（I）选择题： 数学一（5），数学二(7)，数学三（5），数 学四（5） 设A为n阶非零矩阵，E为n阶单位矩阵．若 3 A = 0，则（ ） (A) E − A不可逆，E + A不可逆. (B) E − A不可逆，E + A可逆. (C) E − A可逆，E + A可逆. (D) E − A可逆，E + A不可逆

三维复式格子:晶体由N=N1N2N3个原胞构成,一个原胞中有n个原子。 n个原子的质量分别是:m1,m2,m3…m 第l个原胞的位置:R()=la1+l2a2+l3 原胞中各原子的位置

Plant nitrogen 1.1 Uptake of N 1. N is required by plants in large amounts Most plant material contains 2-4%N and 40%C. 2. N is usually the limiting nutrient in unfertilized systems 3. Most n is taken up from the soil in the form of nht or no a small amount of NH3 can be absorbed through the leaves N2 can be used by legume plant via biological nitrogen fiXation

Types of chemotaxins – C5a attracts neutrophils and monocytes – Made by bacteria Peptide clipped off N-terminus (beginning with N￾formylmethionine) during peptide maturation after protein synthesis

BIBO Stability Condition - A discrete-time LTI system is BIBO stable if the output sequence {y[n]} remains bounded for any bounded input sequence{x[n]}

41.AS a _actor,he can perform, sing, dance and play several kinds of musical instruments. A.flexible灵活的,柔韧的n.flexibility B. versatile形容人多才多艺的,形容工具用途广泛的vers代表随时可以变换的n

一、逆幂法分析 设n阶实方阵A有n个线性无关的特征向量u12…n 相应的特征值分别为,2…n,并按其绝对值的大小排列 即 则由A1=u,可得Au1=u,即A的逆矩阵A的特征值为

设A是n维酉空间V内的线性变换,如果V内的线性变换A满足a,BV,有 (Aa, B)=(a, B) 则称A是A的共轭变换.A为A的共轭变换当且仅当它们在标准正交基下的矩阵互为共轭 转置. 共轭变换的五条性质: 1)E=E 2)(A)=A 3)(kA)*=kA 4)(A+B)=a+B 5)(AB)'=B'A' 如果A=A,则称A是一个厄米特变换

Thus, to find Nash equilibria in IN=N, A(Si)), uin, we use the conditions: for ach player i (1)he is indifferent among all strategies in St,and (2)any strategy in St is at least as good as any strategy in S Example 3.7.(Meeting in an Airport ). Mr Wang and Ms Yang are to meet in an irport. However, they do not know whether they are to meet at door a or door B. The payoffs are specified in the following

1.试证明 Brent定理:令W(n)是某并行算法A在运行时间T(n)内所执行的运算数量, 则A使用p台处理器可在t(n)=((n)p+(n)时间内执行完毕