题目
"用非退化线性变换化下列二次型为标准形,并利用矩阵验算所得结果.(1)-4(x)_(1)(x)_(2)+2(x)_(1)(x)_(3)+2(x)_(2)(x)_(3);(2)({x)_(1)}^2+2(x)_(1)(x)_(2)+2({x)_(2)}^2+4(x)_(2)(x)_(3)+4({x)_(3)}^2;(3)({x)_(1)}^2-3({x)_(2)}^2-2(x)_(1)(x)_(2)+2(x)_(1)(x)_(3)-6(x)_(2)(x)_(3);(4)8(x)_(1)(x)_(4)+2(x)_(3)(x)_(4)+2(x)_(2)(x)_(3)+8(x)_(2)(x)_(4);(5)(x)_(1)(x)_(2)+(x)_(1)(x)_(3)+(x)_(1)(x)_(4)+(x)_(2)(x)_(3)+(x)_(2)(x)_(4)+(x)_(3)(x)_(4);(6)({x)_(1)}^2+2({x)_(2)}^2+({x)_(4)}^2+4(x)_(1)(x)_(2)+4(x)_(1)(x)_(3)+2(x)_(1)(x)_(4)+2(x)_(2)(x)_(3)+2(x)_(2)(x)_(4)+2(x)_(3)(x)_(4);(7)({x)_(1)}^2+({x)_(2)}^2+({x)_(3)}^2+({x)_(4)}^2+2(x)_(1)(x)_(2)+2(x)_(2)(x)_(3)+2(x)_(3)(x)_(4)."
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用非退化线性变换化下列二次型为标准形,并利用矩阵验算所得结果.
(1)$-4{x}_{1}{x}_{2}+2{x}_{1}{x}_{3}+2{x}_{2}{x}_{3}$;
(2)${{x}_{1}}^{2}+2{x}_{1}{x}_{2}+2{{x}_{2}}^{2}+4{x}_{2}{x}_{3}+4{{x}_{3}}^{2}$;
(3)${{x}_{1}}^{2}-3{{x}_{2}}^{2}-2{x}_{1}{x}_{2}+2{x}_{1}{x}_{3}-6{x}_{2}{x}_{3}$;
(4)$8{x}_{1}{x}_{4}+2{x}_{3}{x}_{4}+2{x}_{2}{x}_{3}+8{x}_{2}{x}_{4}$;
(5)${x}_{1}{x}_{2}+{x}_{1}{x}_{3}+{x}_{1}{x}_{4}+{x}_{2}{x}_{3}+{x}_{2}{x}_{4}+{x}_{3}{x}_{4}$;
(6)${{x}_{1}}^{2}+2{{x}_{2}}^{2}+{{x}_{4}}^{2}+4{x}_{1}{x}_{2}+4{x}_{1}{x}_{3}+2{x}_{1}{x}_{4}+2{x}_{2}{x}_{3}+2{x}_{2}{x}_{4}+2{x}_{3}{x}_{4}$;
(7)${{x}_{1}}^{2}+{{x}_{2}}^{2}+{{x}_{3}}^{2}+{{x}_{4}}^{2}+2{x}_{1}{x}_{2}+2{x}_{2}{x}_{3}+2{x}_{3}{x}_{4}$.
题目解答
答案
