Name___ Differential Equations and Linear Algebra 2250 Midterm Exam 2 Version 1, 21 Mar 2013 Instructions: This in-class exam is designed to be completed in 30 minutes. No calculators, notes, tables or books. No answer check is expected. Details count 3/4, answers count 1/4. ( 1. (The 3 Possibilities with Symbols) c denote constants and consider the system of equations Let a, 6 and 0 3 —1 1 0 6+1 —1—b \ 0\(X\ a Y / a 0) b2_b) 0 2 b 1=1 ) 1 Ii’ (a) [40%] Determine a and b such that the system has a unique solution. (b) [30%] Explain why a (c) [30%] Explain why a many solution cases. £ £t.. = 0 and 6 = b = o 0 implies infinitely many solutions. Ignore any other possible infinitely - -4 3 i\\ 0 = 0 \)t1 6 —( ft o 1ArI ft(I—b)* iS cot ,41’ b÷i GLO -- C 0 Ar 1 o, we Yr- A-t a/ 3 v —I / 0 rì j ) 4 ô- / ° I — , ?2 ? 1 bsl-it &4frY 1 ‘ s i.-. 0 I I b’J &O)D. ‘-ir4 Use this page to start your so1uti. ) ilLd’t Q ‘fiyy- 6D IQ-f’n€ C C4-?-% /tL444C 1-4. cc S+eps a , L.J AtZ 3 c & b4-l “i Zru e’j 7i) 0 implies no solution. Ignore any other possible no solution cases. 1 4 ttaclk extra pages as needed A—_e 1 €-n 4 rirtt-i fti-t.-jd, 6 i’p fie oO_ rrj fl)&1)’ONJ. Name. 2. 2250 Exam 2 S2012 [22Mar2012] i<v (Vector Spaces) Do all parts. Details not required for (a)-(d). ) ( ) ( ). S of 1r false: There is a subsce -i-va containing none of the vectorsj 1 2 0 1 I , , (b) [10%] jor false: The set of solutions ii in 7?. of a consistent matrix equ tion Au = b can equal Exl • ? = all vectors in the xy-plane, that is, all vectors of the form = (x, y, 0). C’ 0 12 in subspace define a 0 z = 0, = + y Relations false: + (c) [10%] jor (d) [10%jr false: Equations x = y, z = 2y define a subspace in 7Z. €‘rl (e) [20%] Linear algebra theorems are able to conclude that the set S of all polynomials f(x) = co+cix+ x2) is 2 such that f’(x) + J f ()xdx = 0 is a vector space of functions. Explain why V = span(1, x, cx a vector space, then fully state a linear algebra theorem required to show S is a subspace of V. To save (a) [10%] f 1 time, do not write any subspace proof details. (f) [40%] Find a basis of vectors for the subspace of ‘J?. given by the system of restriction equations 3’l 211 —211 2x + + 213 + 414 13 + + 213 + 414 214 tdk U. e5t oJ \J clIf,cz) 415 1215 aJ y 2 O £i.4Jrre_4 c 1 .— (z) Ok% ee r S5 S’i.c 1 e piLc (.Cc..%2..) 1k i rfcçr C DrQ 2 fV]1ei - U) 415 - i1 S 0, 0, 0, 0. 1015 + + + + (ec* V Vjri. -.- : •1 S rJ! o L-j 2, 2 0 1 2. L.j. (20 12. 44’ /0 1 (iooo.-21o\ 00 1 2 lo 1 C)Jlcc,ooI 1 “\ 1 ) & L / I O\ I D\ ‘‘ j o/ ‘ AtLy = 1 )c: S,CcnJ? cS L rPe d- L4 4 I1AJ &tf, SC&M 3 2t ‘( Xz_2tt_’t3 XqtZ ‘\ ‘ X_..::t 11l 1 1 \.i, - L&i- 1 Use this page to start your solution. Attach extra pages as needed. I /Y444 4.’) v (X’ry L?;-j(3) ! cf : 2250 Exam 2 S2012 [22Mar2012} Name. 3. (Independence and Dependence) Do all parts. (a) [10%] State a dependence test for 3 vectors in 7?. Write the hypothesis and conclusion, not just the name of the test. (b) [10%] State fully an independence test for 3 polynomials. It should apply to show that 1, 1 + x, x(1 ± x) are independent. (c) [10%] For any matrix A, rank(A) equals the number of lead variables for the problem A± = 0. How many non-pivot columns in an 8 x 8 matrix A with rank(A) = 6? , v 3 4 denote the rows of the matrix , v 2 , v 1 (d) [30%] Let v —2 0 2 0 10 10 0 0 0 0 A— — —6 0 5 1 21 30 4 are independent and display the details of the chosen independence ,v 3 Decide if the four rows , v test. (e) [40%] Extract from the list below a largest set of independent vectors. 0 0 0 0 0 0 0 1 3 1 2 0 0 1 3 1 2 0 ,V5 ,V4 2 —1 —1 —2 0 0 0 0 0 0 2 3 5 1 2 0 ‘, — () . k Te.c 1 4 RO.- Vt) Va. 13 ML 3 V €f’ : cf- T1 ro,rLL rts.-t Q,fJ4fJ h& mL< ‘V-t - aOYJç - (\o9erP S’ot tLit 4?) x-= x 7 \x i oIynoL U @ Th &L ‘fle c1 r’. t C.3’j 3) f )e iqL1 L- Q P}isrdA. T Je4cf q( o/€d(/r Y$ (A ‘ A IvkJ Use this page to staiYr lutig A ‘Tie i &x). 74c pfo/f - VV _V aV_ V 3 VV V 1 Att exti2 1 LL iv. ft’ Name_______________________ / Differential Equations and Linear Algebra 2250 Scores Midterm Exam 2 Version 2a, 28 March 2012 4. 5. Instructions: This in-class exam is designed to be completed in under 30 minutes. No calculators, notes, tables or books. No answer check is expected. Details count 3/4, answers count 1/4. 4. (Determinants) Do all parts. (a) [10%] True or alse The value of a determinant is the product of the diagonal elements. (b) [10%] True or False The determinant of the negative of the n x n identity matrix is —1. 2 are elementary matrices E,A and E,, E 2 B=E 2 (c) [20%] Assume given 3 x 3 matrices A, B. Suppose A representing respectively a swap and a multiply by —5. Assume det(B) = 10. Let C = 2A. Find all possible values of det(C). 1 fails to exist, where C equals the transpose of (d) [30%] Determine all values of x for which (I + C) 0 —1 2 0 1 the matrix x—1 x x (e) [30%] Let symbols ü, b, c denote constants. Apply the adjugate [adjoint] formula for the inverse to find the value of the entry in row 3, column 4 of A’, given A below. ( A— — 1100 1 0 0 0 abOl 1 c12 1 ‘e II—J ( JkLo )L—A jaA IAI Mz_IAi=. . ( (- I I 2 = X—2 x — j CiA) 13/)A .1 X—3xt (L) ) Use this page to start your solution. Attach extra pages as needed. r(A) (—i\ 3)/IA) UI (Co C3) 2250 Exam 2 S2013 [28Mar2013] Name. 5. Do all parts. (a) [20%] Solve for the general solution of lSy” + 8y’ + y 0. (r 2r + 10) = 0. Find the general solution y of 3 1) (2r + 2 2 (b) [40%] The characteristic equation is ‘r the linear homogeneous constant-coefficient differential equation. (c) [20%] A fourth order linear homogeneous differential equation with constant coefficients has two X. Write a formula for the general solution. 3 x + 4x and xe 3 particular solutions 2e which cannot be a solution of a linear homogeneous differential the functions (d) [20%] Mark with X equation with constant coefficients. Test your choices against this theorem: (Linear Differential Equations) — The general solution of a linear homogeneous nth order differential equation with constant coefficients is a linear combination of Euler solution atoms. ç eX i eI 2 2 cos(xln 13.71251) =c\ X/5c/ QrQA ThitS t ))c -/z. 2 X/2 c)rhj4,fl’\ ift n+o 4 c-.AL ‘It x. 9( I) X X) E€ 4i _X/ oi. c4-t’mr.. ó r 0 C ill) I c. s(0txI) ivot (6t &(o’AuJ) fY\Ot ch,r,) cL C(\?) (fl&J pQ5ltTh) V(kS’ cr1 ç. 7 .0 1f. frin C - Use this page to start your solution. Attach extra pages as needed. iJi1LS 0 fQW€Y Z-