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MATHEMATICS 201-105-RE
Linear Algebra
Martin Huard
Winter 2006
Review Exercises
SOLUTIONS
⎡2
− 2 3 ⎤
1. Consider the matrices
⎢ ⎥ ⎡−1
2 2⎤
⎡2
−1⎤
A =
⎢
1 5 −1
⎥
, B = ⎢ ⎥ and C = . Evaluate, if
⎢⎣
0 3 1 ⎥
⎣−
3 4 1
⎢ ⎥ ⎦ ⎣1
6 ⎦
⎦
possible. (For f, g, and h, use your answer from (e)).
⎡2 −2 3 ⎤
⎡−1 2 2⎤ 0 18 3
a) BA
⎢
1 5 1
⎥ ⎡ − ⎤
= ⎢
3 4 1
⎥ − = ⎢
⎣−
⎦
⎢ ⎥ −2 29 12
0 3 1
⎣ −
⎥
⎦
⎢⎣
⎥⎦
⎡−1 −3⎤ ⎡2 −2 3 ⎤
T
⎡−1 2 2⎤
b) B B−
3A=
⎢
2 4
⎥
−3 ⎢
1 5 −1
⎥
⎢ ⎥⎢
3 4 1
⎥
2 1
⎣−
⎦
⎢ ⎥
⎢⎣ ⎥⎦ ⎢⎣0 3 1 ⎥⎦
⎡ 10 −14 −5⎤ ⎡6 −6 9 ⎤ ⎡ 4 −8 −14⎤
=
⎢
−14 20 8
⎥
−
⎢
3 15 − 3
⎥
=
⎢
−17 5 11
⎥
⎢ ⎥ ⎢ ⎥ ⎢
⎥
⎢⎣ −5 8 5 ⎥⎦ ⎢⎣0 9 3 ⎥⎦
⎢⎣
−5 −1
2 ⎥⎦
⎡2 −1⎤⎡−1 2 2⎤ ⎡ 1 0 3⎤
c) CB = ⎢
1 6
⎥⎢
3 4 1
⎥ = ⎢
19 26 8
⎥
⎣ ⎦⎣−
⎦ ⎣−
⎦
⎛
⎡−1 −3⎤
⎞
1 2 2 2 1
T ⎜ ⎡−
⎤ ⎡ − ⎤⎟
d) tr
( 3BB
+ C)
= tr 3
⎢
2 4
⎥
⎢
3 4 1
⎥ +
⎢ ⎥ ⎢
1 6
⎥ ⎜ ⎣−
⎦
2 1
⎣ ⎦
⎢ ⎥ ⎟
⎝
⎣ ⎦ ⎠
⎛ ⎡9 13⎤ ⎡2 −1⎤⎞
= tr ⎜3⎢ + 13 26
⎥ ⎢ ⎟ 1 6
⎥
⎝ ⎣ ⎦ ⎣ ⎦⎠
⎛⎡29 38⎤⎞
= tr ⎜⎢
40 84
⎥⎟
⎝⎣
⎦⎠
= 29 + 84=
113
2 −2 3
5 −1 −2 3 −2 3
= 1 5 − 1 = 2 − + 0 = 2⋅8− − 11 + 0=
27
3 1 3 1 5 −1
0 3 1
e) det ( A) ( )
3
f) ( ) ( ) 3 3
det A = ⎡⎣det A ⎤⎦
= 27 = 19683
3
g) det 2A = 2det A = 8⋅ 27 = 216
( ) ( )

Math 105
Semester Review - Solutions
T
T
2
h) ( ) ( ) ( ) ( ) ( )
det
i) det adj( )
AA = det A det A = det A det A = 27 = 729
( ) ⎡ ( A)
T
( ⎣ ⎤⎦
) ( A)
A = det cof = det cof
8 −1 3
= 11 2 −6
−13 5 12
( )
2 −6 11 −6 11 2
= 8 + + 3
5 12 −13 12 −13 5
= 8⋅ 54 + 54 + 3⋅ 81 = 729
T
2. A square matrix A is called skew-symmetric if A = − A.
a) Prove that if A is invertible and skew-symmetric, then
−
To prove: ( A )
−1
T
LS = ( A )
T
−1
= ( A )
( )
−1
1
T
−1
−1
=−A
1
A −
T
= − A since A is skew-symmetric ( A =−A
)
=−A
is skew-symmetric.
= RS
T
b) Prove that A , A + B and kA are skew-symmetric if A and B are skew symmetric.
T
To prove: ( ) T T
A =−A
LS =
T
( A )
T
T
( A) since A is skew-symmetric
= −
=−A
= RS
T
To prove: ( A + B) T
= − ( A+B
)
( )
LS = A + B
T
= +
A
B
T
T
=−A−B since A and B are skew-symmetric
= RS
( A B)
=− +
Winter 2006 Martin Huard 2

Math 105
Semester Review - Solutions
T
To prove: ( kA) =−( kA)
LS =
( kA)
= kA
T
T
( )
( kA)
= k −A since A is skew-symmetric
=−
= RS
c) Prove that A−
A
T is skew symmetric.
T
T
T
To prove: ( A− A ) =−( A−
A )
T
T
LS = ( A − A )
T T
T
= A −( A )
T
= −
A
= RS
A
T
( A A )
=− −
3. Prove that if A and B are n n matrices such that A = B = AB = I then AB = BA.
Since , since A
1
B
− = B .
2
= AA= I
2 2
× ( ) 2
and
2
1
B = BB = I then A and B are invertible with A − = A and
2
−
Since ( AB) = ( AB)( AB)
= I , then AB is invertible and ( AB) 1
To prove: AB = BA
LS = AB
=
−
( AB) 1
= B A
= BA
= RS
−1 −1
= AB.
3
4. Let A be a matrix such that A = I .
a) Prove that A is invertible.
3
A = I
det
3
( A ) = det ( I )
3
⎡⎣det ( A)
⎤⎦
= 1
det ( A)
= 1
Thus, since det ( A)
≠ 0,
then A is invertible.
2
b) To prove: ( 2 )( 2 4 )
2
LS = ( I − 2A)( I + 2A+
4A
)
I − A I + A+ A = I
= + 2 + 4 −2 −4 −8
2 2 2 3
I IA IA AI A A
= I + A+ A − A− A −
= I
= RS
( )
2 2
2 4 2 4 8 0
Winter 2006 Martin Huard 3

Math 105
Semester Review - Solutions
5. Solve the following systems of linear equations, if possible.
a) 2x
− y + 2z
= 1
b)
c)
x +
5x
−
y −
y +
3z
=
z =
4
6
x + 4y
− 11z
= 11
⎡2 −1 2 1⎤ ⎡2 −1 2 1 ⎤
⎢ R2 →2R2 −R1
1 1 −3 4
⎥ ⎢
0 3 −8 7
⎥
⎢ ⎥R3 →2R3−5R
⎢
⎥
1
⎢5 −1 1 6⎥ ⎢0 3 −8 7 ⎥
⎢ ⎥R4 →2R4 −R1
⎢
1 4 −11 11
⎥⎦
⎣ ⎦ ⎣0 9 −24 21
−1 1
⎡2 −1 2 1⎤
⎡1 2
1
2⎤
R3 R3 R
⎢
2
0 3 8 7
⎥ 1
→ − − R
8 7
1
→
⎢
⎥
2
R
⎢
−
3 0 1
⎥
⎢
3 3⎥
R4 → R4 −3R
⎢
2
0 0 0 0⎥
1
R2 3
R ⎢
2
0 0 0 0⎥
⎢
⎥
→
⎢ ⎥
⎣0 0 0 0⎦
⎣0 0 0 0⎦
5 1 7 8
+ t, + t,
t
4x
−
2x
− 5y
+
2x
+ 4y
+
2x
−
Solution: ( )
y + 2z
=
z
=
3
9
z = − 6
3 3 3 3
⎡4 −1 2 3 ⎤ ⎡4 −1 2 3 ⎤
⎢ R2 →2R2 −R1
2 −5 1 9
⎥ ⎢
0 −9 0 15
⎥
⎢ ⎥R3 →2R3−R
⎢
⎥
1
⎢2 4 1 −6⎥
⎣ ⎦ ⎢⎣0 9 0 −15⎥⎦
−1 1 3
⎡4 −1 2 3⎤
⎡1
4 2 4
⎤
1
R3 → R3+ R
⎢
2
0 −9 0 15
⎥ R1 →
4
R1
⎢ −5
⎢
⎥
0 1 0
⎥
−1
3
R2 9
R ⎢
⎥
2
⎢⎣
0 0 0 0⎥
→
⎦ ⎢⎣
0 0 0 0⎥⎦
1 1 5
− t, − , t
Solution: ( )
y +
6x
+ 3y
− 2z
+ 3w
+
z
3 2 3
−
5t
= 12
t =
2x
+ 5y
− 4z
+ 3w
+ 11t
=
1
8
⎡2 −1 1 0 −5 12⎤ ⎡2 −1 1 0 −5 12 ⎤
⎢ 2 2
3
1
6 3 2 3 1 1
⎥R →R − R ⎢
⎢
−
⎥
0 6 5 3 16 35
R3 R3 R ⎢
− −
⎥
→ −
⎥⎥
1
⎢2 5 −4 3 11 8⎥
⎣ ⎦ ⎢⎣0 6 −5 3 16 − 4 ⎦
c
c
c
3
c
c
c
3
= t
= +
7 8
2 3 3
= +
5 1
1 3 3
= t
=
−5
2 3
= −
1 1
1 3 2
t
t
t
⎡2 −1 1 0 −5 12 ⎤
R3 R3 R
⎢
2
0 6 5 3 16 35
⎥
→ −
⎢
− −
⎥
⎢⎣0 0 0 0 0 31 ⎥⎦
No solutions
1
R
1 1
5
1
→
2
R
−
−
1 ⎡1 2 2
0
2
6 ⎤
1 −5 1 8 −35
R2 →
6
R
⎢
2
0 1
⎥
⎢
6 2 3 6 ⎥
1
R 0 0 0 0 0 1
3 31
R ⎢
⎥
→
3⎣ ⎦
Winter 2006 Martin Huard 4

Math 105
Semester Review - Solutions
d)
3x
−
5x
−
2x
+ 3y
−
y + 2z
=
9
y + 3z
= 16
z = 11
⎡3 −1 2 9⎤ ⎡3 −1 2 9⎤
⎢ R2 →3R2 −5R1
5 −1 3 16
⎥ ⎢
0 2 −1 3
⎥
⎢ ⎥R3 →3R3−2R
⎢
⎥
1
⎢2 3 −1 11⎥
⎣ ⎦ ⎢⎣0 11 −7 15⎥⎦
1
⎡3 −1 2 9 ⎤ R
1 2
1
→
3
R
−
1 ⎡1 3 3
3⎤
R3 →2R3−11R
⎢
2
0 2 −1 3
⎥ 1 −1
3
⎢
⎥
R2 →
2
R
⎢
2
0 1
⎥
⎢
2 2⎥
⎢⎣
0 0 −3 −3⎥
−1
⎦ R 0 0 1 1
3 3
R ⎢
⎥
→
3⎣ ⎦
Solution: ( 3, 2,1)
6. For which values of a will the following system of linear equations have
x − y + 2z
= 7
− 2x + ay + − 4z = 5a−24
( )
3x + 2y + 6− a z =
⎡ 1 −1 2 7⎤ ⎡1 −1 2 7⎤
⎢ R2 R2 2R1
2 a 4 5a 24
⎥ → + ⎢
⎢
− − −
⎥
0 a 2 0 5a
R3 R3 3R
⎢
− −10
⎥
→ −
⎥
1
⎢ 3 2 6−a
4⎥
⎣ ⎦ ⎢⎣0 5 −a
−17⎥
⎦
⎡1 −1 2 7⎤
R ↔ R
⎢
0 5 a 17
⎥
⎢
−
⎥
⎢⎣0 a−2 0 5a−10⎥⎦
− ( )
2 3
3 3 2
⎡
1
1 −1 2 7⎤
R2 →
5
R2
⎢ ⎥
−1
−17
0 1
1
5
a
5
R3 →
aa ( 2)
R
⎢
⎥
− − 3
⎢
42a−84
0 0 1 ⎥
⎣
−aa
( −2)
⎦
Illegal if a( a )
− 2 = 0
a = 0 or 2
a) Thus there will be a unique solution if a ≠ 0, 2
b) If a = 0 ,
c) If a = 2 ,
4
⎡1 −1 2 7⎤
⎢
⎥
R →5R − a−2 R
⎢
0 5 −a
−17
⎥
⎢
⎣0 0 −a( a−2)
42a−84⎥
⎦
⎡1 −1 2 7⎤
⎢
⎢
0 5 0 −17
⎥
⎥
so there is no solution
⎢⎣0 0 0 −84⎥⎦
⎡1 −1 2 7⎤
z = t
⎢
0 5 2 17
⎥ −17
⎢
− − ⎥
y =
5
an infinite number of solutions
⎢⎣0 0 0 0⎥
52
⎦ x = − t
5
2
Winter 2006 Martin Huard 5

Math 105
Semester Review - Solutions
7. Solve the following systems of linear equations using (i) Cramer’s rule (ii) the inverse.
3x
− y + = 1
a)
5x
+ 3z
= 14
x + 3y
− z = 4
i – Cramer’s Rule
3 −1 0
0 3 5 3
det ( A) = 5 0 3 = 3 −( − 1)
0 =−27 −8 =− 35
3 −1 1 −1
1 3 −1
1 −1 0
0 3 14 3
det ( A()
1 ) = 14 0 3 = −( − 1)
+ 0 =−9 −26 =− 35
3 −1 4 −1
4 3 −1
3 1 0
14 3 5 3
det ( A ( 2)
) = 5 14 3 = 3 − + 0 =− 78 + 8 =− 70
4 −1 1 −1
1 4 −1
3 −1 1
0 14 5 14 5 0
det ( A) = 5 0 14 = 3 −( − 1)
+ =− 126 + 6 + 15 =− 105
3 4 1 4 1 3
1 3 4
x
z
det ( A()
1 ) −35 1
( A( ))
= = =
y
det ( A)
− 35
( A)
det ( A( 3)
) −105
= = = 3 Solution: ( 1, 2, 3 )
det ( A)
− 35
ii - The inverse.
⎡−9 8 15 ⎤
cof ( A)
=
⎢
1 3 10
⎥
⎢
− − −
⎥
⎢⎣−3 −9 5 ⎥⎦
9 1 3
⎡
35 35 35
⎤
−1 1
−8 3 9
A = adj( A)
=
⎢ ⎥
35 35 35
det ( A)
⎢ ⎥
⎢
−3 2 −1
⎣ ⎥
7 7 7 ⎦
9 1 3
⎡35 35 35
⎤⎡ 1⎤ ⎡1⎤
−1 −8 3 9
X = A b= ⎢ ⎥⎢
35 35 35
14
⎥ ⎢
2
⎥
⎢ ⎥⎢ ⎥
=
⎢ ⎥
⎢−
3 2 −1
⎣ ⎥⎢
7 7 7 ⎦⎣ 4⎥⎦ ⎢⎣3⎥⎦
det 2 −70
= = = 2
det −35
⎡−9 −−1
−3⎤
adj( A)
=
⎢
8 3 9
⎥
⎢
− −
⎥
⎢⎣
15 −10 5 ⎥⎦
Winter 2006 Martin Huard 6

Math 105
Semester Review - Solutions
b)
2x + y + 4z
= 8
2x
− y + =−16
3x
+ 5z
= 2
i – Cramer’s Rule
2 1 4
1 4 2 1
det ( A ) = 2 − 1 0 = 3 − 0 + = 12 − 4 = 8
−1 0 2 −1
3 0 5
8 1 4
1 4 8 1
det ( A () 1 ) =−16 − 1 0 = 2 − 0 + = 8 + 8 = 16
−1 0 −16 −1
2 0 5
2 8 4
8 4 2 4
det ( A ( 2)
) = 2 − 16 0 =−2 − 16 + 0 =− 64 + 32 =− 32
2 5 3 5
3 2 5
2 1 8
1 8 2 1
det ( A ( 3)
) = 2 −1 − 16 = 3 − 0 + 2 =−24 − 8 =−32
−1 −16 2 −1
3 0 2
x
z
det ( A()
1 ) 16
( A( ))
−
= = = 2
y
det ( A)
8
( A)
det ( A( 3)
) −32
= = =− 4 Solution: ( − 6, 4, 4)
det ( A)
8
det 2 32
= = =− 4
det 8
ii - The inverse.
⎡−5 −10 3 ⎤
⎡ −5 −5 4 ⎤
cof ( A
⎢
) = −5 −2 3
⎥
⎢
⎥
adj( A)
=
⎢
10 2 8
⎥
⎢
− −
⎥
⎣⎢
4 8 −4⎦⎥
⎢⎣
3 3 −4⎥⎦
5 5 −1
⎡
8 8 2
⎤
−1 1
5 1
A = adj( A)
=
⎢
4 4
1
⎥
det ( A)
⎢
−
⎥
⎢
−3 −3 1
⎣ ⎥
8 8 2 ⎦
5 5 −1
⎡
8 8 2
⎤⎡ 8 ⎤ ⎡−6⎤
−1 5 1
X = A b= ⎢
4 4
1
⎥⎢
16
⎥ ⎢
4
⎥
⎢
−
⎥⎢
−
⎥
=
⎢ ⎥
⎢−
3 −3 1
⎣ ⎥⎢
8 8 2 ⎦⎣ 2 ⎥⎦ ⎢⎣ 4 ⎥⎦
Winter 2006 Martin Huard 7

Math 105
Semester Review - Solutions
⎡3
−1⎤
8. Consider the matrix A = ⎢ ⎥ .
⎣5
4 ⎦
a) Find the inverse of A using the adjoint.
⎡
( ) 4 − 5 ⎤ ⎡
cof A
adj( A)
4 1 ⎤
= ⎢ =
1 3
⎥ ⎢
⎣ ⎦ ⎣ − 5 3
⎥
⎦
4 1
−1
1
⎡ 17 17 ⎤
A = adj( A)
=
−5 3
det ( A)
⎢ ⎥
⎣ 17 17 ⎦
−1
b) Express A as a product of elementary matrices.
−1 1
1
⎡3 −1 1 0⎤
⎡1 1
3 3
0⎤
⎡ 3
0⎤
⎢
R1 →
3
R1
5 4 0 1 ⎥ ⎢
5 4 0 1 ⎥ E1
= ⎢
⎣ ⎦ ⎣ ⎦
0 1 ⎥
⎣ ⎦
−1 1
⎡1 3 3
0⎤
⎡ 1 0⎤
R2 →R2 −5R1⎢ 17 −5
0
3 3
1
⎥ E =
⎣
⎦
2 ⎢
−5 1 ⎥
⎣ ⎦
−1 1
⎡1 3
3 3
0⎤
⎡1 0⎤
R2 →
17
R2⎢ −5 3
0 1
⎥ E =
⎣
17 17 ⎦
3 ⎢
0 ⎥
⎣
3
17 ⎦
4 1
1
⎡1 0
1
17 17 ⎤ ⎡1
3⎤
R1 → R1+ 3
R2⎢ −5 3
0 1
⎥ E4
= ⎢
⎣
17 17 ⎦
0 1 ⎥
⎣ ⎦
−1
A = E E E E
4 3 2 1
4 1 1 1
⎡17 17 ⎤ ⎡1 3⎤⎡1 0⎤⎡
1 0⎤⎡3
0⎤ ⎢−5 3 ⎥ = ⎢ 3
0
17 17
0 1
⎥⎢ ⎥⎢
17
5 1
⎥⎢
⎣ ⎦ ⎣ ⎦⎣
⎦⎣−
⎦⎣0 1 ⎥⎦
c) Express A as a product of elementary matrices.
−1 −1 −1 −1
A=
E E E E
1 2 3 4
−1
⎡3 −1⎤ ⎡3 0⎤⎡1 0⎤⎡1 0⎤⎡1
3 ⎤
⎢ 17
5 4
⎥ = ⎢
0 1
⎥⎢
5 1
⎥⎢ 0
⎥⎢
3
0 1
⎥
⎣ ⎦ ⎣ ⎦⎣ ⎦⎣
⎦⎣ ⎦
⎡2
0 1⎤
9. Consider the matrix A =
⎢ ⎥
⎢
4 1 2
⎥
.
⎢⎣
3 3 0⎥⎦
a) Find the inverse of A using the adjoint.
⎡−6 6 9 ⎤ ⎡−6 3 −1⎤ cof ( A) =
⎢
3 −3 − 6
⎥
adj( A)
=
⎢
6 −3 0 ⎥⎥⎥⎦
⎢ ⎥ ⎢
⎢⎣−1 0 2 ⎥⎦ ⎢⎣
9 −6 2
2 0 1
1 2 4 1
det ( A ) = 4 1 2 = 2 − 0 + =− 12 + 9 =− 3
3 0 3 3
3 3 0
det ( A ) = 17
Winter 2006 Martin Huard 8

Math 105
Semester Review - Solutions
1
⎡ 2 −1
3
⎤
−1
1
A = adj( A)
=
⎢
2 1 0
⎥
det ( A)
⎢
− ⎥
⎢
−2
⎣−3 2 ⎥
3 ⎦
−1
b) Express A and A as a product of elementary matrices.
1 1
1
⎡2 0 1 1 0 0⎤
⎡1 0
2 2
0 0⎤
⎡2
0 0⎤
⎢
1
4 1 2 0 1 0
⎥
R1 2
R
⎢
1
4 1 2 0 1 0
⎥
⎢ ⎥
→ ⎢ ⎥
E1
=
⎢
0 1 0
⎥
⎢ ⎥
⎢⎣3 3 0 0 0 1⎥⎦ ⎢⎣ 3 3 0 0 0 1⎥⎦
⎢⎣0 0 1⎥⎦
1 1
⎡1 0
2 2
0 0⎤
⎡ 1 0 0⎤
R2 →R2 −4R1⎢ 0 1 0 2 1 0
⎥
R3 R3 3R ⎢
−
⎥
E2
=
⎢
5 1 0
⎥
→ −
⎢
− ⎥
1
⎢
−3 −3
⎣0 3
2 2
0 1⎥⎦
⎢⎣
0 0 1⎥⎦
⎡ 1 0 0⎤
E3
=
⎢
0 1 0
⎥
⎢⎢ ⎥
⎣ − 3 0 1 ⎥ ⎦
1 1
⎡1 0
2 2
0 0⎤
⎡1 0 0⎤
R3 R3 3R
⎢
2
0 1 0 2 1 0
⎥
→ −
⎢
−
⎥
E4
=
⎢
0 1 0
⎥
⎢ ⎥
⎢
−3 9
⎣0 0
2 2
−3 1⎥⎦
⎢⎣0 −3 1⎥⎦
1 1
⎡1 0
2 2
0 0⎤
⎡1 0 0⎤
−2
R3 3
R
⎢
3
0 1 0 2 1 0
⎥
→
⎢
−
⎥
E5
=
⎢
0 1 0
⎥
⎢ ⎥
⎢
−2
⎣0 0 1 −3 2 ⎥
−2
3 ⎦
⎢⎣0 0 ⎥
3 ⎦
1 1
−1
⎡1 0
2
2 −1
3
⎤ ⎡1 0
2 ⎤
1
R1 R1 2
R
⎢
3
0 1 0 2 1 0
⎥
→ −
⎢
−
⎥
E6
=
⎢
0 1 0
⎥
⎢ ⎥
⎢
−2
⎣0 0 1 −3 2 ⎥
3 ⎦ ⎢⎣0 0 1⎥⎦
−1
A = E E E E E E
6 5 4 3 2 1
1 −1 1
⎡ 2 −1 3 ⎤ ⎡1 0
2 ⎤⎡1 0 0⎤⎡1 0 0⎤⎡ 1 0 0⎤⎡ 1 0 0⎤⎡
2
0 0⎤
⎢
− 2 1 0
⎥
=
⎢
0 1 0
⎥⎢ 0 1 0
⎥⎢ 0 1 0
⎥⎢
0 1 0
⎥⎢
−5 1 0
⎥⎢
0 1 0
⎥
⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢
⎥
⎢
−2 −2
⎣−3 2 ⎥
3 ⎦ ⎢⎣0 0 1⎥⎦⎢⎣0 0 ⎥
3 ⎦⎢⎣0 −3 1⎥⎢ ⎦⎣−3 0 1⎥⎢ ⎦⎣ 0 0 1⎥⎦⎢⎣0 0 1⎥⎦
−1 −1 −1 −1 −1 −1
A=
E E E E E E
1 2 3 4 5 6
1
⎡2 0 1⎤ ⎡2 0 0⎤⎡1 0 0⎤⎡1 0 0⎤⎡1 0 0⎤⎡1 0 0⎤⎡1 0
2 ⎤
⎢
4 1 2 ⎥ ⎢
0 1 0 ⎥⎢
5 1 0 ⎥⎢
0 1 0 ⎥⎢
0 1 0 ⎥⎢
0 1 0 ⎥⎢
0 1 0 ⎥
⎢ ⎥
=
⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥
⎢
−3
⎣3 3 0 ⎥ ⎦ ⎢ ⎣0 0 1 ⎥⎢ ⎦⎣0 0 1 ⎥⎢ ⎦⎣3 0 1 ⎥⎢ ⎦⎣0 3 1 ⎥ ⎦⎣ ⎢ 0 0
2 ⎥⎢ ⎦⎣0 0 1 ⎥ ⎦
Winter 2006 Martin Huard 9

Math 105
Semester Review - Solutions
10. A coin bank has only nickels, dimes and quarters. The value of the coins is $2. There are
twice as many nickles as dimes and one more dime than quarters. Find the number of each
coin in the bank.
5x+ 10y+ 25z
= 200
x− 2y
= 0
y− z = 1
⎡5 10 25 200⎤ ⎡5 10 25 200 ⎤
⎢
1 2 0 0
⎥
R2 5R2 R
⎢
⎢
−
⎥
→ −
1⎢
0 −20 −25 −200
⎥
⎥
⎢⎣0 1 −1 1 ⎥⎦ ⎢⎣0 1 −1 1 ⎥⎦
1
⎡5 10 25 200 ⎤ R1 →
5
R1
⎡1 2 5 40⎤
R3 → 20R3+ R
⎢
2
0 −20 −25 −200
⎥ −1
5
⎢
⎥
R2 →
20
R
⎢
2
0 1
4
10
⎥
⎢
⎥
⎢⎣
0 0 −45 −180⎥
−1
⎦ R 0 0 1 4
3 45
R ⎢
⎥
→
3 ⎣ ⎦
z = 4, y = 5, x = 10
Thus there are 10 nickels, 5 dimes and 4 quarters in the bank.
11. Suppose a man has three modes of transportation to work: he can walk, drive his car, or take
the bus. If he walks one day, then he will either take the car the next day with a probability
of 2 or take the bus with a probability of 1 3 3
. If he drove one day, then he will walk the next
day with a probability of 1 and take the bus with a probability of 1 2 2
. If he took the bus one
day, then he will walk the next day with a probability of 1 , drive with a probability of 1 and
3 3
take the bus with a probability of 1 3 .
a) Find the transition matrix.
A
⎡0
⎤
1 1
2 3
2 1
=
⎢
3
0
⎥
⎢ 3 ⎥
⎢
1 1 1⎥
3 2 3
⎣ ⎦
b) If he drives on Monday, find the probability that he will walk, drive or take the bus on
Wednesday.
1 1 1
1 1 1 1
⎡0 2 3⎤⎡0⎤
⎡ 2⎤
⎡0
2 3⎤⎡2⎤
⎡6⎤
2 1
X
⎢
1
=
3
0
⎥⎢ 3
1
⎥
=
⎢
0
⎥
2 1
1
⎢ ⎥⎢ ⎥ ⎢ ⎥
X
2
=
⎢
3
0
⎥⎢ 3
0
⎥
=
⎢ ⎥
⎢ ⎥⎢ ⎥ ⎢2⎥
⎢1 1 1⎥ 1
⎣ ⎢
3 2 3⎦⎣0⎥⎦ ⎢⎣ ⎥
1 1 1 1 1
2⎦
⎢⎣ ⎥
3 2 3⎦⎢⎣ ⎥ ⎢
2⎦
⎣ ⎥
3⎦
Walk: 1 Drive: 1 Bus: 1 6 2 3
c) In the long run, how often will he walk, drive and take the bus
I − A X =
( ) 0
−1 −1 −1 −1
⎡1 2 3
0⎤ ⎡1 2 3
0⎤
⎢ R
2 1 2
3R2 2R
−
−
1 5
3
1
3
0
⎥ → + ⎢
−
0 2
3
0
⎥
⎢ ⎥R ⎥
1 1 2 3
→ 3R3+
R ⎢
− −
1
⎢
5
3 2 3
0⎥
⎣ ⎦ ⎢⎣0 −2
3
0⎥⎦
−1 −1
⎡1 2 3
0⎤
5
R3 R3 R
⎢
−
2
0 2
3
0
⎥
→ +
⎢
⎥
⎢⎣0 0 0 0⎥⎦
Winter 2006 Martin Huard 10

Math 105
Semester Review - Solutions
−1 −1
⎡1 2 3
0⎤
1
5
R2 2
R
⎢
−
2
0 1
6
0
⎥
→ ⎢ ⎥
⎢⎣0 0 0 0⎥⎦
t+ t+ t = 1
3 5
4 6
Walk: 9 31
t =
12
31
Drive:
10
31
z = t
y =
x =
5
6
3
4
Thus X = ( 9 , 10 ,
12 )
Bus:
12
31
t
t
31 31 31
12. Find the equation of the parabola passing through the points 1, 4 ,
C ( −3,32).
2
y = a+ bx+
cx
4= 1+ x+
x
12 = 1+ 2x
+ 4x
2
2
A ( ) ( 2,12)
B and
2
32 = 1− 3x
+ 9x
⎡1 1 1 4 ⎤ 1 1 1 4
R2 2 1
1 2 4 12 → ⎡ ⎤ ⎡1 1 1 4 ⎤
⎢ ⎥ R −R
⎢
0 1 3 8 ⎥⎥⎥⎦
⎢ ⎥R3 R3 R ⎢
R3 R3 4R
⎢
1
0 1 3 8
⎥
→ −
→ +
⎢
⎥
1
⎢⎣1 −3 9 32⎥
⎦ ⎢⎣0 −4 8 28
⎢⎣0 0 20 60⎥⎦
⎡1 1 1 4⎤
c = 3
1
R3 20
R
⎢
3
0 1 3 8
⎥
→
b = −1
⎢ ⎥
⎢⎣0 0 1 3⎥⎦
a = 2
2
Thus the parabola is y = 2− x+ 3x
.
13. Let ABC be a triangle and E a point on the segment BC dividing it in a ration of 1 to 3. Let
D be the midpoint of AC. Join A to E and B to D, and let P be the point on intersection of
the segments AE and B D. In what ration does P divide AE and BD
Let AP=
kAG and FP = lFE .
B
We have
1
1
E
BE =
4
BC
1
AD =
2
AC
P 3
k
l
A 1 D 1
Let us express AP
C
in terms of AB and AC in two different ways.
Winter 2006 Martin Huard 11

Math 105
Semester Review - Solutions
AP = k AE
= k( AB+
BE)
1
= kAB+
4
kBC
k
= kAB+ 4
BA+
AC
= AB + AC
3k
k
4 4
( )
By the basis theorem, we have the equations
3k
4
l
Combining these equations, we have
1 1 1 3k
k = −
4 2 2
k =
5 1
8 2
k =
4
5
4
AP = AD + DP
1
=
2
AC + lDB
1
=
2
AC + l DA + AB
1 1
=
2
AC + l
2
AC + AB
= − l AC+
lAB
1 1
( )
2 2
1 1 1
=
4k
=
2
−
2
( )
−
( )
4
3
thus k =
5
and l =
5
.
Hence, P divides AE in a ration of 4 to 1 and divides DB in a ration of 3 to 2.
14. Let ABC be a triangle and M, N and P the midpoints of AB, BC and CA respectively. Prove
that if O is any point (inside or outside the triangle) then
OA + OB + OC = OM + ON + OP
OA + OB + OC = ( OM + MA) + ( ON + NB) + ( OP + PC)
= OM + ON + OP + MA + NB + PC
1 1 1
= OM + ON + OP +
2
BA +
2CB +
2
AC
1
= OM + ON + OP +
2 ( AC + CB + BA)
1
OM ON OP 0
= + + +
2
= OM + ON + OP
15. A Boeing 737 aircraft maintains a constant airspeed of 500 miles per hour in the direction
due south. The velocity of the jet stream is 80 miles per hour in a northeasterly direction
(N45 o E). Find the actual speed and direction of the aircraft relative to the ground.
2 2 2
J
A+ J = A + J + 2 A J cosθ
φ
45 o
l
= + − ⋅ ⋅
2 2
500 80 2 500 80cos 45
= 199831
A+ J = 447
sin 45 sinφ
A
= φ = 7.3°
447 80
Thus the plane has a speed of 447 miles per hour in a S7.3 o E direction.
°
Winter 2006 Martin Huard 12

Math 105
Semester Review - Solutions
16. A river flows from west to east. There are ferry terminals on the north and south shores, the
north dock being 15° east of north (i.e.N15°E) from the south dock. The ferry captain knows
from experience that in order to reach the dock on the north shore from the south shore dock,
she has to steer N30°W.
a) If the ferry travels at 12 km/h, what is the speed of the current
sin 45° sin 75°
=
B
T
60 W 12
o 75 o
W = 8.78
45 o Speed of current is 8.78 km/h
W
b) If the trip takes ¼ hour, how far apart are the dock
sin 60° sin 75°
=
T = 10.76 km/h
T 12
Thus the distance between the docks is 1 4 10.76
= 2.69 km
17. Consider the vectors u = ( − 2,5,5)
, v = ( 1,
−1,2
) and = ( 5,
−1,2
)
w .
a) Evaluate 2u
− 3v
2u− 3v
= 2 −2,5,5 −3 1, − 1,2 = −7,13,4
( ) ( ) ( )
b) Find the vector projection of u onto w .
uw i −5 ⎛−5
1 −1⎞
proj
w
u = w= ( 5, − 1,2 ) = ⎜ , , ⎟
ww i 30 ⎝ 6 6 3 ⎠
c) Find the angle between u and w .
uw i −5
cosθ
= =
θ ≈ 97.1°
u w 54 30
d) Find the area of the triangle having sides u and v .
i j k
1
1 1
3 35
A= 2
u× v =
2
− 2 5 5 =
2 ( 15,9, − 3)
=
2
1 −1 2
e) Find the volume of the parallelepiped having sides u , v and w .
−2 5 5
−1 2 1 2 1 −1
V = ui( v× w)
= 1 − 1 2 = −2 − 5 + 5 = 0+ 40+ 20 = 60
−1 2 5 2 5 −1
5 −1 2
18. Consider the points A(2,-1,3), and B(3,-1,5), C(-2,2,3) and D(-1,0,5).
a) Find the vector projection of AB onto AC .
ABiAC
( 1, 0, 2) i( −4, 3, 0)
−4 ⎛16 −12
⎞
proj
AB = AC = − 4,3,0 = − 4,3,0 = , ,0
AC
⎜ ⎟
ACiAC
−4,3,0 i −4,3,0 25 ⎝25 25 ⎠
( ) ( ) ( ) ( )
Winter 2006 Martin Huard 13

Math 105
Semester Review - Solutions
b) Find the angle between
AB and AC .
ABiAC
−4
cosθ
= =
θ ≈ 111
AB AC 5 25
c) Find the volume of the tetrahedron ABCD.
1 0 2
3 0 −4 3
1 1 1 1
8
V =
6
ABi ( AC× AD)
=
6
− 4 3 0 =
6
1 − 0+ 2 =
6
6+
10 =
1 2 −3 1 3
−3 1 2
d) Find the equation of the line (in parametric form) passing through D and parallel to
AB .
⎧x
= − 1+
t
⎪
u = AB=
(1, 0, 2)
l: ⎨ y = 0
⎪ ⎩z
= 5 + 2t
e) Find the equation of the plane (in general form) parallel to AB and AC , and passing
through D.
i j k
n = AB× AC = 1 0 2 = −6, −8,3
−4 3 0
( ) ( ) ( )
−6x− 8x+ 3z
=−6 −1 − 8 0 + 3 5 = 21
π :6x+ 8y− 3z
=− 21
(
f) Find the equation of the plane perpendicular to AC and passing through D.
n = AC = −4,3,0
− 4x+ 3y =−4 − 1 + 3 0 + 0 5 = 4
( )
)
( ) ( ) ( )
π :4x− 3y
=− 4
x −1
2y
+ 1
19. Consider the plane π : 2x
+ y − 5z
+ 1 = 0 and the line L : = = 3 − z
3 4
a) Find the equation of the line (in symmetric form) perpendicular to π and passing
through P(1,1,-3).
x− 1 z+
3
u = (2,1, −5)
= y − 1 =
2 − 5
b) Find the equation of the line (in parametric form) parallel to L and passing through
P(1,1,-3).
⎧x
= 1+
3t
⎪
u = (3, 2, −1)
⎨y
= 1 + 2t
⎪
⎩z
= − 3 − t
Winter 2006 Martin Huard 14

Math 105
Semester Review - Solutions
c) Find the distance between the line L and the line found in (b).
Since the lines are parallel, we have
−1
P1 ( 1,
2
, 3)
PP ( 3
1 2=
0,
2
, −6)
P 1,1, −3
2
( )
i j k
3
0
2
−6
PP
21 9
( )
3
1 2× u
−
3 2 −1 2
, −18,
2 2
202 3
d = = = = =
u 14 14 14 14
d) Find the intersection, if possible, of the plane π and the line L.
−1
⎧x
= 1+
3t
21 ( + 3t) + ( 2
+ 2t) −53 ( − t)
+ 1=
0
⎪
−1
27
L:
= ⎨y
=
2
+ 2t
13t
=
2
⎪
⎩z
= 3 −
27
t
t =
26
27 − 1 27 27 101 37 53
1+ 2 , + 2 ,3 − = , ,
Intersection: ( ) ( )
26 2 26 26 26 26 26
e) Find the equation of the plane π in vector form.
z = t
y = s
x =− − s+
t
1 1 5
2 2 2
707
1 1
5
( xyz) = ( ) + s( ) + t( )
− −
π
2 2 2
: , , ,0,0 ,1,0 ,0,1
f) Find the equation of the plane π
2
(in general form) perpendicular to L and passing
through P(1,1,-3).
n
2
= ( 3, 2, −1)
3x+ 2y− z = 3( 1) + 2( 1) −( − 3)
= 8
π :3x+ 2y− z = 8
2
g) Find the angle between the plane π and the plane π
2
found in (f).
n 1 2
13
cos θ = in
n1 n
=
θ ≈ 50.6°
2 30 14
h) Find the point Q on the plane π that is closest to the point P(1,1,-3).
PRin
R ( 0, −1,0
)
PQ = proj
n
PR = n
nn i
PR = ( −1, −2, 3)
−19
( x−1, y− 1, z+ 3) = ( 2,1, −5)
n = ( 2,1, −5)
30
−19
−
xyz , , = 2,1, − 5 + 1,1, − 3 = , ,
30
Q − 4 , 11 ,
1
( )
15 30 6
4 11 1
( ) ( ) ( ) ( )
15 30 6
Winter 2006 Martin Huard 15

Math 105
Semester Review - Solutions
i) Find the point Q on the line L that is closest to the point P(1,1,-3).
RPiu
−1
R( 1,
2
, 3)
RQ = proj
u
RP = u
uu i
3
RP = ( 0,
2
, −6)
9
( x−1, y− 1, z+ 3) = ( 3, 2, −1)
u = ( 3, 2, −1)
14
9
41 11
xyz , , = 3,2, − 1 + 1,1, − 3 = , ,
14
41 11 33
Q , ,
( )
14 14 14
33
( ) ( ) ( ) ( )
14 14 14
j) Find the distance from the point P(1,1,-3) to the plane π .
R ( 0, −1,0
)
PRin
−19 19 30
d = = =
PR = ( −1, −2, 3)
n 30 30
k) Find the distance from the point P(1,1,-3) to the line L.
i j k
−1
−3 −21
9
R( 1,
2
, 3)
PR× u = 0
2
6 = ( 2
,18,
2 )
−3
PR = ( 0,
2
,6)
3 2 −1
PR×
u −21 9
( )
3
2
,18,
2 2
202 3 707
d = = = =
u 14 14 14
l) Find the equation of the plane (if possible), in general form containing the lines L and
x + 5 y − 2
L2 : = = −z
.
3 2
u = ( 3, 2, −1)
∴L
L2
u = 3, 2, −1
2
( )
1
−
1 5
P( 1, 1
2
2
2
, 3)
∈ L, Since
+ −
− ≠ ≠− 3 then P∉
L
3 2
2
Thus L and L 2 are parallel and distinct.
i j k
P2
( −5, 2,0)
−
n = u× PP = 3 2 − 1 = ,15,
5
PP2 = ( −6, 2
, −3)
5
−6 −3
( 7 39 )
−7 39 −7
2 2 2
( ) ( )
2 2 2
x+ 15y− z = − 5 + 15 2 − 0 = 95 2
Plane: 7x−30y− 39z
=−95
m) Find the equation of the plane (if possible), in general form containing the lines L and
x 4
5
+ 10
L
3
: + = y + =
z .
2
2
3
u = ( 3, 2, −1)
∴L L 3
u = 2,1,3
3
To find
( )
L∩
L 3
, we have
2
Winter 2006 Martin Huard 16

Math 105
Semester Review - Solutions
−1
:( , , ) = ( 1,
2
,3) + ( 3,2, −1)
−5
3
:( , , ) = ( −4, 2
, − 10) + ( 2,1,3)
1
5
( 1, − 2
,3) + t( 3, 2, − 1) = ( − −
4,
2
, − 10) + s( 2,1,3)
( 3, 2, − 1) + s( −2, −1, − 3) = ( −5, −2, −13)
L x y z t
L x y z s
t
⎡ 3 −2 −5 ⎤ 3 2 5
R2 3
2
2
1
2 1 2 → R R
⎡ − − ⎤ ⎡3 −2 −5⎤
⎢ ⎥ − ⎢
0 1 4 ⎥⎥⎥⎦
⎢
− −
⎥R3 → 3R3+
R ⎢
R3 11R3 R
⎢
2
0 1 4
⎥
→ +
⎢ ⎥
1
⎢⎣−1 −3 −13⎥
⎦ ⎢⎣0 −11 −44
⎢⎣0 0 0 ⎥⎦
−2
−5
⎡1
3 3
⎤
1
R1 3
R
⎢
1
0 1 4
⎥ s = 4
→ ⎢ ⎥ t = 1
⎢⎣
0 0 0⎥⎦
−1
3
xyz , , = 1, ,3 + 3,2, − 1 = 4, ,2
( ) ( ) ( ) ( )
2 2
3
Thus L∩ L = P( 4, , 2), so the lines are nonparallel and intersecting.
3 2
i j k
n = u × u
3
= 3 2 − 1 = 7, −11, −1
2 1 3
19
Plane: 7x−11y− z =
2
( )
20. Is the set V a vector space with the following operations
2
a)
u , u ⊕ v , v = u −v , u −v
b)
( ) ( )
9
7x−11y− z = 7 4 −11 − 2=
V = ( 1 2) ( 1 2) ( 1 1 2 2)
k
( u1, u2) = ( ku1,
ku2)
No, Axiom 2 fails: Counter example with u = ( 1, 2)
and v = ( 3, 4)
( 1,2) ⊕ ( 3,4) = ( 1−3,2− 4) = ( −2, −2)
( 3,4) ⊕ ( 1,2) = ( 3−1,4− 2) = ( 2,2)
Thus ( 1,2) ⊕( 3,4) ≠( 3,4) ⊕ ( 1,2)
, that is u⊕ v ≠v⊕u
.
2
V = ( u , u ) ⊕ ( v , v ) = ( u + v + 1, u + v + 1)
1 2 1 2 1 1 2 2
( ) (
( u , u ) v = ( v , v )
k u1, u2 = k+ ku1− 1, k+ ku2
−1
2
Yes. Let u =
1 2
,
1 2
and w=
( w1,
w2)
be in .
2
1. u
⊕ v = ( u1+ v1+ 1, u2 + v2
+ 1)
∈
u ⊕ v = u + v + 1, u + v + 1 = v + u + 1, v + u + 1 = v ⊕u
2. ( ) ( )
1 1 2 2 1 1 2 2
)
3 1
2 2
Winter 2006 Martin Huard 17

Math 105
Semester Review - Solutions
3. u⊕( v⊕ w) = ( u1, u2) ⊕ ( v1+ w1+ 1, v2 + w2
+ 1)
= ( u1+ ( v1+ w1+ 1) + 1, u2 + ( v2 + w2
+ 1)
+ 1)
= (( u1+ v1+ 1) + w1+ 1, ( u2 + v2 + 1)
+ w2
+ 1)
= ( u1+ v1+ 1, u2 + v2 + 1 ) ⊕( w1,
w2)
= ( u⊕v)
⊕w
4. 0= 0 ( u1, u2) = ( −1, −1)
u⊕ 0 = ( u ) ( ) ( ) ( ) u
1, u2 ⊕ −1, − 1 = u1− 1+ 1, u2 − 1+ 1 = u1,
u2
=
5. − u = ( − 1 ) ( u1, u2) = ( −1−u1−1, −1−u2 − 1) = ( −2 −u1, −2−u1)
u⊕− u = ( u1, u2) ⊕( −2 −u1, −2−u1)
= ( u −2− u + 1, u −2− u + 1) = ( −1, − 1)
= 0
1 1 2 2
k
u1, u2 = k+ ku1− 1, k+ ku2
−1
∈
k+ l u = k+ l u , u = k+ l + k+ l u − 1, k+ 1 + k+ l u −1
6. ( ) ( )
2
( )
)
7. ( ) ( ) ( ) ( ) ( ) ( ) ( )
= (( k+ ku1− ) + ( l+ lu1− ) + ( k+ ku2 − ) + ( l+ lu2
− ) +
= ( k+ ku1− 1, k+ ku2 −1) ⊕ ( l+ lu1− 1, l+ lu2
−1)
= ( ku) ⊕( lu)
8. k( u⊕ v) = k( u1+ v1+ 1, u2 + v2
+ 1)
= ( k+ k( u1+ v1+ 1) − 1, l+ l( u2 + v2
+ 1)
−1)
(( k ku1 1) ( k kv1 1) 1, ( k ku2 1) ( k kv2
1)
= ( k+ ku1− 1, k+ ku2 −1) ⊕ ( k+ kv1− 1, k+ kv2
−1)
= ( ku) ⊕( kv)
9. ( kl ) u = ( kl + klu1− 1, kl + klu2
−1)
= ( k+ k( l+ lu1−1) − 1, k+ k( l+ lu2
−1)
−1)
= k
( l+ lu1− 1, l+ lu2
−1)
= k( lu)
10. 1u = 1 ( u , u ) = ( 1+ u − 1,1+ u − 1 ) = ( u , u ) = u
1 2 1 2
1 1 1, 1 1 1
= + − + + − + + − + + − +
1 2 1 2 1 2
1)
Winter 2006 Martin Huard 18

Math 105
Semester Review - Solutions
21. Is the set W a subspace of V Support your answer.
⎧ ⎡ a a+
b⎤
⎫
a) W = ⎨⎢
: a,
b∈
⎬
a−
b b
⎥ V = M2,2
⎩⎣
⎦ ⎭
⎡0 0⎤ Yes W is nonempty since ⎢ ∈W
0 0
⎥ .
⎣ ⎦
⎡ a a+
b⎤
Let A = ⎢
a−
b b ⎥
⎣ ⎦ and ⎡ r r+
s⎤
B = ⎢
∈W
r−
s s
⎥ .
⎣ ⎦
⎡ a+ r a+ b+ r+
s⎤
⎡ a+ r ( a+ r) + ( b+
s)
⎤
1. A + B = ⎢
a b r s b s
⎥ = ⎢
∈
( a r) ( b s)
⎥ W
⎣ − + − + ⎦ ⎣ + − + b+
s ⎦
2.
( + )
⎡ ka k a b ⎤ ⎡ ka ka + kb⎤
kA = ⎢ ⎥ =
W
k( a−
b)
kb
⎢
ka − kb kb
⎥∈
⎣
⎦ ⎣
⎦
b) W { A: A is nilpotent, A M2,2}
2
= ∈ V = M2,2
(A is nilpotent if A = 0 )
0 1
No. Let A ⎡ ⎤ 0 0
= ⎢⎣ and
0 0
⎥ B = ⎡ ⎤
⎢ ⎦ 1 0 ⎥
⎣ ⎦ . Then 2
2
A = 0 and B = 0 , so A,
B∈ V but
0 1
A B ⎡ ⎤
+ = ⎢ ∉V
since
1 0
⎥ ( ) 2 ⎡1 0⎤
A+ B = ⎢ ≠0
⎣ ⎦
0 1
⎥
⎣ ⎦
W = ax 3 −b: a,
b∈ V = P3
c) { }
Yes W is nonempty since 0∈ W .
3
p x = ax −b and q( x) = cx −d∈
W
Let ( )
3
p x + q x = ax − b+ cx − d = a+ c x − b+ d ∈ W
3 3 3
1. ( ) ( ) ( ) ( )
3 3
2. ( ) ( )
{ : 1 2 , ( ) 2 }
kp x = k ax − b = akx −bk ∈W
d) W = p( x) p( ) = p( ) p x ∈ P V = P2
Yes V is nonempty since p( x) = 0∈ V since p( 1) = 0= p(
2).
Let p( x) , q( x)
∈ V . Then p( 1) = p( 2 ) and q( 1) = q( 2)
.
1. ( p + q)( x)
∈ V since ( p+ q)( 1) = p( 1) + q( 1) = p( 2) + q( 2) = ( p+ q)( 2 )
2. ( kp)( x)
∈ V since ( kp)( 1) = kp ( 1) = kp ( 2) = ( kp)( 2 )
3
e) W = ( x, y, z)
:2x− 3y+ z = 0, x, y,
z∈
V =
{ }
Yes W is nonempty since 0∈W
Let u = x , y , z ∈W
and v = x , y , z ∈W
.
( )
1 1 1
( )
2 2 2
Then 2x1− 3y1+ z1
=0 and 2x2 − 3y2 + z2
= 0
u+ v = x + x , y + y , z + z ∈W
1. ( 1 2 1 2 1 2)
since 2( x + x ) − 3( y + y ) + ( z + z ) = ( 2x − 3y + z ) + ( 2x − 3y + z )
1 2 1 2 1 2 1 1 1 2 2 2
= 0+ 0=
0
Winter 2006 Martin Huard 19

Math 105
Semester Review - Solutions
2. ku = ( kx1, ky1,
kz1)
∈W
f) ( )
No. If = ( 4,0,0)
∈
then ku ( 0,0,0)
{ , , :2 3 8 0, , , }
2 kx − 3 ky + kz = k 2x − 3y + z = k0=
0
since ( ) ( ) ( )
W= xyz x− y+ z− = xyz∈ V
u W and k = 0 ,
= ∉W
since
Geometrically, W is the plane 2x− 3y+ z = 8.
{ }
g) W = ( a,3a− 4, a+ 1 ):
a∈
No. If u = ( 1, −1, 2)
∈W
then ku ( 0,0,0)
and
= ∉W
3
V =
k = 0 ,
since if
1 1 1 1 1 1
3
=
20 ⋅ −30 ⋅ + 0−8≠
0
( 0,0,0 ) ( a,3a 4, a 1)
= − + , then
4
we have a = 0, a= and a =− 1 which is impossible.
3
22. For each of the subsets W in ,
i) Find a basis for W.
ii) Find the dimension of W
iii) Give a geometrical interpretation of W.
W = 2 a+ b, a, a−2 b : a,
b∈
{ }
a) ( )
Since ( 2 a+ b, a, a− 2b) = a( 2,1,1) + b( 1,0, − 2)
Then if B = ( 2,1,1 ), ( 1, 0. − 2 we have W span ( B)
{ )}
3
= .
Since B is linearly independent (the two vectors are not multiples of each other)
then B is a basis for W.
dim W = 2
( )
i j k
n = × − = = − −
( 2,1,1) ( 1,0, 2) 2 1 1 ( 2,5, 1)
1 0 −2
Ergo, W is the plane π :2x− 5y+ z = 0.
{ }
b) W= ( a− 2b+ 3 c,3a+ b+ 2 ca , + 4b−3 c)
: abc , , ∈
Since ( a b c a b c a b c) a( ) b( ) c(
then if B = {( 1,3,1 ),( −2,1, 4 ),( 3, 2, − 3)
} we have W = span ( B)
c ( 1,3,1) + c ( − 2,1, 4) + c ( 3, 2, − 3) = ( 0, 0, 0)
If
− 2 + 3 ,3 + + 2 , + 4 − 3 = 1,3,1 + − 2,1, 4 + 3, 2, − 3)
1 2 3
⎡1 −2 3 0⎤ ⎡1 −2 3 0⎤
⎢ 2 2
3
1
3 1 2 0
⎥R →R − R ⎢
0 7 −7 0
⎥
⎢ ⎥R3 →R3−R
⎢
⎥
1
⎢1 4 −3 0⎥
⎣ ⎦ ⎢⎣0 6 −6 0⎥⎦
⎡1 −2 3 0⎤
⎡1 −2 3 0⎤
R3 →7R3−6R
⎢
1
0 7 −7 0
⎥ 1
⎢
⎥
R2 7
R
⎢
2
0 1 1 0
⎥
→
⎢
−
⎥
⎢⎣
0 0 0 0⎥⎦
⎢⎣
0 0 0 0⎥⎦
1, then 3, 2, − 3 = 1,3,1 − − 2,1, 4
t = ( ) ( ) ( )
c
c
c
3
2
1
= t
= t
= −t
Winter 2006 Martin Huard 20

Math 105
Semester Review - Solutions
{ }
By the +/- theorem, if B = ( 1, 3,1 ), ( − 2,1, 4)
, then span ( B ) span ( B)
W
w
= =W .
Since B
W
is linearly independent (the two vectors are not multiples of each other),
if forms a basis for W.
dim W = 2
( )
i j k
n = × − = = −
( 1, 3,1) ( 2,1, 4) 1 3 1 ( 11, 6, 7)
−2 1 4
Ergo, W is the plane π :11x− 6y+ 7z
= 0.
{ }
c) W = ( 2 a− b+ c, a+ b+ c,3a+ 2 c)
: a, b,
c∈
Since( 2 a− b+ c, a+ b+ c,3a+ 2c) = a( 2,1,3) + b( − 1,1,0 ) + c(
1,1,2 )
then if B = {( 2,1, 3 ), ( − 1,1, 0 ), ( 1,1, 2)
} we have W span ( B)
c ( 2,1, 3) + c ( − 1,1, 0) + c ( 1,1, 2) = ( 0, 0, 0)
1 2 3
= .
⎡2 −1 1 0⎤ ⎡2 −1 1 0⎤
⎢ 2
2
2 1
1 1 1 0
⎥R → R −R
⎢
0 3 1 0
⎥
⎢ ⎥R3 →2R3−3R
⎢
⎥
1
⎢3 0 2 0⎥
⎣ ⎦ ⎢⎣0 3 1 0⎥⎦
−1 1
⎡2 −1 1 0⎤
⎡1 2 2
0⎤
1
R3 →R3−R
⎢
2
0 3 1 0
⎥ R1 →
2
R1
⎢ 1
⎢
⎥
0 1
1
3
0
⎥
R2 3
R ⎢
⎥
2
⎢⎣
0 0 0 0⎥
→
⎦ ⎢ ⎣0 0 0 0 ⎥ ⎦
1, then 1,1, 2 = 2 2,1, 3 + 1 − 1,1, 0
If t = ( ) 3( ) 3( )
By the +/- theorem, if B = ( 2,1, 3 ), ( − 1,1, 0)
, then span ( B ) span ( B)
W
{ }
Since B
W
is linearly independent, if forms a basis for W.
w
c
c
c
3
= t
=
−1
2 3
=
−2
1 3
= =W .
23. Find a basis and the dimension of each of the following subspaces W,
⎧ ⎡ a a+
b⎤
⎫
a) W = ⎨⎢
: a,
∈ ⎬
a−
b b
⎥ b
⎩⎣
⎦ ⎭
⎡ a a+
b⎤ ⎡1 1⎤ ⎡ 0 1⎤
Since ⎢ a b ⎥ then
a b b
⎥ = ⎢ +
1 0
⎥ ⎢
⎣ − ⎦ ⎣ ⎦ ⎣−1 1⎦
1 1 0 1
if B = ⎧ ⎨ ⎡ ,
⎫
⎢ ⎤ ⎡ ⎤ we have
1 0
⎥ ⎢ ⎬
−1 1
⎥ W = span ( B)
.
⎩⎣ ⎦ ⎣ ⎦⎭
Since B is linearly independent (the two matrices are not multiples of each other)
dim W = 2
then B is a basis for W and ( )
t
t
Winter 2006 Martin Huard 21

Math 105
Semester Review - Solutions
b)
⎧ ⎡ a+ b+ 2c 2a+ 3b+
3c⎤
⎫
W = ⎨⎢
: a, , ∈ ⎬
2a+ 3b+ 3c − a+ b−4c
⎥ b c
⎩⎣
⎦ ⎭
⎡ a+ b+ 2c 2a+ 3b+
3c⎤ ⎡1 2 ⎤ ⎡1 3⎤
⎡2 3 ⎤
Since ⎢ a b c⎢
⎥ then
2a 3b 3c a b 4c
⎥ = ⎢
2 1
⎥+ ⎢
3 1
⎥+
⎣ + + − + − ⎦ ⎣ − ⎦ ⎣ ⎦ ⎣3 −4⎦
1 2 1 3 2 3 ⎤
if B = ⎨ ⎧ , ,
⎫
⎢ ⎡ ⎥ we have
2 1
⎥ ⎤ ⎢ ⎡ ⎤ ⎡
⎬
3 1
⎥ ⎢
W = span ( B ) .
⎩⎣ − ⎦ ⎣ ⎦ ⎣3 −4⎦⎭
⎡1 2 ⎤ ⎡1 3⎤ ⎡2 3 ⎤ ⎡0 0⎤
c1⎢ c2 c3
2 1
⎥+ ⎢
3 1
⎥+ ⎢ =
3 4
⎥ ⎢
0 0
⎥
⎣ − ⎦ ⎣ ⎦ ⎣ − ⎦ ⎣ ⎦
⎡ 1 1 2 0⎤ ⎡1 1 2 0⎤
⎢ 2 2
2
1
2 3 3 0
⎥R →R − R ⎢
0 1 −1 0
⎥
⎢ ⎥R3 →R3 −2R
⎢
⎥
1
⎢ 2 3 3 0⎥ ⎢0 1 −1 0⎥ ⎢ ⎥R4 → R4 + R1
⎢
−1 1 −4 0
⎥⎦
⎣ ⎦ ⎣0 2 −2 0
⎡1 1 2 0⎤
R3 R3 R
⎢
2
0 1 1 0
⎥ c3
= t
→ −
⎢
−
⎥ c2
= t
R4 → R4 −2R
⎢
2
0 0 0 0⎥
⎢
⎥ c1 =−3t
⎣0 0 0 0⎦
If t =1, then ⎡ 2 3 ⎤ 1 2 1 3
3
⎡ ⎤ ⎡ ⎤
⎧⎡1 2 ⎤ ⎡1 3⎤⎫
⎢ = − , so if
3 −4 ⎥ ⎢
2 −1 ⎥ ⎢
⎣ ⎦ ⎣ ⎦ ⎣3 1
⎥ B W
= ⎨⎢ , ⎬
⎦
2 −1 ⎥ ⎢
3 1
⎥ , then
⎩⎣ ⎦ ⎣ ⎦⎭
span B = span B = W .
by the +/- theorem ( ) ( )
W
Since B is linearly independent (the two matrices are not multiples of each other)
dim W = 2
then B is a basis for W and ( )
2
c) W = ax + ( b− a)
x+ b: a,
b∈
{ }
Since ax 2 + ( b − a) x + b = a ( x 2 − x) + b( x + 1)
then if B { x 2 1, x 1}
have W = span ( )
multiples of each other) then B is a basis for W and dim( W ) = 2
24. Find all values of t for which S is linearly independent.
S = 2,3,5 , −1, t, −1 , −1, −1,
t
= − + we
B . Since B is linearly independent (the two polynomials are not
{ }
a) ( ) ( ) ( )
c ( 2,3,5) + c ( −1, t, − 1) + c ( −1, − 1, t) = ( 0,0,0)
1 2 3
⎡2 −1 −1 0⎤ ⎡2 −1 −1 0⎤
⎢ R2 2R2 3R1
3 t 1 0
⎥ → − ⎢
⎢
−
⎥
0 2t
3 1
R3 2R3 5R
⎢
+ 0
⎥
→ −
⎥
1
⎢5 − 1 t 0⎥
⎣ ⎦ ⎢⎣0 3 2t+
5 0⎥⎦
Winter 2006 Martin Huard 22

Math 105
Semester Review - Solutions
⎡2 −1 −1 0⎤
R2 R
⎢
3
0 3 2t
5 0
⎥
↔
⎢
+
⎥
⎢⎣0 2t
+ 3 1 0⎥⎦
⎡2 −1 −1 0⎤
R3 3R3 ( 2t 3)
R
⎢
2
0 3 2t
5 0
⎥
→ − +
⎢
+
⎥
2
⎢⎣
0 0 −4t
−16t−12 0⎥⎦
1
R
1 1
1
→
2
R
− −
1 ⎡1 2 2
0⎤
1
2t+
5
R1 →
3R
⎢
1
0 1
3
0
⎥
⎢
⎥
1
R
0 0 1 0
1
→ 2 R ⎢
⎥
−4t
−16t−12
1
⎣ ⎦
Illegal if
2
4 16 12 0
− t − t− ≠
t
2
+ 4t+ 3=
0
( t )( t )
+ 3 + 1 = 0
t = −3, −1
Thus if t ≠−3, −1
then the solution is c3 = c2 = c1 = 0 so S is linearly independent.
−1 −1
⎡2 −1 −1 0⎤
⎡1 1
2 2
0⎤
1 2 1
1
If t =−3 , then
⎢
0 3 1 0
⎥R
→ R ⎢ −
0 1
1
3
0
⎥
⎢
−
⎥R2 → R ⎢
⎥
⎢
3 2
0 0 0 0 ⎥
⎣ ⎦ ⎢ ⎣0 0 0 0 ⎥ ⎦
so S is linearly dependent
−1 −1
⎡2 −1 −1 0⎤
⎡1 1
2 2
0⎤
1 2 1
If t =−1, then
⎢
0 3 3 0
⎥R
→ R ⎢
0 1 1 0
⎥
⎢ ⎥ 1
R2 → R ⎢
⎥
⎢
3 2
0 0 0 0 ⎥
⎣ ⎦ ⎢ ⎣0 0 1 0 ⎥ ⎦
so S is linearly dependent
S = 3x 2 + x+ 4,2 x 2 − x,
x 2 + tx+2
t
b) { }
c ( 2 ) ( 2 ) ( 2
1
3x x 4 c2 2x x c3
x tx 2t)
+ + + − + + + =0
⎡3 2 1 0⎤ ⎡3 2 1 0⎤
⎢ R2 3R2 R1
1 1 t 0
⎥ → − ⎢
⎢
−
⎥
0 5 3t
1 0
R3 3R3 4R
⎢
− −
⎥
→ −
⎥
1
⎢4 0 2t
0⎥
⎣ ⎦ ⎢⎣0 −8 6t−4 0⎥⎦
⎡3 2 1 0⎤
R3 5R3 8R ⎢
2
0 5 3t
1 0
⎥
→ −
⎢
− −
⎥
⎢⎣
0 0 6t
−12 0⎥⎦
1
R
2 1
1
→
3
R1
⎡1 3 3
0⎤
−1
3t−1
R2 →
5
R
⎢
2
0 1
5
0
⎥
⎢
− ⎥
1
R 0 0 1 0
3 2
R ⎢
⎥
→
6 t− 1 3⎣ ⎦
c
c
c
3
= t
=
1
2 3
=
7
1 6
c
c
3
t
t
= t
=−t
2
1
= 0
c
Winter 2006 Martin Huard 23

Math 105
Semester Review - Solutions
Illegal if 6t
− 12=
0
t = 2
Thus if t ≠ 2 , then the solution is c3 = c2 = c1 = 0 so S is linearly independent.
2 1
⎡3 2 1 0⎤
⎡1 1
3 3
0⎤
1 3 1
If t = 2 then
⎢
0 5 5 0
⎥R
→ R ⎢
0 1 1 0
⎥
⎢
−
⎥
−
−1
R2 → R ⎢
⎥
5 2
⎢0 0 0 0⎥
⎣ ⎦ ⎢⎣ 0 0 0 0⎥⎦
thus S is linearly dependent.
25. Do the following sets S span V
S = 2, −4,1 , 1, 2, −3 , 5, − 14,6
{ )}
3
a) ( ) ( ) (
V =
3
Let ( abc , , ) ∈ .
c( 2, − 4,1) + c( 1, 2, − 3) + c( 5, − 14,6 ) = ( abc , , )
1 2 3
c
c
c
3
2
1
= t
= t
=−t
⎡ 2 1 5 a⎤ ⎡2 1 5 a ⎤
⎢ R2 R2 2R1
4 2 14 b
⎥ → + ⎢
⎢
− −
⎥
0 4 4 b 2
R3 2R3 R ⎢
− + a
⎥
→ −
⎥
1
⎢ 1 −3 6 c⎥
⎣ ⎦ ⎢⎣0 −7 7 2c−a⎥
⎦
⎡2 1 5 a ⎤
R3 4R3 7R ⎢
2
0 4 4 b 2a
⎥
→ +
⎢
− +
⎥
⎢⎣
0 0 0 10a+ 7b+
8c⎥⎦
3
There is no solution if 10a+ 7b+ 8c≠ 0 . Thus S does not span .
⎧⎡ 2 −1⎤ ⎡3 2⎤ ⎡4 1⎤⎫
b) S = ⎨⎢ , ,
V
−1 3
⎥ ⎢
2 1
⎥ ⎢
1 4
⎥⎬ = S 2,2
(The set of symmetric 2×
2 matrices)
⎩⎣ ⎦ ⎣ ⎦ ⎣ ⎦⎭
⎡a
b⎤
Let A= ⎢ ∈
2,2
b c
⎥ S
⎣ ⎦
⎡ 2 −1⎤ ⎡3 2⎤ ⎡4 1⎤ ⎡a
b⎤
c1⎢ c2 c3
1 3
⎥+ ⎢
2 1
⎥+ ⎢ =
1 4
⎥ ⎢
b c
⎥
⎣−
⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦
⎡ 2 3 4 a⎤ ⎡2 3 4 a ⎤
⎢ R2 2R2 R1
1 2 1 b
⎥ → + ⎢
⎢
−
⎥
0 7 6 a 2
R3 2R3 3R
⎢
+ b
⎥
→ −
⎥
2
⎢ 3 1 4 c⎥
⎣ ⎦ ⎢⎣0 −7 −4 − 3a+
2c⎥⎦
1
⎡2 3 4 a ⎤ R
3 1
1
→
2
R1
⎡1 2
2
2a
⎤
R3 → R3+ R
⎢
2
0 7 6 a+
2b
⎥ 1 6 1 2
⎢
⎥
R2 →
7
R
⎢
2
0 1
7 7a
7b
⎥
⎢
+
⎥
⎢⎣0 0 2 − 2a+ 2b+
2c⎥
1
⎦ R 0 0 1 a b c
3 2
R ⎢ − + + ⎥
→
3 ⎣ ⎦
4 6 8
c =− a+ b+ c, c = a− b− c,
c = a− b− 5 c
3 2 7 7 1 7 7
Thus S spans S2,2
= − + + + + + V = P3
c) S { x 3 2x 1, x 3 x 2 , x 3 x 2 x 1}
Winter 2006 Martin Huard 24

Math 105
Semester Review - Solutions
Since n( S) ( P)
= 3< dim
3
= 4, then S does not span P3
Winter 2006 Martin Huard 25

Math 105
Semester Review - Solutions
26. Are the following sets S bases for the vector space V
3
a) S = 2,
−1,3 , 1,1,7 , − 2,4,1 , V =
{( ) ( ) ( )}
( 2, − 1,3) + ( 1,1,7) + ( − 2,4,1) = ( 0,0,0)
c c c
1 2 3
⎡ 2 1 −2 0⎤ ⎡2 1 −2 0⎤
⎢ R2 → 2R2 + R1
−1 1 4 0
⎥ ⎢
0 3 6 0
⎥
⎢ ⎥R3 →2R3−3R
⎢
⎥
1
⎢ 3 7 1 0⎥
⎣ ⎦ ⎢⎣0 11 8 0⎥⎦
1
⎡2 1 −2 0⎤
R
1
1
→
2
R1
⎡1 2
−1
0⎤
R3 →3R3−11R
⎢
2
0 3 6 0
⎥ 1
⎢
⎥
R2 →
3
R
⎢
2
0 1 2 0
⎥
⎢
⎥
⎢⎣
0 0 −42 0⎥
−1
⎦ R 0 0 1 0
3 42
R ⎢
⎥
→
3⎣ ⎦
c1 = c2 = c3 = 0
3
Since S is linearly independent and ( ) 3 dim( )
{( ) ( )}
b) S = − 3,5,1 , 2, −7,12
, V
3
=
3
No since n( S ) = 2≠ dim( ) = 3.
2 2 2
2
c) = { x + 1, x −1,
x + x + 1, x − x −1}
S , V = P2
n S = 4≠ dim P = 3
No since ( ) ( )
2
n S = = , then S is a basis for
⎧⎡2
−1⎤
⎡4
−1⎤
⎡ 3 3⎤
⎡5
5 ⎤⎫
d) S = ⎨⎢
⎥,
⎢ ⎥,
⎢ ⎥,
⎢ ⎥⎬
, V = M 2, 2
⎩⎣2
3 ⎦ ⎣4
3 ⎦ ⎣−1
1⎦
⎣1
− 2⎦⎭
⎡2 −1⎤ ⎡4 −1⎤ ⎡ 3 3⎤ ⎡5 5 ⎤ ⎡0 0⎤
c1⎢ c2 c3 c4
2 3
⎥+ ⎢
4 3
⎥+ ⎢ + =
1 1
⎥ ⎢
1 2
⎥ ⎢
⎣ ⎦ ⎣ ⎦ ⎣−
⎦ ⎣ − ⎦ ⎣0 0
⎥
⎦
⎡ 2 4 3 5 0⎤ ⎡2 4 3 5 0⎤
⎢ R2 → 2R2 + R1
−1 −1 3 5 0
⎥ ⎢
0 2 9 15 0
⎥
⎢ ⎥R3 →R3 −R
⎢
⎥
1
⎢ 2 4 −1 1 0⎥ ⎢0 0 −4 −4 0⎥
⎢ ⎥R4 →2R4 −3R1⎢
3 3 1 −2 0
⎥
⎣ ⎦ ⎣0 −6 −7 −19 0⎦
⎡2 4 3 5 0⎤
⎡2 4 3 5 0⎤
⎢
0 2 9 15 0
⎥
⎢
R4 R4 3R
⎢
⎥
0 2 9 15 0
⎥
→ +
2
R4 R4 5R
⎢
⎥
3
⎢0 0 −4 −4 0⎥
→ +
⎢0 0 −4 −4 0⎥
⎢
⎥
⎢
⎥
⎣0 0 20 26 0⎦
⎣0 0 0 6 0⎦
1
R
3 5
1
→
2
R1
⎡1 2
2 2
0⎤
1
R
9 15
2
→
2
R
⎢
2 0 1
2 2
0
⎥
⎢
⎥
c
−1
4
= c3 = c2 = c1 = 0
R3 →
4
R ⎢
3
0 0 1 1 0⎥
⎢
⎥
1
R 0 0 0 1 0
4
→
6
R4
⎣ ⎦
n S = 4= dim M , then S is a basis for
Since S is linearly independent and ( ) ( 2,2 )
M
2,2
3
Winter 2006 Martin Huard 26

Math 105
Semester Review - Solutions
27. Add or subtract vectors to the set S to so that it forms a basis for
a) S = {( 1, 3, 5 ), ( 2, − 1, 3)}
v = 1, 0, 0 , then
If ( )
c ( 1, 3, 5) + c ( 2, − 1, 3) = ( 1, 0, 0)
1 2
⎡1 2 1⎤ ⎡1 2 1⎤ ⎡1 2 1⎤
⎢ 2 2
3
1
3 1 0
⎥R →R − R ⎢
0 7 3 ⎥
⎢
−
⎥
−
R3 R3 5R
⎢
− ⎥⎥
R3 R3 R
⎢
2
0 7 3
⎥
→ −
→ −
⎢
− −
⎥
1
⎢⎣5 3 0⎥
⎦ ⎢⎣0 −7 −5⎦
⎢⎣0 0 −2⎥⎦
No solution, so ( 1, 0, 0) ∉ span ( S )
Since S is linearly independent (the two vectors are not multiples of each other) ,
B = 1, 3, 5 , 2, − 1, 3 , 1, 0, 0 is linearly independent.
{ }
then by the +/- theorem ( ) ( ) ( )
3
Since ( ) 3 dim( )
{ 1, 2, 4 , 1, 3, 9 , 1, 0, 6)}
b) ( ) ( ) (
n B = = , then B is a basis for
S = −
Let us verify independence.
c 1,2,4 + c 1,3,9 + c 1,0, − 6 = 0,0,0
( ) ( ) ( ) ( )
1 2 3
3
⎡1 1 1 0⎤ ⎡1 1 1 0⎤
⎢ 2 2
2
1
2 3 0 0
⎥R →R − R ⎢
0 1 −2 0
⎥
⎢ ⎥R3 →R3−4R
⎢
⎥
1
⎢4 9 −6 0⎥
⎣ ⎦ ⎢⎣0 5 −10 0⎥⎦
⎡1 1 1 0⎤
R3 R3 5R
⎢
1
0 1 2 0
⎥
→ −
⎢
−
⎥
⎢⎣
0 0 0 0⎥⎦
1, then 1,0, − 6 = 3 1, 2, 4 − 2 1,3,9
3
c
c
3
2
1
= t
= 2t
c =−3t
If t = ( ) ( ) ( )
By the +/- theorem, if B = {( 1, 2, 4 ), ( 1, 3, 9)
} , then span ( S) span (
If v = ( 1, 0, 0)
, then
c ( 1,2,4) + c ( 1,3,9) = ( 1,0,0)
1 2
⎡1 1 1⎤ ⎡1 2 1⎤ ⎢ 2 2
2
1
2 3 0
⎥R → R − R ⎢
0 1 2 ⎥
⎢ ⎥
−
R3 →R3−4R
⎢ ⎥
1
⎢4 9 0⎥
⎣ ⎦ ⎢⎣0 5 −4⎥⎦
No solution, so ( 1, 0, 0) ∉ span ( B)
= B).
⎡1 2 1⎤
R3 R3 5R
⎢
2
0 1 2
⎥
→ −
⎢
−
⎥
⎢⎣0 0 6⎥⎦
Since B is linearly independent (the two vectors are not multiples of each other) ,
B = 1, 2, 4 , 1, 3, 9 , 1, 0, 0 is linearly independent.
{ }
then by the +/- theorem ( ) ( ) ( )
3
Since ( S ) 3 dim( )
S
3
n B = = , then B
S
is a basis for .
Winter 2006 Martin Huard 27

Math 105
Semester Review - Solutions
( )
a) S = {( −1,1, − 1 ), ( 2,1,1 ), ( 1, 5,1)}
c ( −1,1, − 1) + c ( 2,1,1) + c ( 1, 5,1) = ( 0, 0, 0)
28. Find a basis for span S if
1 2 3
⎡−1 2 1 0⎤ ⎡−1 2 1 0⎤
⎢ 2 2 1
1 1 5 0
⎥R → R + R ⎢
0 3 6 0
⎥
⎢ ⎥R3 →R3−R
⎢
⎥
1
⎢−1 1 1 0⎥
⎣ ⎦ ⎢⎣
0 −1 0 0⎥⎦
⎡−1 2 1 0⎤
R1 →−R1
⎡1 −2 −1 0⎤
R3 → 3R3+
R
⎢
2
0 3 6 0
⎥ 1
⎢
⎥
R2 →
3
R
⎢
2
0 1 2 0
⎥
⎢ ⎥
⎢⎣
0 0 6 0⎥
1
⎦ R 0 0 1 0
3 6
R ⎢
⎥
→
3 ⎣ ⎦
span S . Since
Since S is linearly independent, then it is a basis for ( )
3
3
dim( span ( S )) = 3 = dim( ) , then span ( S ) =
b) S = {( 1,3, −2,2,6, ) ( −4, ) ( −3, − 9,6)}
c ( 1,3, − 2) + c ( 2, 6, − 4) + c ( −3, − 9, 6) = ( 0, 0, 0)
1 2 3
⎡ 1 2 −3 0 1 2 −3 0
⎢ ⎤ 2 2
3
1
3 6 9 0
⎥R → R − R
⎡ ⎢
⎤ 0 0 0 0 ⎥⎥⎥⎦
⎢
−
⎥R3 → R3+
2R
⎢
1
⎢−2 −4 6 0⎥
⎣ ⎦ ⎢⎣0 0 0 0
If t = 1 and 0 then −3, − 9, 6 =−3 1,3, − 2
s = ( ) ( )
If t = 0 and s = 1 then ( 2,6, − 4) = 2( 1,3, − 2)
By the +/- theorem, if B = ( 1, 3, − 2)
, then span ( B ) = span ( )
{ }
c
c
3
2
= t
= s
c
c
3
2
1
= 0
= 0
c = 0
c2 =− 2s+
3t
S and since B is
linearly independent (only a single vector) then B is linearly independent, thus B is a
basis for span S .
( )
y z
Geometrically, span ( S ) is the line x = = .
3 − 2
S = 1, 2, −1 , 2,3,1 , 4,7, − 1 , 1,1, 2
{ }
c) ( ) ( ) ( ) ( )
c ( 1, 2, − 1) + c ( 2,3,1) + c ( 4,7, − 1) + c ( 1,1, 2) = ( 0,0,0)
1 2 3 4
⎡ 1 2 4 1 0⎤ ⎡1 2 4 1 0⎤
⎢ 2 2
2
1
2 3 7 1 0
⎥R →R − R ⎢
0 −1 −1 −1 0
⎥
⎢ ⎥R3 → R3+
R ⎢
⎥
1
⎢−1 1 −1 2 0⎥
⎣ ⎦ ⎢⎣0 3 3 3 0⎥⎦
⎡1 2 4 1 0⎤
R3 R3 3R
⎢
1
0 1 1 1 0
⎥
→ +
⎢
− − −
⎥
⎢⎣
0 0 0 0 0⎥⎦
Winter 2006 Martin Huard 28

Math 105
Semester Review - Solutions
⎡1 2 4 1 0⎤
R2 R
⎢
2
0 1 1 1 0
⎥
→−
⎢
⎥
⎢⎣
0 0 0 0 0⎥⎦
c
c
c
4
3
= t
= s
=−t−s
2
1
= −2
c t s
If t = 1 and s = 0 then ( 1,1, 2) =−( 1, 2, − 1) + ( 2, 3,1)
If t = 0 and s = 1 then ( 4,7, − 1) = 2( 1, 2, − 1) + ( 2,3,1)
By the +/- theorem, if B = ( 1, 2, − 1 ),( 2,3,1)
, then span ( ) span (
{ }
B = S ) and since
B is linearly independent (the two vectors are not multiples of each other), thus B
is a basis for span S .
( )
i j k
n = − × = − = − −
( 1,2, 1) ( 2,3,1) 1 2 1 ( 5, 3, 1)
( )
2 3 1
Geometrically, span S is the line 5x− 3y− z = 0.
S = { 1,2,3,4 , 4,3,2,1 , 1,1,1,1 )}
d) ( ) ( ) (
⎡1 2 3 4⎤ ⎡1 2 3 4 ⎤
⎢ 2 2
4
1
4 3 2 1
⎥R →R − R ⎢
0 5 10 15
⎥
⎢ ⎥
− − −
R ⎥
3
→R3−R
⎢
1
⎢1 1 1 1⎥
⎣ ⎦ ⎣⎢0 −1 −2 −3
⎦⎥
⎡1 2 3 4 ⎤ ⎡1 2 3 4⎤
R3 →5R3−R
⎢
1
0 −5 −10 −15
⎥ −1
⎢
⎥
R2 5
R
⎢
2
0 1 2 3
⎥
→
⎢ ⎥
⎢⎣
0 0 0 0 ⎥⎦
⎢⎣
0 0 0 0⎥⎦
B = 1, 2,3, 4 , 0,1, 2,3
{ }
Thus a basis for span( S ) is
S ( ) ( )
Note: If you used the +/- theorem, then you obtain B = ( 1, 2,3, 4 )( , 4,3, 2,1)
29. Find the coordinate vector for w in the vector space V relative to the basis S.
w 3
= − 5,
−6,24
, and S = 2,
−1,4
, 1,2,4 , − 3, −3,4
.
a) ( ) V = {( ) ( ) ( )}
c ( 2, − 1,4) + c ( 1,2,4) + c ( −3, − 3,4) = ( −5, − 6,24)
1 2 3
S
{ }
⎡ 2 1 −3 −5⎤ ⎡2 1 −3 −5
⎤
⎢ R2 → 2R2 + R1
−1 2 −3 −6 ⎥ ⎢
0 5 −9 −17
⎥
⎢ ⎥R3 →R3−2R
⎢
⎥
1
⎢ 4 4 4 24⎥
⎣ ⎦ ⎢⎣0 2 10 34 ⎥⎦
1
⎡2 1 −3 −5
⎤ R
1 3 5
1
→
2
R
− −
1 ⎡1
2 2 2 ⎤
R3 →5R3−2R
⎢
2
0 5 −9 −17
⎥ 1
−9 −17
⎢
⎥
R2 →
5R
⎢
2
0 1
⎥
⎢
5 5 ⎥
⎢⎣
0 0 68 204⎥
1
⎦ R 0 0 1 3
3 68
R ⎢
⎥
→
3⎣ ⎦
c3 = 3, c2 = 2, c1
= 1
w = (1, 2,3)
S
Winter 2006 Martin Huard 29

Math 105
Semester Review - Solutions
b) w 3
= ( 12,
−7,10)
, V = and = {( 3,
−1,5 ),
( 1,2,3 ),
( 2, −1,
−1)
}
c ( 3, − 1, 5) + c ( 1, 2, 3) + c ( 2, −1, − 1) = ( 12, − 7,10)
1 2 3
S .
⎡ 3 1 2 12⎤ ⎡3 1 2 12 ⎤
⎢ R2 → 3R2 + R1
−1 2 −1 −7 ⎥ ⎢
0 7 −1 −9
⎥
⎢ ⎥R3 →3R3−5R
⎢
⎥
1
⎢ 5 3 −1 10⎥
⎣ ⎦ ⎢⎣0 4 −13 −30⎥
⎦
1
⎡3 1 2 12 ⎤ R
1 2
1
→
3
R1
⎡1 3 3
4⎤
R3 →7R3−4R
⎢
1
0 7 −1 −9
⎥ 1 −1
−9
⎢
⎥
R2 →
7
R
⎢
2
0 1
⎥
⎢
7 7 ⎥
⎢⎣
0 0 −87 −174⎥
−1
⎦ R 0 0 1 2
3 87
R ⎢
⎥
→
3⎣ ⎦
c3 = 2, c2 =− 1, c1
= 3
w = 3, −1, 2
S ( )
c) w = ( 4,
−3,5,3
), = Span( {( 3, −1,4,0
),
( 0,1, −2,4) ,( 2,2,1,1 )})
S = {( 3,
−1,4,0
),
( 0,1, −2,4) ,( 2,2,1,1 )}.
c ( 3, − 1, 4,0) + c ( 0,1, − 2, 4) + c ( 2, 2,1,1) = ( 4, − 3,5,3 )
V and
1 2 3
⎡ 3 0 2 4 ⎤ ⎡3 0 2 4 ⎤
⎢
1 1 2 3
⎥
R2 → 3R2 + R
⎢
⎢
− −
⎥
1
⎢
0 3 8 −5
⎥
⎥⎥
⎢ 4 −2 1 5 ⎥R3 →3R3 −4R
⎢
1
0 −6 5 1
⎢ ⎥ − −
⎢
⎥⎦
⎣ 0 4 1 3 ⎦ ⎣0 4 1 3
⎡3 0 2 4 ⎤
⎡3 0 2 4 ⎤
R3 → R3 + 2R
⎢
2
0 3 8 −5
⎥
⎢
⎢
⎥
0 3 8 5
⎥
R4 11R4 29R
⎢
−
⎥
3
R4 →3R4 −4R
⎢
2
0 0 11 −11⎥
→ +
⎢0 0 11 −11⎥
⎢
⎥
⎢
⎥
⎣0 0 −29 29 ⎦
⎣0 0 0 0 ⎦
2 4
1
⎡1 0
3 3 ⎤
R1 →
3
R1
⎢ 8 −5
0 1
⎥ c3
= −1
1
3 3
R2 3
R ⎢
⎥
→
2
c2
= 1 w
S
= ( 2,1, −1)
⎢0 0 1 −1⎥
1
R3 →
11
R3⎢ ⎥ c1
= 2
⎣0 0 0 0 ⎦
d) w 9 2 2 2 2
= x − x +11, V = P2
and S = { x + x,
x −1,2
x − 3x
+ 4}
.
( ) ( ) ( )
c x + x + c x − 1 + c 2x − 3x+ 4 = 9x − x+11
2 2 2 2
1 2 3
⎡1 1 2 9 ⎤ ⎡1 1 1 9 ⎤
⎢
1 0 −3 −1 ⎥
R2 →R2 −R
⎢
1
0 −1 −5 −10
⎥
⎢ ⎥
⎢
⎥
⎢⎣0 −1 4 11⎥⎦ ⎢⎣0 −1 4 11 ⎥⎦
⎡1 1 2 9 ⎤ ⎡1 1 2 9⎤
R3 →R3−R
⎢
2
0 −1 −5 −10
⎥ R2 →−R2⎢ ⎢
⎥
0 1 5 10
⎥
1
R3 9
R ⎢
⎥
3
7
⎢⎣0 0 9 21 ⎥
→
⎦ ⎢ ⎣0 0 1
3 ⎥ ⎦
−
w = 6, ,
S
5 7
( 3 3
)
c
c
7
3 3
−5
2 3
1
= 6
c
=
=
Winter 2006 Martin Huard 30

Math 105
Semester Review - Solutions
30. Find a basis, and the dimension, for the solution space of AX = 0 .
a) x − 2y + z − w=
0
3x + y + w=
0
4x + 6y − 2z + 4w=0
⎡1 −2 1 −1 0⎤ ⎡1 −2 1 −1 0⎤
⎢ 2 2
3
1
3 1 0 1 0
⎥R →R − R ⎢
0 7 −3 4 0
⎥
⎢ ⎥R3 →R3−4R
⎢
⎥
1
⎢4 6 −2 4 0⎥
⎣ ⎦ ⎢⎣0 14 −6 8 0⎥⎦
⎡1 −2 1 −1 0⎤
⎡1 −2 1 −1 0⎤
R3 →R3−2R
⎢
2
0 7 −3 4 0
⎥ 1 3 4
⎢
⎥
R2 7
R
⎢
−
2
0 1
7 7
0
⎥
→ ⎢ ⎥
⎢⎣
0 0 0 0 0⎥⎦
⎢⎣0 0 0 0 0⎥⎦
3 4 −1
w= t, z = s, y = s− t,
x= s− 1 t
7 7 7 7
−
( ) ( 1 1 3 4 −
x, y, z, t = ) ( 1 2 ) ( 1 4
7
s− 7t, 7s− 7t, s, t = s
7
,
7,1,0 + t
− 7
, −
7
,0,1)
= {( 1 3 ) ( 1 4
7
,
7,1,0 ,
7
,
7
,0,1)}
dim( SS ) = 2
− − −
BSS
b) x − y + 3z
= 0
2x + 5y + 6z
= 0
x − 8y + 3z
= 0
2x + y + 6z
= 0
⎡1 −1 3 0⎤ ⎡1 −1 3 0⎤
⎢ 2 2
2
1
2 5 6 0
⎥R →R − R ⎢
0 8 0 0
⎥
⎢ ⎥R3 →R3 −R
⎢
⎥
1
⎢1 −8 3 0⎥ ⎢0 −7 0 0⎥
⎢ ⎥R4 →R4 −2R4
⎢
⎣2 1 6 0
⎥
⎦ ⎣0 2 0 0⎦
⎡1 −1 3 0⎤
⎡1 −1 3 0⎤
R3 → 8R3+
7R
⎢
2
0 8 0 0
⎥ ⎢
⎢
⎥
0 1 0 0
⎥
1
R2 8
R ⎢
⎥
2
R4 →4R4 −R
⎢
2
0 0 0 0⎥
→ ⎢ 0 0 0 0 ⎥
⎢
⎥ ⎢
⎥
⎣0 0 0 0⎦
⎣0 0 0 0⎦
z = t, y = 0, x =−3t
( xyz , , ) = t( − 3,0,1)
B = ( − 3, 0,1)
( )
SS
{ }
dim SS = 1
Winter 2006 Martin Huard 31

Math 105
Semester Review - Solutions
31. Find the minimum or maximum values of the given objective function, subject to the
indicated constraints.
a) Objective function: f = 3x+ 5y
Constraints: x+ y≥2
2x+ 3y≤12
3x+ 2y≤12
x≥0, y ≥ 0
0, 2 f = 10
At A( )
B( )
12 12
C( , )
D( )
E( )
0, 4 f = 20
96
5 5 5
4,0 f = 12
2,0 f = 6
f
=
Thus the minimum value of f is 6 when x = 2 and y = 0, and the maximum value is
20 when x = 0 and y = 4 .
b) Objective function: f = 5x+ 2y
Constraints: x+ y≤10
2x+ y≥10
x+ 2y≥10
x≥0, y ≥0
At A( )
B( )
C( , )
0,10 f = 20
10,0 f = 50
10 10 70
3 3 3
f
Thus the minimum value of f is 20
when x = 0 and y = 10 , and the
maximum value is 50 when x = 10 and
y = 0.
=
Winter 2006 Martin Huard 32

Math 105
Semester Review - Solutions
c) Objective function: z = 2x+ 5y
Constraints: 2x+ y≥8
− 4x+ y≤2
2x−3y≤0
x≥0, y ≥0
At A( )
B( )
1, 6 f = 32
3, 2 f = 16
Thus there is no maximum, and the
minimum value is 16 when x = 3 and
y = 2
32. An entrepreneur is having a design group produce at least six samples of a new kind of
fastener that he wants to market. It cost $9.00 to produce each metal fastener and $4.00 to
produce each plastic fastener. He wants to have at least two of each version of the fastener
and needs to have all the samples 24 hours from now. It takes 4 hours to produce each metal
sample and 2 hours to produce each plastic sample. To minimize the cost of the samples,
how many of each kind should the entrepreneur order What will be the cost of the samples
Minimize C = 9x+ 4y
Constraints: x+ y≥6
At A( )
B( )
C( )
D( )
x ≥ 2
y ≥ 2
4x+ 2y≤24
2, 4 C = 34
2,8 C = 50
5, 2 C = 53
4, 2 D=
44
Thus minimal cost of the samples is
$34 when 2 metal and 4 plastic
samples are ordered.
Winter 2006 Martin Huard 33