Math 3C
Practice Final Problems
Solutions
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1a) (find a general solution)
y  2 y  7
First we solve the homogeneous equation:
y  2 y  0
separate... int egrate...
yh  Ce 2t
Now find a particular solution – it will be a constant, say yp=A.
Plug this into the equation and solve for A:
( A )  2( A )  7
0  2A  7
A
7
2
yp  72
y general  yh  yp
y general  Ce 2t  72
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1b) (find a general solution)
ty  4y  t 2
y 
4
yt
t
First we solve the homogeneous equation:
y  4t y  0
dy
dt

  4t y 
dy
y
dy
y
 4 dtt
   4 dtt
ln y  4 ln t  C
C
t4
Now we use variation of parameters
Quotient
to find a particular solution:
v
rule
yp  4
t

vt 4  4vt 3 4v
 v  4 v 
 5  t  vt 4  4vt 3  4vt 3  t 9
 4   4t
8
t
t
t  t t 
yh  e
ln t  C

Multiply both sides by t8
vt 4  t 9  v  t 5  v  61 t 6
yp 
1 t6
6
4
t
y general 
 61 t 2
C 1 2
 t
t4 6
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1c) (solve the initial value problem)
y  3y  2e3t ; y(0)  1
First we solve the homogeneous equation:
y  3 y  0
yh  Ce3 t
Now we use variation of parameters
to find a particular solution:
y p  v  e3 t
v  e   3v  e   2e
3t
3t
3t
v  e3 t  3v  e3 t  3v  e3 t  2e3 t
v  2
v  2t
y p  2t  e 3 t
y general  Ce3 t  2t  e3 t
Now plug in the initial value to find C.
1  Ce30   20   e30 
1 C
Solution to IVP
y( t )  e 3 t  2 t  e 3 t
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1d) (solve the initial value problem)
y  tan( t )y  sin( t ); y( )  1
First we solve the homogeneous equation:
y  tan( t )y  0
dy
dt
 tan( t )y  
dy
  tan( t )dt
y
ln y   ln cos(t )  C
yh  C  sec(t )
Now we use variation of parameters
to find a particular solution:
yp  v  sec(t )
v  sec(t )  tan( t )v  sec(t )  sin( t )
v  sec(t )  v  sec(t )  tan( t )  tan( t )v  sec(t )  sin( t )
v  sec(t )  sin( t )  v  sin( t )  cos(t )
v   sin( t )  cos(t )dt  21 sin2 ( t )
yp  21 sin2 ( t )  sec(t )
y general  C  sec(t )  21 sin2 ( t )  sec(t )
Now plug in the initial value to find C.
1  C  sec()  21 sin2 ( )  sec()
1  C   1 
1
2
0    1  1  C
y( t )   sec(t )  21 sin2 ( t )  sec(t )
Solution to IVP
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2) Farmer Fred opens a bank account with an initial deposit of $10,000. The account earns a monthly interest
rate of 0.5%, compounded continuously. In addition, Fred will deposit $100 each month. How many months
will it take for the account value to double?
Define variables: y(t) = account value at time t (months)
dy
We also have an initial value y(0)=10,000
 0.005y  100
The differential equation is
dt
We could solve this equation by separation, but let’s use the excellent guess method instead.
First we solve the homogeneous equation:
y  0.005y  0  yh  Ce0.005 t
For the particular solution, notice that we should get a constant. Plug in and solve:
0  0.005A  100  A   0100
 20,000
.005
yp  20,000
y general  Ce0.005 t  20,000
Now plug in the initial value to find C.
10,000  Ce0.005 0   20,000  C  30,000
y( t )  30,000  e0.005 t  20,000
Finally we can use our formula to find the answer. Just set y(t)=20,000 and solve for t.
30,000  e0.005 t  20,000  20,000
0.005 t
e

4
3
 0.005t 
ln( 34 ) 
t
ln( 34 )
0.005
 57.5months
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3) A lukewarm beverage (initially at 70°F) is placed into a cold (40°F) refrigerator. Five minutes later the
temperature of the still-not-cold-enough beverage is 60°F. It is replaced in the fridge to continue the chilling
process. How many minutes will it take for the beverage to reach the temperature of optimal refreshment
(43°F)? Assume that the temperature follows Newton’s law of cooling, i.e. the rate of change of the
temperature is proportional to the difference between the temperature of the beverage and the temperature of
the surroundings.
Define variables: T(t)=temperature of beverage (in°F) at time t (minutes)
dT
 k(T  40) Here k is a constant of proportionality
The differential equation is
dt
Let’s solve this one by separation.
dT
dT
 k(T  40)  
  k  dt
dt
T  40
ln T  40  kt  C
T  Cekt  40
We need to find the 2 constants C and k. From the given information we know y(0)=70 and y(5)=60.
70  Cek 0   40  C  30
60  30  e5k  40  e5k 
k 
ln( 32 )
 0.081
5
Now we can rewrite our formula with the constants we just found, then use it to solve the problem.
2
3
T( t )  30  e0.081t  40
1
43  30  e0.081t  40  e0.081t  10
t
1)
ln( 10
 0.081
 28.4 min utes
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4) Consider the following autonomous differential equation:
y  y 2  y  2  0
a) Sketch a slope field for this equation.
b) Find any equilibrium solutions and classify them as stable or unstable.
c) Find an explicit formula for the solution that passes through the initial value y(0) = 1.
a) The slope field shows the
equilibrium solutions, and their
stability. We will calculate them
algebraically as well.
b) To find equilibrium solutions,
set y’=0:
y2  y  2  0
( y  2)( y  1)  0
y  2; y  1
Equilibrium solutions are the
horizontal lines y=2 and y=-1
To assess the stability, plug
in values on either side to
find if the slope is positive or
negative.
y(3)  2; y(0)  2; y( 2)  4
 y  2 unstable
 y  1 stable
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4) Consider the following autonomous differential equation:
y  y 2  y  2  0
c) Find an explicit formula for the solution that passes through the initial value y(0) = 1.
We will solve this by separation:
dy
dy
 y2  y  2  
  dt
dt
( y  2)( y  1)
1  1  1 dy  t  C
3   y  2 y  1


We use partial fractions to simplify the integral:
1
A
B


( y  2)( y  1) y  2 y  1
1  A( y  1)  B( y  2)
y 2 A  1
ln y  2  ln y  1  3t  C
3
y  1  B   1
y2
ln
 3t  C
y 1
3
1
1
1
3

 3
( y  2)( y  1) y  2 y  1
y2
 Ce3 t
y 1
Use the initial value to find C
1 2
 Ce30   C  1
2
1 1
This is an implicit solution – we can
do some algebra and solve for y to
get an explicit solution
y  2 1 3t
 e
y 1 2
Careful with the absolute values – in this case
we have y(0)=1, so we can drop the abs value
bars on the bottom, but we have to switch the
top of the fraction.
This solution is
valid for -1<y<2
2  1 e3t
2  y 1 3t
3
t
2
 e  2  y  1 e y  1  y 
2
2
1
y 1
1  e3t
2
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5) Consider the following system of differential equations:
dx
a) Find and graph the nullclines of this system.
 2x  xy
dt
b) What are the equilibria of this system?
dy
c) Classify any equilibria as stable or unstable.
 3 y  0.5 xy
d) Sketch a reasonable solution to this system of differential equations.
dt
Set x’=0 to find v-nullcline:
dx
 2x  xy  0  x  0; y  2
dt
h-nullclines in yellow
Set y’=0 to find h-nullcline:
v-nullclines in red
dy
 3y  0.5xy  0  x  6; y  0
dt
Solution curve
Equilibria at intersections
Equlibrium Points are
(0,0) – unstable
Unstable equil.
Stable equil.
(6,2) - stable
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6) Solve the following system of linear equations:
xyz 1
x  y  2z  5
3 x  y  4z  2
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6) Solve the following system of linear equations:
xyz 1
x  y  2z  5
3 x  y  4z  2
Augmented matrix form
1  1 1 1


 1 1 2 5
3  1 4 2
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6) Solve the following system of linear equations:
xyz 1
x  y  2z  5
3 x  y  4z  2
Augmented matrix form
1  1 1 1


 1 1 2 5
3  1 4 2
Row reduction:
1  1 1 1 
1  1 1 1 


R2* R2R1
2 1 4 R3*  R3  R2  0 2 1 4 
R3* R3 3R1  0
0 2 1 1
0 0 0  5
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6) Solve the following system of linear equations:
xyz 1
x  y  2z  5
3 x  y  4z  2
Augmented matrix form
1  1 1 1


 1 1 2 5
3  1 4 2
Row reduction:
1  1 1 1 
1  1 1 1 


R2* R2R1
2 1 4 R3*  R3  R2  0 2 1 4 
R3* R3 3R1  0
0 2 1 1
0 0 0  5
The 3rd row represents the equation 0x+0y+0z = -5. This equation has no solution, so the original
system of equations is inconsistent (not solvable).
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7) Consider the following matrix:
6 22 0  3
0  1 0
4 

A
0 0 13 0 


4
0 0 0
a)Find the determinant of this matrix.
b)Is this matrix invertible? If so, find the inverse.
c)What is the determinant of the inverse of this matrix?
d)Using this information, solve the following system of linear equations:
6 x  22y  3 w  2
4w  y  3
13z  26
4 w  12
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7) Consider the following matrix:
6 22 0  3
0  1 0
4 

A
0 0 13 0 


4
0 0 0
a)Find the determinant of this matrix.
b)Is this matrix invertible? If so, find the inverse.
c)What is the determinant of the inverse of this matrix?
d)Using this information, solve the following system of linear equations:
6 x  22y  3 w  2
4w  y  3
13z  26
4 w  12
a) Since A is a triangular matrix, the determinant is the product of the diagonal elements – det(A) = -312
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7) Consider the following matrix:
6 22 0  3
0  1 0
4 

A
0 0 13 0 


4
0 0 0
a)Find the determinant of this matrix.
b)Is this matrix invertible? If so, find the inverse.
c)What is the determinant of the inverse of this matrix?
d)Using this information, solve the following system of linear equations:
6 x  22y  3 w  2
4w  y  3
13z  26
4 w  12
a) Since A is a triangular matrix, the determinant is the product of the diagonal elements – det(A) = -312
b) Since det(A) is not zero, A is invertible, and we can use row reduction to find the inverse.
6 22 0  3  1 0 0
0  1 0
4  0 1 0

0 0 13 0  0 0 1


4  0 0 0
0 0 0
0
6 22 0  3   1 0
0 1 0  4 0  1
0 R2*  R2


*
1


0 R4  4 R4
0 0 13 0  0 0


 0 0
1
0
0
0
1

 
6 22 0 0  1 0
0 1 0 0 0  1
R1*  R1  3R4


*
0 0 13 0 0 0
R2  R2  4R4


0 0 0 1 0 0
3
4
6

0
0 1 *
R1  R1  22R2  
0
1 0


0 41 
0
0
0 0
0 0 
1 0

0 41 

0  1 22 0  85
4


1 
1 0 0 0  1 0
0 13 0 0 0 1 0 


1
0 0 1 0 0 0

4 
0
0

0  85
 1 0 0 0  61 11
3
24


0 1 0 0 0  1 0
R1*  61 R1
1



*
1
1

0 0 1 0 0 0 13
0 
R3  13 R3




1
0
0
0
0
0
0
1


4 
 

Identity
A 1
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7) Consider the following matrix:
6 22 0  3
0  1 0
4 

A
0 0 13 0 


4
0 0 0
a)Find the determinant of this matrix.
b)Is this matrix invertible? If so, find the inverse.
c)What is the determinant of the inverse of this matrix?
d)Using this information, solve the following system of linear equations:
6 x  22y  3 w  2
4w  y  3
13z  26
4 w  12
c) Since A-1is a triangular matrix, the determinant is the product of the diagonal elements: det(A-1) = -1/312
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7) Consider the following matrix:
6 22 0  3
0  1 0
4 

A
0 0 13 0 


4
0 0 0
a)Find the determinant of this matrix.
b)Is this matrix invertible? If so, find the inverse.
c)What is the determinant of the inverse of this matrix?
d)Using this information, solve the following system of linear equations:
6 x  22y  3 w  2
4w  y  3
13z  26
4 w  12
c) Since A-1is a triangular matrix, the determinant is the product of the diagonal elements: det(A-1) = -1/312
d) Notice that we have the inverse for the coefficient matrix, so we just need to multiply:
1 11
 x  6 3 0
 y  0  1 0
 
1
 z   0 0 13
  
 w   0 0 0
 85   2 
24 
1  3
 
0  26

1  12 
4   
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7) Consider the following matrix:
6 22 0  3
0  1 0
4 

A
0 0 13 0 


4
0 0 0
a)Find the determinant of this matrix.
b)Is this matrix invertible? If so, find the inverse.
c)What is the determinant of the inverse of this matrix?
d)Using this information, solve the following system of linear equations:
6 x  22y  3 w  2
4w  y  3
13z  26
4 w  12
c) Since A-1is a triangular matrix, the determinant is the product of the diagonal elements: det(A-1) = -1/312
d) Notice that we have the inverse for the coefficient matrix, so we just need to multiply:
0
 x   61 11
3
 y  0  1 0
 
1
 z  0 0 13
  
 w  0 0 0
 2
 85
24

1  3
 
0  26
  
1
 12
4 

 x   187
6

y 
  9 
z  2 

  
w   3 
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8) Consider the following set of vectors in R3

 1
0
 1
 1 





 
 
 
 
a1  0, a2   1, a3   1, a4  2 

 1
 1
2
3 

a)What is the dimension of the span of this set of vectors?
b)Is the span of this set of vectors R3?
c)Is this a basis for R3?
d)Is each of the following vectors within the span of this set?
If so, express them as a linear combination of the vectors in the set.
 2
2


v1  0, v 2  2
3
4
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
8) Consider the following set of vectors in R3

 1
0
 1
 1 





 
 
 
 
a1  0, a2   1, a3   1, a4  2 

 1
 1
2
3 

a)What is the dimension of the span of this set of vectors?
b)Is the span of this set of vectors R3?
c)Is this a basis for R3?
d)Is each of the following vectors within the span of this set?
If so, express them as a linear combination of the vectors in the set.
 2
2


v1  0, v 2  2
a) Put the column vectors in a matrix and
3
4
perform elementary row operations:
1 0 1 1 
1 0 1 1 
1 0 1 1 
0 1 1 2R3*  R3  R1  0 1 1 2R3*  R3  R2  0 1 1 2 






1 1 2 3
0 1 1 2
0 0 0 0
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
8) Consider the following set of vectors in R3

 1
0
 1
 1 





 
 
 
 
a1  0, a2   1, a3   1, a4  2 

 1
 1
2
3 

a)What is the dimension of the span of this set of vectors?
b)Is the span of this set of vectors R3?
c)Is this a basis for R3?
d)Is each of the following vectors within the span of this set?
If so, express them as a linear combination of the vectors in the set.
 2
2


v1  0, v 2  2
a) Put the column vectors in a matrix and
3
4
perform elementary row operations:
1 0 1 1 
1 0 1 1 
1 0 1 1 
0 1 1 2R3*  R3  R1  0 1 1 2R3*  R3  R2  0 1 1 2 






1 1 2 3
0 1 1 2
0 0 0 0
This has 2 pivot columns, so there are 2 linearly independent vectors in the set.
In other words, the span of this set is 2-dimensional.
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
8) Consider the following set of vectors in R3

 1
0
 1
 1 





 
 
 
 
a1  0, a2   1, a3   1, a4  2 

 1
 1
2
3 

a)What is the dimension of the span of this set of vectors?
b)Is the span of this set of vectors R3?
c)Is this a basis for R3?
d)Is each of the following vectors within the span of this set?
If so, express them as a linear combination of the vectors in the set.
 2
2


v1  0, v 2  2
a) Put the column vectors in a matrix and
3
4
perform elementary row operations:
1 0 1 1 
1 0 1 1 
1 0 1 1 
0 1 1 2R3*  R3  R1  0 1 1 2R3*  R3  R2  0 1 1 2 






1 1 2 3
0 1 1 2
0 0 0 0
This has 2 pivot columns, so there are 2 linearly independent vectors in the set.
In other words, the span of this set is 2-dimensional.
b)
No, the span is 2-dimensional, and R3 is 3-dimensional.
c)
No, this is not a basis because is does not span R3.
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
8) Consider the following set of vectors in R3

 1
0
 1
 1 





 
 
 
 
a1  0, a2   1, a3   1, a4  2 

 1
 1
2
3 

a)What is the dimension of the span of this set of vectors?
b)Is the span of this set of vectors R3?
c)Is this a basis for R3?
d)Is each of the following vectors within the span of this set?
If so, express them as a linear combination of the vectors in the set.
 2
2


v1  0, v 2  2
a) Put the column vectors in a matrix and
3
4
perform elementary row operations:
1 0 1 1 
1 0 1 1 
1 0 1 1 
0 1 1 2R3*  R3  R1  0 1 1 2R3*  R3  R2  0 1 1 2 






1 1 2 3
0 1 1 2
0 0 0 0
This has 2 pivot columns, so there are 2 linearly independent vectors in the set.
In other words, the span of this set is 2-dimensional.
b)
No, the span is 2-dimensional, and R3 is 3-dimensional.
c)
No, this is not a basis because is does not span R3.
2
 1
d) We have several options for this part. One way is to make a good guess. 2  2   1
 
If you notice that v2 is just 2 times a3 then we have our answer for that one:  
4
2
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
8) Consider the following set of vectors in R3

 1
0
 1
 1 





 
 
 
 
a1  0, a2   1, a3   1, a4  2 

 1
 1
2
3 

a)What is the dimension of the span of this set of vectors?
b)Is the span of this set of vectors R3?
c)Is this a basis for R3?
d)Is each of the following vectors within the span of this set?
If so, express them as a linear combination of the vectors in the set.
 2
2


v1  0, v 2  2
a) Put the column vectors in a matrix and
3
4
perform elementary row operations:
1 0 1 1 
1 0 1 1 
1 0 1 1 
0 1 1 2R3*  R3  R1  0 1 1 2R3*  R3  R2  0 1 1 2 






1 1 2 3
0 1 1 2
0 0 0 0
This has 2 pivot columns, so there are 2 linearly independent vectors in the set.
In other words, the span of this set is 2-dimensional.
b)
No, the span is 2-dimensional, and R3 is 3-dimensional.
c)
No, this is not a basis because is does not span R3.
2
 1
d) We have several options for this part. One way is to make a good guess. 2  2   1
 
If you notice that v2 is just 2 times a3 then we have our answer for that one:  
4
2
For v1 there is no obvious answer, so we need
to try to solve for it. Since we know that our
span is only 2-dimensional, we only need two
basis vectors. Choose any 2 independent
vectors from our set, say a1 and a2, and try to
write v1 as a linear combination of them.
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
8) Consider the following set of vectors in R3

 1
0
 1
 1 





 
 
 
 
a1  0, a2   1, a3   1, a4  2 

 1
 1
2
3 

a)What is the dimension of the span of this set of vectors?
b)Is the span of this set of vectors R3?
c)Is this a basis for R3?
d)Is each of the following vectors within the span of this set?
If so, express them as a linear combination of the vectors in the set.
 2
2


v1  0, v 2  2
a) Put the column vectors in a matrix and
3
4
perform elementary row operations:
1 0 1 1 
1 0 1 1 
1 0 1 1 
0 1 1 2R3*  R3  R1  0 1 1 2R3*  R3  R2  0 1 1 2 






1 1 2 3
0 1 1 2
0 0 0 0
This has 2 pivot columns, so there are 2 linearly independent vectors in the set.
In other words, the span of this set is 2-dimensional.
b)
No, the span is 2-dimensional, and R3 is 3-dimensional.
c)
No, this is not a basis because is does not span R3.
2
 1
d) We have several options for this part. One way is to make a good guess. 2  2   1
 
If you notice that v2 is just 2 times a3 then we have our answer for that one:  
4
2


For v1 there is no obvious answer, so we need 
v1  c1a1  c 2a2
to try to solve for it. Since we know that our
2  c1 
span is only 2-dimensional, we only need two 2 c1  0 
0   0   c   0  c Inconsistent
basis vectors. Choose any 2 independent
2 
     2
vectors from our set, say a1 and a2, and try to
3 c1 c 2  3  c1  c 2 
write v1 as a linear combination of them.
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
8) Consider the following set of vectors in R3

 1
0
 1
 1 





 
 
 
 
a1  0, a2   1, a3   1, a4  2 

 1
 1
2
3 

a)What is the dimension of the span of this set of vectors?
b)Is the span of this set of vectors R3?
c)Is this a basis for R3?
d)Is each of the following vectors within the span of this set?
If so, express them as a linear combination of the vectors in the set.
 2
2


v1  0, v 2  2
a) Put the column vectors in a matrix and
3
4
perform elementary row operations:
1 0 1 1 
1 0 1 1 
1 0 1 1 
0 1 1 2R3*  R3  R1  0 1 1 2R3*  R3  R2  0 1 1 2 






1 1 2 3
0 1 1 2
0 0 0 0
This has 2 pivot columns, so there are 2 linearly independent vectors in the set.
In other words, the span of this set is 2-dimensional.
b)
No, the span is 2-dimensional, and R3 is 3-dimensional.
c)
No, this is not a basis because is does not span R3.
2
 1
d) We have several options for this part. One way is to make a good guess. 2  2   1
 
If you notice that v2 is just 2 times a3 then we have our answer for that one:  
4
2


For v1 there is no obvious answer, so we need 
v1  c1a1  c 2a2
to try to solve for it. Since we know that our
2  c1 
span is only 2-dimensional, we only need two 2 c1  0 
0   0   c   0  c Inconsistent
basis vectors. Choose any 2 independent
2 
     2
vectors from our set, say a1 and a2, and try to
3 c1 c 2  3  c1  c 2 
write v1 as a linear combination of them.
We get an inconsistent set of equations, so v1 is not in the span.
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
9) Consider the following set of vectors in P2:
a

2 
2 

1

x

x
,
a

1

x

x
, a3  2  x
1
2
a)What is dimension of the span of this set?
b)Does this set span P2?
c)Is this set a basis for P2?
d)Express each of the standard basis vectors {1, x, x2} as
a linear combination of the vectors in the set.
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
9) Consider the following set of vectors in P2:
a

2 
2 

1

x

x
,
a

1

x

x
, a3  2  x
1
2
a)What is dimension of the span of this set?
b)Does this set span P2?
c)Is this set a basis for P2?
d)Express each of the standard basis vectors {1, x, x2} as
a linear combination of the vectors in the set.
a) These are 3 independent vectors. We can see this if we try to form a linear combination of them,
and find that there is no way to cancel them out:
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
9) Consider the following set of vectors in P2:
a

2 
2 

1

x

x
,
a

1

x

x
, a3  2  x
1
2
a)What is dimension of the span of this set?
b)Does this set span P2?
c)Is this set a basis for P2?
d)Express each of the standard basis vectors {1, x, x2} as
a linear combination of the vectors in the set.
a) These are 3 independent vectors. We can see this if we try to form a linear combination of them,
and find that there is no way to cancel them out: c  a  c  a  c  a  0 ?
1
1
2
2
3
3
c1(1  x  x 2 )  c 2 (1  x  x 2 )  c 3 (2  x )  0 ?
(c1  c 2 )x 2  (c1  c 2  c 3 )x  (c1  c 2  2c 3 )  0 ?
c1  c 2  0


c1  c 2  c 3  0 c1  c 2  c 3  0  independent
c1  c 2  2c 3  0
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For Campus Learning
Assistance Services at UCSB
9) Consider the following set of vectors in P2:
a

2 
2 

1

x

x
,
a

1

x

x
, a3  2  x
1
2
a)What is dimension of the span of this set?
b)Does this set span P2?
c)Is this set a basis for P2?
d)Express each of the standard basis vectors {1, x, x2} as
a linear combination of the vectors in the set.
a) These are 3 independent vectors. We can see this if we try to form a linear combination of them,
and find that there is no way to cancel them out: c  a  c  a  c  a  0 ?
1
b) P2 is a 3-dimensional space, and we have 3
independent vectors in P2 – yes they span P2.
1
2
2
3
3
c1(1  x  x 2 )  c 2 (1  x  x 2 )  c 3 (2  x )  0 ?
(c1  c 2 )x 2  (c1  c 2  c 3 )x  (c1  c 2  2c 3 )  0 ?
c1  c 2  0


c1  c 2  c 3  0 c1  c 2  c 3  0  independent
c1  c 2  2c 3  0
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Assistance Services at UCSB
9) Consider the following set of vectors in P2:
a

2 
2 

1

x

x
,
a

1

x

x
, a3  2  x
1
2
a)What is dimension of the span of this set?
b)Does this set span P2?
c)Is this set a basis for P2?
d)Express each of the standard basis vectors {1, x, x2} as
a linear combination of the vectors in the set.
a) These are 3 independent vectors. We can see this if we try to form a linear combination of them,
and find that there is no way to cancel them out: c  a  c  a  c  a  0 ?
1
b) P2 is a 3-dimensional space, and we have 3
independent vectors in P2 – yes they span P2.
c) P2 is a 3-dimensional space, and we have 3
independent vectors in P2 – yes they form a
basis for P2.
1
2
2
3
3
c1(1  x  x 2 )  c 2 (1  x  x 2 )  c 3 (2  x )  0 ?
(c1  c 2 )x 2  (c1  c 2  c 3 )x  (c1  c 2  2c 3 )  0 ?
c1  c 2  0


c1  c 2  c 3  0 c1  c 2  c 3  0  independent
c1  c 2  2c 3  0
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
9) Consider the following set of vectors in P2:
a

2 
2 

1

x

x
,
a

1

x

x
, a3  2  x
1
2
a)What is dimension of the span of this set?
b)Does this set span P2?
c)Is this set a basis for P2?
d)Express each of the standard basis vectors {1, x, x2} as
a linear combination of the vectors in the set.
a) These are 3 independent vectors. We can see this if we try to form a linear combination of them,
and find that there is no way to cancel them out: c  a  c  a  c  a  0 ?
1
b) P2 is a 3-dimensional space, and we have 3
independent vectors in P2 – yes they span P2.
c) P2 is a 3-dimensional space, and we have 3
independent vectors in P2 – yes they form a
basis for P2.
1
2
2
3
3
c1(1  x  x 2 )  c 2 (1  x  x 2 )  c 3 (2  x )  0 ?
(c1  c 2 )x 2  (c1  c 2  c 3 )x  (c1  c 2  2c 3 )  0 ?
c1  c 2  0


c1  c 2  c 3  0 c1  c 2  c 3  0  independent
c1  c 2  2c 3  0
d) Form a linear combination of a1, a2 and a3:
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
9) Consider the following set of vectors in P2:
a

2 
2 

1

x

x
,
a

1

x

x
, a3  2  x
1
2
a)What is dimension of the span of this set?
b)Does this set span P2?
c)Is this set a basis for P2?
d)Express each of the standard basis vectors {1, x, x2} as
a linear combination of the vectors in the set.
a) These are 3 independent vectors. We can see this if we try to form a linear combination of them,
and find that there is no way to cancel them out: c  a  c  a  c  a  0 ?
1
b) P2 is a 3-dimensional space, and we have 3
independent vectors in P2 – yes they span P2.
c) P2 is a 3-dimensional space, and we have 3
independent vectors in P2 – yes they form a
basis for P2.
1
2
2
3
3
c1(1  x  x 2 )  c 2 (1  x  x 2 )  c 3 (2  x )  0 ?
(c1  c 2 )x 2  (c1  c 2  c 3 )x  (c1  c 2  2c 3 )  0 ?
c1  c 2  0


c1  c 2  c 3  0 c1  c 2  c 3  0  independent
c1  c 2  2c 3  0
d) Form a linear combination of a1, a2 and a3:



c1  a1  c 2  a2  c 3  a3  1  (c1  c 2 )x 2  (c1  c 2  c 3 )x  (c1  c 2  2c 3 )  0 x 2  0 x  1
 c1  c 2  0

  c1  c 2  c 3  0  c1   21 ; c 2   21 ; c 3  1
c  c  2c  1
2
3
 1



1   21 a1  21 a2  a3
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
9) Consider the following set of vectors in P2:
a

2 
2 

1

x

x
,
a

1

x

x
, a3  2  x
1
2
a)What is dimension of the span of this set?
b)Does this set span P2?
c)Is this set a basis for P2?
d)Express each of the standard basis vectors {1, x, x2} as
a linear combination of the vectors in the set.
a) These are 3 independent vectors. We can see this if we try to form a linear combination of them,
and find that there is no way to cancel them out: c  a  c  a  c  a  0 ?
1
b) P2 is a 3-dimensional space, and we have 3
independent vectors in P2 – yes they span P2.
c) P2 is a 3-dimensional space, and we have 3
independent vectors in P2 – yes they form a
basis for P2.
1
2
2
3
3
c1(1  x  x 2 )  c 2 (1  x  x 2 )  c 3 (2  x )  0 ?
(c1  c 2 )x 2  (c1  c 2  c 3 )x  (c1  c 2  2c 3 )  0 ?
c1  c 2  0


c1  c 2  c 3  0 c1  c 2  c 3  0  independent
c1  c 2  2c 3  0
d) Form a linear combination of a1, a2 and a3:



c1  a1  c 2  a2  c 3  a3  1  (c1  c 2 )x 2  (c1  c 2  c 3 )x  (c1  c 2  2c 3 )  0 x 2  0 x  1
 c1  c 2  0

  c1  c 2  c 3  0  c1   21 ; c 2   21 ; c 3  1
c  c  2c  1
2
3
 1



1   21 a1  21 a2  a3
Similar analysis yields the other 2 combinations.
 

x  a1  a2  a3



x2  21 a1  21 a2  0a3
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB