Question

приведите квадратичную форму к каноническому виду.Укажите базис , в котором квадратичная форма имеет канонический вид. 6x1^2 +5x2^2+7x3^2 -4x1x2 +4x1x3

Answer 1

$A = 6x_1^2 + 5x_2^2 + 7x_3^2 - 4x_1x_2 + 4x_1x_3 = 6(x_1^2 - \frac23x_1x_2 + \frac23x_1x_3) + 5x_2^2 + 7x_3^2\\x_1^2 - \frac23x_1x_2 + \frac23x_1x_3 = x_1^2 - 2x_1\left(\frac{x_2}3 - \frac{x_3}3\right) +\left(\frac{x_2}3 - \frac{x_3}3\right)^2 - \left(\frac{x_2}3 - \frac{x_3}3\right)^2 =\\ = (x_1 - \left(\frac{x_2}3 - \frac{x_3}3\right))^2 - \left(\frac{x_2}3 - \frac{x_3}3\right)^2 = (x_1 - \left(\frac{x_2}3 - \frac{x_3}3\right))^2 - \frac{x_2^2}9 + \frac{2x_2x_3}9 - \frac{x_3^2}{9}$

$\widetilde{x}_1 = x_1 - \frac{x_2}3 + \frac{x_3}3\\A = 6\widetilde{x}_1^2 - \frac{2x_2^2}{3} + \frac{4x_2x_3}{3} - \frac{2x_3^2}{3} + 5x_2^2 + 7x_3^2 = 6\widetilde{x}_1^2 + \frac{13x_2^2}{3} + \frac{4x_2x_3}{3} + \frac{19x_3^2}{3} =\\6\widetilde{x}_1^2 + \frac{13}{3}\left(x_2^2 + \frac{4x_2x_3}{13} \right) + \frac{19x_3^2}{3}$

$x_2^2 + \frac{4x_2x_3}{13} = \left(x_2 + \frac{2x_3}{13} \right)^2 - \frac{4x_3^2}{169}\\\widetilde{x}_2 = x_2 + \frac{2x_3}{13}$

$A = 6\widetilde{x}_1^2 + \frac{13}3\widetilde{x}_2^2 - \frac{4x_3^2}{39} + \frac{19x_3^2}{3} = 6\widetilde{x}_1^2 + \frac{13}3\widetilde{x}_2^2 + \frac{243x_3^2}{39}\\\widetilde{x}_3 = x_3$

$A = 6\widetilde{x}_1^2 + \frac{13}3\widetilde{x}_2^2 + \frac{243}{39}\widetilde{x}_3^2\\\begin{cases}\widetilde{x}_1 = x_1 - \frac{x_2}3 + \frac{x_3}3\\\widetilde{x}_2 = x_2 + \frac{2x_3}{13}\\\widetilde{x}_3 = x_3\end{cases}$