1、Following parts of ordinary bolt connection: wall-beams, purlins, knee. Design of drainage, electrical, heating, ventilation, air conditioning, power should be subject to safety requirements for steel structures of other professions, it is forbidden to weld hanging load and increased roof and column

2、s. Construction prepared and the construction layout first section material prepared and the processing production prepared according to Zhejiang more Palace steel limited and Wuhan intelligent elevator limited by signed of contract file and the Institute provides of steel design blueprint, I compan

3、y and the Organization Technology Department, production department, procurement Department, bid do Engineering Department, related production sector for drawings triage, and according to owners requirements timely design out production and the processing detail and for has related professional sect

4、or of strictly triage, specific work process following: Bid do bid notice Technology Department add up do procurement department production department quality Department Engineering Department Technology Department production and the processing detail proposed Hou, by add up do coordination the sect

5、or of supporting work, and by figure line by varieties, and specifications, and quality for classified, and notification Procurement Department for material of market survey and procurement, by quality Department according to by purchased material of specifications varieties for quality review and t

6、est, not qualified of material varieties and specifications strongly ban for the engineering in the, After passing inspection informed the workshop of production of timber according to strict production details of production, workshop organized by the General Manager before the production technology

7、 and security sector technical safety training, safety education company to strictly three-level disclosure, training of qualified personnel is strictly prohibited to post production, production personnel must be listed to work, you must have the corresponding certificate. Road workers must have the

8、 certificate, and regular assessments should be carried out, not a skilled operator posts is strictly prohibited. Production of various items of equipment, appliances and other, non-professional is strictly prohibited actions and strict regulations, strong rewards, be eliminated. Production line set

9、 on person, title, responsibilities, and performance and safety comprehensive assessment. Material procurement process strictly handling the裝訂線華南農(nóng)業(yè)大學(xué)期末考試試卷（A卷）2010-2011學(xué)年第2學(xué)期 考試科目：線性代數(shù) 試類型：（閉卷）考試 考試時(shí)間：120分鐘學(xué)號(hào) 姓名 年級(jí)專業(yè) 題號(hào)一二三四五總分得分評(píng)閱人得分一、選擇題（本大題共5小題，每小題3分，共15分）在每小題的選項(xiàng)中，只有一項(xiàng)符合要求，把所選項(xiàng)前的字母填在題中括號(hào)內(nèi)1. 設(shè)矩陣A,

10、B, C能進(jìn)行乘法運(yùn)算，那么（B）成立(A) AB = AC，A 0，則B = C(B)AB = AC，A可逆，則B = C(C)A可逆，則AB = BA(D)AB = 0，則有A = 0，或B = 02. 設(shè)A為n(n2)階矩陣，且A2= I，其中I為單位陣（下同），則必有（C）(A)A的行列式等于1(B)A的逆矩陣等于I(C)A的秩等于n(D)A的特征值均為13設(shè)向量組線性相關(guān)，則向量組中（A）(A) 必有一個(gè)向量可以表為其余向量的線性組合(B)必有兩個(gè)向量可以表為其余向量的線性組合(C) 必有三個(gè)向量可以表為其余向量的線性組合(D) 每一個(gè)向量都可以表為其余向量的線性組合4設(shè)n元齊次線性

11、方程組的系數(shù)矩陣A的秩為r，則有非零解的充分必要條件是（B）(A) (B) (C) (D) 5. 設(shè)A為n階方陣，是A的伴隨矩陣。則：等于 (C)(A) (B) (C) (D) 得分二、填空題（本大題共5小題，每小題4分，滿分20分）6. 已知行列式，則數(shù)a =3.7. 設(shè)向量組, 線性相關(guān)，則數(shù)k =.8. 設(shè), ,則與的距離為9，內(nèi)積為37.9. 設(shè)n階實(shí)對(duì)稱矩陣A的特征值分別為1, 2, , n，則使為正定矩陣的數(shù)t取值范圍是10. 設(shè)矩陣A和B相似，其中A = ，B = 則0，.得分1.5CM三、計(jì)算題11.(滿分8分) 設(shè)矩陣 ，計(jì)算.解答：= = = 計(jì)算正確2分， 2分12.（滿

12、分8分）計(jì)算行列式 D = 的值。解：將D的各列全部加到第一列，得D = (x+1+2+n) （3分）= (x+) （ 2分） = (x+)(x-1)(x-2)(x-n) （ 3分）13.(滿分7分) 設(shè) , 求 . 解：(A I)= （ 2分） , （ 4分）所以 A-1 = . （ 1分）用公式法酌情給步驟分得分1.5CM四、解答題14.(滿分10分) 已知方程組有無(wú)窮多解，求a以及方程組的通解。解法一：由方程組有無(wú)窮多解，得，因此其系數(shù)行列式。即或。（3分）當(dāng)時(shí)，該方程組的增廣矩陣于是，方程組有無(wú)窮多解。分別求出其導(dǎo)出組的一個(gè)基礎(chǔ)解系，原方程組的一個(gè)特解，故時(shí)，方程組有無(wú)窮多解，其通解為

13、， （4分）當(dāng)時(shí)增廣矩陣，此時(shí)方程組無(wú)解。 （3分）解法二：首先利用初等行變換將其增廣矩陣化為階梯形。由于該方程組有無(wú)窮多解，得。因此，即。求通解的方法與解法一相同。15.（滿分10分）求向量組， 的一個(gè)最大無(wú)關(guān)組，且將不屬于最大無(wú)關(guān)組的向量用最大無(wú)關(guān)組線性表示出來(lái)1 所以為向量組的一個(gè)最大無(wú)關(guān)組. （6分） （2分） （2分） 16.(滿分6分) A，B為4階方陣，AB+2B=0，矩陣B的秩為2且|I+A|=|2I-A|=0。（1）求矩陣A的特征值；（2）A是否可相似對(duì)角化？說(shuō)明原因。（3）求|A+3I|。解：（1）由知-1，2為的特征值。，故-2為的特征值，又的秩為2，即特征值-2有兩個(gè)線性無(wú)關(guān)的特征向量，故的特征值為-1，2，-2，-2。 （2分）（建議只求出-1，2也給2分）（2）能相似對(duì)角化。因?yàn)閷?duì)應(yīng)于特征值-1，2各有一個(gè)特征向量，對(duì)應(yīng)于特征值-2有兩個(gè)線性無(wú)關(guān)的特征向量，所以有四個(gè)線性無(wú)關(guān)的特征向量，故可相似對(duì)角化。（2分）（3）的特征值為2，5，1，1。故=10。（2分）17.（滿分10分）求一個(gè)正交變換，化二次型 為標(biāo)準(zhǔn)型。解：設(shè)： ，則： ， 1分首先求 A 的特征根和特征向量令：