如果实对称矩阵 [tex=0.786x1.0]b4HkKtHXeHofHX/gJc8Agg==[/tex] 与矩阵[tex=7.786x3.5]UwKAx/AIX9lGAocRZ6Xn8zO00lJi2f9RK3BkpKQ/gyxx7j32eb24p3FXvU/uG/wF2HCwa3BU+4YQ2Y0jRU4zkuI3Z0sv4hyzbe/RLRcEobA97DijWRYbGZyq2y6Ofn36[/tex]合同，则二次型 [tex=2.357x1.214]spLCz7RzhAETjEn9//GZbA==[/tex]的规范形为
未知类型：{'options': ['[tex=4.357x1.5]n7llpFRGrz65GgqdHQsAQaUuLahpLoBbdFNLO3WkfUc=[/tex]', '[tex=4.357x1.5]n7llpFRGrz65GgqdHQsAQTFdUhGvrJdWBFd0od9P2Wo=[/tex]', '[tex=4.357x1.5]K5qe6sVvINOYMHFLsiYZ8LCKpz7A/VqPIDl+4DdxLV8=[/tex]', '[tex=2.643x1.5]n7llpFRGrz65GgqdHQsAQWbni/clYBhhP7BD0gPebsM=[/tex]'], 'type': 102}
未知类型：{'options': ['[tex=4.357x1.5]n7llpFRGrz65GgqdHQsAQaUuLahpLoBbdFNLO3WkfUc=[/tex]', '[tex=4.357x1.5]n7llpFRGrz65GgqdHQsAQTFdUhGvrJdWBFd0od9P2Wo=[/tex]', '[tex=4.357x1.5]K5qe6sVvINOYMHFLsiYZ8LCKpz7A/VqPIDl+4DdxLV8=[/tex]', '[tex=2.643x1.5]n7llpFRGrz65GgqdHQsAQWbni/clYBhhP7BD0gPebsM=[/tex]'], 'type': 102}
举一反三
- 设 [tex=0.786x1.0]b4HkKtHXeHofHX/gJc8Agg==[/tex]是三阶实对称矩阵，且满足[tex=8.857x1.357]LXtcz8hY+gk4rolY95FMakoPU6LabB+mkJ/Jg0ZX/nc=[/tex]则二次型 [tex=5.071x1.5]KpZPhGD3OkExOkKRfj1dIlflwSVp+KDco3hnPWs/ve0=[/tex] 的规范形为 未知类型：{'options': ['[tex=4.357x1.5]n7llpFRGrz65GgqdHQsAQaUuLahpLoBbdFNLO3WkfUc=[/tex]', '[tex=4.357x1.5]K5qe6sVvINOYMHFLsiYZ8LCKpz7A/VqPIDl+4DdxLV8=[/tex]', '[tex=4.357x1.5]n7llpFRGrz65GgqdHQsAQTFdUhGvrJdWBFd0od9P2Wo=[/tex]', '[tex=5.071x1.5]oUes5HZpdNqw/0CiiOd0HPNAoaw11/6yKHY0ZqGu3caRQtCnE71DzrfplD/HRrYQ[/tex]'], 'type': 102}
- set1 = {x for x in range(10)} print(set1) 以上代码的运行结果为？ A: {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} B: {0, 1, 2, 3, 4, 5, 6, 7, 8, 9,10} C: {1, 2, 3, 4, 5, 6, 7, 8, 9} D: {1, 2, 3, 4, 5, 6, 7, 8, 9,10}
- 【计算题】5 ×8= 6×4= 7×7= 9×5= 2×3= 9 ×2= 8×9= 7×8= 5×5= 4×3= 5+8= 6 ×6= 3×7= 4×8= 9×3= 1 ×2= 9×9= 6×8= 8×0= 4×7=
- >>>x= [10, 6, 0, 1, 7, 4, 3, 2, 8, 5, 9]>>>print(x.sort()) 语句运行结果正确的是( )。 A: [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10] B: [10, 6, 0, 1, 7, 4, 3, 2, 8, 5, 9] C: [10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0] D: ['2', '4', '0', '6', '10', '7', '8', '3', '9', '1', '5']
- 求解下列矩阵对策,其中赢得矩阵 [tex=0.786x1.0]b4HkKtHXeHofHX/gJc8Agg==[/tex] 为$\left[\begin{array}{llll}2 & 7 & 2 & 1 \\ 2 & 2 & 3 & 4 \\ 3 & 5 & 4 & 4 \\ 2 & 3 & 1 & 6\end{array}\right]$