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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.
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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
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Y
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0)
b2_b)
0
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1
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(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.
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0 and 6
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=
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2.
2250 Exam 2 S2012 [22Mar2012]
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(Vector Spaces)
Do all parts. Details not required for (a)-(d).
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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
+
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213
+
414
214
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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
‘,
—
()
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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
(
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jaA
IAI
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.
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(-
I
I
2
=
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j
CiA) 13/)A
.1
X—3xt
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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.
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