线性代数基础二次型线性代数做题本(一)解答版选择题不看本题本题讨论已掌握查看解析收入错题#8591二次型 f(x1,x2,x3)=x12+4x22+4x32−4x1x2+4x1x3−8x2x3f\left({{x}_{1},{x}_{2},{x}_{3}}\right) = {x}_{1}^{2} + 4{x}_{2}^{2} + 4{x}_{3}^{2} - 4{x}_{1}{x}_{2} + 4{x}_{1}{x}_{3} - 8{x}_{2}{x}_{3}f(x1,x2,x3)=x12+4x22+4x32−4x1x2+4x1x3−8x2x3 的规范形为‾\underline{\qquad}.A. f=z12f = {z}_{1}^{2}f=z12B. f=z12−z22f = {z}_{1}^{2} - {z}_{2}^{2}f=z12−z22C. f=z12+z22+z32f = {z}_{1}^{2} + {z}_{2}^{2} + {z}_{3}^{2}f=z12+z22+z32D. f=z12+z22−z32f = {z}_{1}^{2} + {z}_{2}^{2} - {z}_{3}^{2}f=z12+z22−z32