MATH 271 Midterm: MATH 271 Amherst F14M271 28Benedetto 29
Math 271, Section 01, Fall 2014
Final Exam, Friday, December 19, 2014
Instructions: Do all twelve numbered problems. If you wish, you may also attempt the optional bonus
problems. Show all work, including scratch work. Little or no credit may be awarded, even when
your answer is correct, if you fail to follow instructions for a problem or fail to justify your answer.
If you have time, check your answers.
WRITE LEGIBLY. NO CALCULATORS.
1. (10 points). Take my word for it that
1 0 −1 0 0 2
0 1 0 0 1 −1
0 0 0 1 −21
0 0 0 0 0 0
is an echelon form of
1 0 −1−1 2 1
−3−133−7−2
0−103−74
2 0 −2−5 10 −1
.
Find the general solution of the following system of equations:
x1−x3−x4+ 2x5= 1
−3x1−x2+ 3x3+ 3x4−7x5=−2
−x2+ 3x4−7x5= 4
2x1−2x3−5x4+ 10x5=−1
2. (15 points). Let C=
−2−5 2
130
3 7 −6
. Compute C−1.
3. (15 points). Let V, W be vector spaces, let T:V→Wbe a linear transformation, and let
S={~v1, . . . , ~vn} ⊆ Vbe a finite set of vectors in V.
Suppose that dim(W) = nand also that Span({T~v1, . . . T~vn}) = W. Prove that Sis linearly indepen-
dent.
4. (20 points). Solve the system of differential equations
x′(t) = −2x+ 5y
y′(t) = 2x+y
subject to the initial conditions x(0) = 7, y(0) = 0.
5. (20 points). Let W= Span
1
1
1
1
,
−1
0
3
2
,
2
−2
4
4
Find an orthonormal basis for W.
6. (20 points) A sequence of real numbers z0, z1, z2, . . . is defined by the formula
z0= 0, z1= 1,and zk=zk−1+ 6zk−2for k≥0.
Find a simple formula for zk, for each integer k≥0.
Document Summary
If you wish, you may also attempt the optional bonus problems. Little or no credit may be awarded, even when your answer is correct, if you fail to follow instructions for a problem or fail to justify your answer. Find the general solution of the following system of equations: x1. 3x1 x2 + 3x3 + 3x4 7x5 = 2. Let v, w be vector spaces, let t : v w be a linear transformation, and let. , ~vn} v be a nite set of vectors in v . Suppose that dim(w ) = n and also that span({t ~v1, . Prove that s is linearly indepen- dent: (20 points). Solve the system of di erential equations subject to the initial conditions x(0) = 7, y(0) = 0. x (t) = 2x + 5y y (t) = 2x + y: (20 points).