Download First Midterm Answer Key

Math 265 Midterm 1
Oct 2nd, 2008
Name:
Section:
This exam consists of 8 pages including this front page.
Ground Rules
1. No calculator is allowed.
2. Show your work for every problem unless otherwise stated.
3. You may use one 4-by-6 index card, both sides.
Score
1
20
2
20
3
10
4
10
5
20
6
10
7
10
Total
100
1
1. The following are true/false questions. You don’t have to justify your answers. Just write down either T or F in the table below. A, B, C, X, b are
always matrices here.
(a) If A = AT then A is a square matrix.
(b) If AB = AC and A 6= 0 then B = C.
(c) Consider a system of linear equations AX = b where A is a square
matrix. If the system is inconsistent then det(A) = 0.
(d) Let L be a straight line in the plane R2 . Then L is a subspace of R2 .
(e) If A2 = 0 then A is singular.
(a)
Answer
T
(b) (c)
F
T
(d)
(e)
F
T
Explain:
1. If A is a m × n-matrix then AT is a n × m-matrix. So if A = AT then
m = n and A is a square matrix.
µ
¶
µ
¶
µ
¶
1 0
0 0
0 0
2. Counterexample: A =
,B=
,C=
.
0 0
0 1
0 0
3. If det(A) 6= 0 then A−1 exists and AX = b has a unique solution
X = A−1 b. Thus if AX = b is inconsistent then det(A) must be 0.
4. A straight line in R2 is a subspace if and only if the line passing through
the origin.
5. A2 = 0 implies that det(A)2 = 0, hence det(A) = 0 and A is singular.
2
2. Quick Questions, A, B, C, X, b are always matrices here:
(a) Suppose that A is nonsingular and A2 = A. det(A) =?
det(A2 ) = (det(A))2 = det(A). Since A is nonsingular, det(A) 6= 0. So
det(A) = 1.
(b) Suppose AX = 2X and A−1 X = aX. Then a =?
Since A exists, times A−1 to AX = 2X, we get X = A−1 2X. Hence
A−1 X = 21 X and a = 12 .
½µ
2×2
(c) Let R
be the vector space of 2×2-matrices and W =
¶
¾
a b
|abcd = 0 .
c d
Is W the subspace of R2×2 ? why?
¶
µ
0 1
No. Because it is not closed under addition. For example, A =
0 1
¶
¶
µ
µ
1 1
1 0
are not in W .
are in W . But A + B =
and B =
1 1
1 0
 
1
2

(d) Find a matrix A such that Y = 
 0  is in the null space N (A) of A
−1
(Recall that N (A) = {X|AX = 0}).
¡
¢
We may assume that A = a1 a2 a3 a4 a 4×1-matrix. If Y ∈ N (A)
then AY = 0. Hence we must have a1 · 1 + a2 · 2 + a3 · 0 + a4 · −1 = 0.
Equivalently, a1 + 2a2 − a4 = 0. So we may take a1 = 1, a2 = 2, a3 = 3
and a4 = a1 + 2a2 = 5. Thus A = (1, 2, 3, 5).
3
3. Let
·
1 0 −1
A=
2 1 −2
¸
·
¸
0 −2 1
and B =
.
1 −2 1
(a) Compute 3A − 2B.
µ
Solution: 3A − 2B =
¶
3 4 −5
4 7 −8
(b) Compute AB T .
Solution:
µ
AB T =


¶
¶
µ
0
1
−1
0
1 0 −1 
· −2 −2 =
−4 −2
2 1 −2
1
1
4
4. In the following linear system, determine all values of a for which the resulting
linear system has
(a) no solution.
(b) a unique solution
(c) infinitely many solutions
x+y−z = 2
x + 2y + z = 3
x + y + (a2 − 5)z = a
Solution:
We have a augmented matrix


1 1
−1
2
 1 2
1
3 
2
1 1 a −5 a
Use row operations r2 − r1 → r2 and r3 − r1 → r3 , we get


1 1
−1
2
 0 1
2
1 
2
0 0 a −4 a−2
If a2 −4 = 0 and a−2 6= 0 then the last leading on locates in the last column,
and the system has no solution. Note that a2 − 4 = 0 if and only if a = ±2,
and a − 2 6= 0 if and only if a 6= 2. That is, the system has no solution if and
only if a = −2.
If a2 − 4 6= 0 then the coefficients matrix is nonsingular and the system has
a unique solution. That is if a 6= ±2 then the system has a unique solution.
Finally, if a2 − 4 = a − 2 = 0 then the system has infinitely many solution.
This only happens only a = 2.
5
5. (a) Compute
¯
¯−2
¯
¯5
¯
¯3
¯
¯8
0
3
0
0
0
5
2
2
¯
0¯¯
7¯¯
.
1¯¯
2¯
Solution: Use cofactor formula for the first
¯
¯3
¯
1+1 ¯
det(A) = −2(−1) ¯0
¯0
row, we have
¯
5 7¯¯
2 1¯¯ .
2 2¯
Use cofactor formula for the first column,
¯
¯
¯2 1 ¯
¯
¯ = −2 · 3 · 2 = −12.
det(A) = −2 · 3 · ¯
2 2¯
(b) Compute A−1 , where


1 0 2
A = 0 1 2
2 1 5
Solution: Using Gauss-Jordan method, we have A−1
6


−3 −2 2
= −4 −1 2 .
2
1 −1
6. Solve the following linear system using Cramer’s rule.
2x1 + 4x2 + 6x3 = 2
x1 + 2x3 = 0
2x1 + 3x2 − x3 = −5
Solution:
By Cramer’s rule, set



2 4 6
2 4 6
A = 1 0 2  , A 1 =  0 0 2  ,
2 3 −1
−5 3 −1





2 2
6
2 4 2
2  and A3 = 1 0 0 
A2 = 2 0
2 −5 −1
2 3 −5
Hence
x1 =
−52
det(A1 )
=
= −2.
det(A)
26
x2 =
det(A2 )
0
=
= 0.
det(A)
26
x3 =
det(A3 )
26
=
= 1.
det(A)
26
7
7. Let V be the set of all positive real numbers; define u⊕v = uv (⊕ is ordinary
multiplication) define ¯ by c ¯ u = uc . Prove that V is a vector space by
the following steps:
(a) Verify that V is closed under addition and scalar multiplication.
Let u, v ∈ V and c ∈ R. u⊕v = uv is positive thus u⊕v ∈ V . c¯u = uc
is still positive and c ¯ u ∈ V . Thus V is closed under addition and
scalar multiplication.
(b) Verify u ⊕ v = v ⊕ u and (u ⊕ v) ⊕ w = u ⊕ (v ⊕ w).
u ⊕ v = uv = vu = v ⊕ u.
(u ⊕ v) ⊕ w = (uv)w = u(vw) = u ⊕ (v ⊕ w).
(c) What is zero vector in the set V , explain?
The zero vector is 1 because for any u ∈ V , 1 ⊕ u = 1u = u.
(d) For any u ∈ V , find −u.
(−u) = u1 because (−u) ⊕ u = u1 u = 1, which is the zero vector proved
in last question.
(e) For any real number c, d ∈ R, and u, v ∈ V , Verify c ¯ (u ⊕ v) =
c ¯ u ⊕ c ¯ v; (c + d) ¯ u = c ¯ u ⊕ d ¯ u; c ¯ (d ¯ u) = (c · d) ¯ u and
1 ¯ u = u.
c ¯ (u ⊕ v) = c ¯ (uv) = (uv)c = uc v c = (c ¯ u)(c ¯ v) = c ¯ u ⊕ c ¯ v.
(c + d) ¯ u = uc+d = uc ud = (uc ) ⊕ (ud ) = c ¯ u ⊕ d ¯ u
c ¯ (d ¯ u) = c ¯ (ud ) = (ud )c = ucd = (c · d) ¯ u
1 ¯ u = u1 = u.
8