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

$A = 6x_1^2 + 5x_2^2 + 7x_3^2 - 4x_1x_2 + 4x_1x_3 = 6(x_1^2 - frac23x_1x_2 + frac23x_1x_3) + 5x_2^2 + 7x_3^2\x_1^2 - frac23x_1x_2 + frac23x_1x_3 = x_1^2 - 2x_1left(frac{x_2}3 - frac{x_3}3right) +left(frac{x_2}3 - frac{x_3}3right)^2 - left(frac{x_2}3 - frac{x_3}3right)^2 =\ = (x_1 - left(frac{x_2}3 - frac{x_3}3right))^2 - left(frac{x_2}3 - frac{x_3}3right)^2 = (x_1 - left(frac{x_2}3 - frac{x_3}3right))^2 - frac{x_2^2}9 + frac{2x_2x_3}9 - frac{x_3^2}{9}$

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

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

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

$A = 6widetilde{x}_1^2 + frac{13}3widetilde{x}_2^2 + frac{243}{39}widetilde{x}_3^2\begin{cases}widetilde{x}_1 = x_1 - frac{x_2}3 + frac{x_3}3\widetilde{x}_2 = x_2 + frac{2x_3}{13}\widetilde{x}_3 = x_3end{cases}$