & ax^2 + 2bxy + cy^2 \\ by & cy & bxy + cy^2 \\ ax + by & bx + cy & 0\end{vmatrix}$

$\begin{array}{ccc}R_1-R_3\\R_2\div xy\\ .\end{array}\;\begin{vmatrix}0 & 0 & ax^2+2bxy+cy^2\\ \frac{b}{x} & \frac{c}{x} & \frac{bx + cy}{x} \\ ax + by & bx + cy & 0\end{vmatrix}$

Now we "expand by minors" (using the top row):

$(ax^2 + 2bxy + cy^2)\,\begin{vmatrix}\frac{b}{x} & \frac{c}{x} \\ ax + by & bx + cy\end{vmatrix}$

. . $=\;(ax^2+2bxy+cy^2)\,\left[\frac{b}{x}(bx+cy) - \frac{c}{x}(ax + by)\right]$

. . $= \;(ax^2 + 2bxy + cy^2)\,\left(b^2 + \frac{bcy}{x} - ac - \frac{bcy}{x}\right)$

. . $= \;(ax^2 + 2bxy + cy^2)\,(b^2 - ac)$ . . . ta-DAA!