Name
/<EY
/ Differential Equations and Linear Algebra 2250
I Scores
II 4.
Midterm Exam 2
Version 2a, Thu 29 March 2012
Instructions: This in-class exam is 50 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) [20%] State four different determinant rules for n x ri matrices.
, E
2
E,A and E,, E
2
3
3 are elementary
(b) [20%] Assume given 3 x 3 matrices A, B. Suppose AB = E
det(A) = 13.
multiply
Assume
by
—1/3.
a
and
a
combination,
swap,
matrices representing respectively a
Find det(2B).
(c) [20%] Determine all values of x for which B’ fails to exist, where B equals the transpose of the
5x 0”
0
2
1001
0
3x
1 x—1 7 0
2
x
3
x
4
x
matrlx(
(d) [40%] Apply the adjugate [adjointj formula for the inverse to find the value of the entry in row 3,
column 4 of A’, given A below. Other methods are not acceptable.
A—
—
”
ThR
(
7
(it)
1
1
i\
—1 1
120
1 2 0
21
fi
1
COh
)L44) L#f
1
0
iJ
flAL{,
A4) c*—
f’m
AAc;7L
)L)
()
irut4i1
7
p
j
.)
2
)
3
fE
.
iMfI
1-SI=.
4
b
1 pk-J ?-a
1iI iM k)y
6a< IJ-,Y iE/, 1EI..—’i
”- I81*
71
£
—
(
—
Qcj
t”..
4..
j-)
±-,‘
‘)(=
(
1/1
j8)ø.
c)
Cct-(A)43)
‘-ow 3ci q.
0_I I
1
1
.i
l 2. )
If’\... (‘)(‘)
12,
I
‘
t
I
I
2
I
II
I
c
2..
\
fa*p.
37
I 2.
I IJ
+—1’(÷O 1).-\
,j
(H)
s’ !
_((K_..I)
x- 7
(?(-t)(9 3x)c
0
—
>
4or(A
—
itt
2.
J
1%ift -[Jv 1Lot if)
c,
.
i
I
C.DoS
Use this page to start your solution. Attach extra pages as needed.
ff1
p-
.. coI(A,)
CIeJL-4 17k. 47
‘443
HI’)
Name.
2250 Exam 2 S2012 [29Mar2012]
E ‘/
5. (Linear Differentia’ Equations) Do all parts.
(a) [20%] Solve for the general solution of 12y” + 7y’ + y = 0.
(b) [40%] The characteristic equation is r
(2r 2
2
(r 2r + 5) = 0. Find the general solution y of the
3)
linear homogeneous constant-coefficient differential equation.
(c) [20%] A third order linear homogeneous differential equation with constant coefficients has two
x + 4 sin 2x and
3
particular solutions 2e
What are the roots of the characteristic equation?
(d) [20%] Circle the functions which can be a solution of a linear homogeneous differential equation
with constant coefficients. For example, you would circle cos
2 x because cos
2 x = ± cos(2x) is a linear
combination of the two solutions 1 and cos(2x) of a third order equation whose characteristic equation
has roots 0, 2i, —2i.
—
—
2
eX
121
e”
tanx
cos(1nIx)
x_e_xsin(7rx)
(a
2rt+7 +1
)
2
cos(x
l
—
lL4-i
j
7\
e
)Y3
tior.
6)
34
/
3
010)
)
Z)
I
=
cO)e.
1))L,
C)
CS
j
A-
L)
AooJ
Ly
f.c.
e
jb,c)
gJ
z,vrn,
a-
j
-
J
=1
ft
i.
.
3• 7/
I
4
M
L/-t’IIlrj
(XL)
.
Use this page to start your solution. Attach extra pages as needed.
4) fl,
,Z:/C(