MATHEMATIC
Short Exam 1 14-15 Spring.pdf

Short Exam 1 14-15 Spring.pdf - METU NCC LINEAR ALGEBRA...

• Test Prep
• 2

This preview shows pages 1–2. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: METU - NCC LINEAR ALGEBRA SHORT EXAM 1 Code : MAT 260 Last Name: AcadYeari 2014-2015 Name ' Semester I SPRING Student #I Date : 17.03.2014 Signatilic : Time I 12:30 3 QUESTIONS ON 2 PAGES Duration - 05 min TOTAL 10 POINTS 1. (319113) Let E— — {(1, 2, 3, —1),( (,1 0,2,1),,(11.,1,1)} Show that E IS linearly independent. l(i:£:3,-i)+)0,®,1,i) + &(I,1,I,I ()1: +34)”; =0 “91, + 9 :0 3} +35 +19 5'0 JHJV +y+93'oll'ﬂ}:oy 9‘30) j/VZ/O 2.(3pts) Let S = {a, b, c}, V = Fun(S) and E = {2Xa+3Xb“Xc: 2Xa+Xb+3Xca 2Xa+2Xb+Xc}' Find a linearly independent subset of E. (2)3}hl) 3. (31.93.))<°L)‘21l)- .3-(3253,-~I) +33 (we) +53 (23%?) :0. Cay-33.361901}; . 1, 3, 3:: : e 3 =3 ~0 «33::- , ,3), +3r-k923ﬂ. ‘£),9.¢O:28:’azj' lye) . (333m) 3 (3,132») .2. 333mm) U“ (£310, 4-“ 343—3: Xmi‘ﬁbtiﬂlag c/‘K'S‘ g +£‘4r.-0 X911”... ‘3‘)“.53 9:5; (SJ-leg 3“"? j ‘1‘ :«efr tax-i»? 5m :2 J 3. (4pm) Let V = 733(R) be the vector space of all polynomials of their degrees at most 3, and let U = {a0 + alx + (12532 + (132:3 E V : a0 + a1 —.— 2c2 — a3 = O} E V be the subspace. Find a linearly independent subset E of vectors from U such that Span(E) = U. 1. —-a , J. # I ) “'- a—Q 5 z ”I 3' ”’3‘! 5'" I' {Goiqwahe} “ it? ’ rﬂqe .. F a a. a, I!“ a "_:__ “3 W J21, §\§.53 V) é~ 3?‘ J! ...
View Full Document

{[ snackBarMessage ]}

What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern