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Problem: Find a reduced form equivalent to $7x^2 + 25xy + 23y^2$

I know there are two forms to do it, when $c<a$ and $|b|>a$ I did the process but I don't know how to do the matrix.

$$7x^2 + 25xy + 23y^2\quad (|b|>a)$$ and $D=b^2 -4ac =-19$ doing the formula I have $k=-2$, $b'=9$ and $c'=1$

The result it will be $7x^2 -3xy + y^2$, I have all the sketch with the formulas (Next, $x^2 + 3xy + 7y^2$ and $x^2 + xy + 5y^2$), but none of the matrix

Thanks for your help

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I prefer to write the matrices on the right...

The mapping $$ \langle a, b, c \rangle \mapsto \langle a, \; b + 2at, \; c +bt +at^2 \rangle $$ is brought about by the matrix $$ \left( \begin{array}{rr} 1 & t \\ 0 & 1 \end{array} \right) $$

The mapping $$ \langle a, b, c \rangle \mapsto \langle c, -b , a \rangle $$ is brought about by the matrix $$ \left( \begin{array}{rr} 0 & -1 \\ 1 & 0 \end{array} \right) $$

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$$ \langle 7, 25, 23 \rangle $$ $$ R_1 = \left( \begin{array}{rr} 1 & -2 \\ 0 & 1 \end{array} \right) $$

$$ \langle 7, -3, 1 \rangle $$ $$ R_2 = \left( \begin{array}{rr} 0 & -1 \\ 1 & 0 \end{array} \right) $$

$$ \langle 1, 3, 7 \rangle $$ $$ R_3 = \left( \begin{array}{rr} 1 & -1 \\ 0 & 1 \end{array} \right) $$

$$ \langle 1, 1, 5 \rangle $$

$$ R = R_1 R_2 R_3 = \left( \begin{array}{rr} -2 & 1 \\ 1 & -1 \end{array} \right) $$

$$ H = \left( \begin{array}{rr} 14 & 25 \\ 25 & 46 \end{array} \right) $$ $$ G = \left( \begin{array}{rr} 2 & 1 \\ 1 & 10 \end{array} \right) $$ does give $$ R^T H R = G $$

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  • $\begingroup$ Thanks,now it's clear. $\endgroup$ – Ck27 May 31 '17 at 19:15

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