Lecture 15: Linear Combinations:
Marvin and Marilyn Get Their Meals, Bees,
Chemical Reaction Systems, and Consistency
Winfried Just
Department of Mathematics, Ohio University
September 28, 2015
Winfried Just, Ohio University
MATH3200, Lecture 15: Linear Combinations
A health-conscious couple
Marilyn and Marvin decide to go on Dr. Losit’s new scientifically
proven diet. According to this expert, they should eat exactly 50
grams of sugar, exactly 300 grams of protein, and exactly 100
grams of fat per day, and restrict themselves to one meal per day
prepared from a mixture of a wide range of products offered by his
company.
Being on a budget and having somewhat different tastes, they
decide to purchase Losit-Quick (strawberry taste, which they both
like), Losit-Fast (Marylin’s beloved watermelon taste that Marvin
cannot stand), and Losit-Easy (beer-flavored for Marvin, detested
by Marylin).
Ohio University – Since 1804
Winfried Just, Ohio University
Department of Mathematics
MATH3200, Lecture 15: Linear Combinations
The products are in, now what?
After receiving the products in the mail and reading the fine print
on the labels, they discover that:
One serving of Losit-Quick contains 20 grams of sugar, 200
grams of protein, and 20 grams of fat.
One serving of Losit-Fast contains 15 grams of sugar, 50
grams of protein, and 40 grams of fat.
One serving of Losit-Easy contains 25 grams of sugar, 150
grams of protein, and 60 grams of fat.
How should they compose their meals if they want to strictly
follow the guidelines?
Each of our protagonists sits down with a pencil and paper and
tries to figure it out.
Ohio University – Since 1804
Winfried Just, Ohio University
Department of Mathematics
MATH3200, Lecture 15: Linear Combinations
Marilyn’s reasoning
Marilyn reasons as follows: I will need x1 servings of Losit-Quick
and x2 servings of Losit-Fast.
The numbers x1 , x2 need to be chosen in such a way that the total
amounts of sugar, protein, and fat add up to exactly the
recommended values.
They must be solutions of the following system of linear equations,
where all quantities are expressed in grams:
20x1 + 15x2 = 50
200x1 + 50x2 = 300
20x1 + 40x2 = 100
(the total amount of sugar)
(the total amount of protein)
(the total amount of fat)
She is really good at Gaussian elimination and uses it to quickly
solve the system.
Ohio University – Since 1804
Winfried Just, Ohio University
Department of Mathematics
MATH3200, Lecture 15: Linear Combinations
Marilyn’s solution




20 15 50 subtract 10(row 1) from row 2 20
15
50
200 50 300
 0 −100 −200
−→
20 40 100
20
40
100




20
15
50
20
15
50
1 from row 3 
 0 −100 −200 subtract row
0 −100 −200
−→
20
40
100
0
25
50

20
15
 0 −100
0
25


50
1
divide row 1 by 20 
−200
0
−→
50
0
Ohio University – Since 1804
Winfried Just, Ohio University
0.75
−100
25

2.5
−200
50
Department of Mathematics
MATH3200, Lecture 15: Linear Combinations
Marilyn’s solution, continued




1 0.75
2.5
1 0.75 2.5
2 by -100 
0 −100 −200 divide row
0
1
2
−→
0
25
50
0 25
50




1 0.75 2.5
1 0.75 2.5
divide row 3 by 25 
0
1
2
0
1
2
−→
0 25
50
0
1
2




1 0.75 2.5
1 0.75 2.5
subtract row 2 from row 3 
0
1
2
0
1
2
−→
0
1
2
0
0
0
Ohio University – Since 1804
Winfried Just, Ohio University
Department of Mathematics
MATH3200, Lecture 15: Linear Combinations
Marilyn’s starts cooking
Marilyn has found the augmented matrix


1 0.75 2.5
~
1
2
Ab = 0
0
0
0
of the equivalent system
x1 + 0.75x2 = 2.5
x2 = 2
0=0
This tells her that she needs to mix x2 = 2 servings of Losit-Fast
with x1 = 1 serving of Losit-Quick.
She starts preparing her meal.
Ohio University – Since 1804
Winfried Just, Ohio University
Department of Mathematics
MATH3200, Lecture 15: Linear Combinations
In the meantime Marvin ...
who isn’t a big fan of Gaussian elimination tried randomly guessing
an appropriate mixture of Losit-Fast and Losit-Easy, without much
success.
He likes vector notation though, and reasons as follows:
Let ~vQ ,~vE ,~vM be the vectors of amounts of sugar, protein, and fat
in one serving of Losit-Quick, Losit-Easy, and in the meal that I
need to compose. Then


20
~vQ = 200
20


25
~vE = 150
60

~vM

50
= 300
100
I need to find scalars dQ , dE such that ~vM is the linear combination
dQ ~vQ + dE ~vE = ~vM ,
where dQ , dE are the numbers of servings of Losit-Quick and
Losit-Easy that I need for my meal.
Ohio University – Since 1804
Winfried Just, Ohio University
Department of Mathematics
MATH3200, Lecture 15: Linear Combinations
Marvin is getting hungry
It remains to find the actual numbers dQ , dE such that ~vM is the
linear combination
dQ ~vQ + dE ~vE = ~vM .
Hmmm.
????
Eventually hunger prevails over pride and Marvin decides that his
notation and the phrase linear combination look sufficiently
impressive so that it wouldn’t be too embarassing to ask:
“Hey, sweethart, I have practically figured it out. Could you just
give me a hand with these boring calculations?”
Marilyn tells him: “Sorry, bud. You will need to add some of my
Losit-Fast.”
Homework 29: Is Marilyn right or just being nasty?
Homework 30: If Marilyn is right, what is the minimum amount
of Losit-Fast that Marvin will need to digest? What is the
maximum amount of Losit-Easy that he can add?
Ohio University – Since 1804
Winfried Just, Ohio University
Department of Mathematics
MATH3200, Lecture 15: Linear Combinations
Solution for Homework 29
Marilyn argues as follows: You will need x1 servings of Losit-Quick
and x2 servings of Losit-Easy.
The numbers x1 , x2 need to be chosen in such a way that the total
amounts of sugar, protein, and fat add up to exactly the
recommended values.
They must be solutions of the following system of linear equations,
where all quantities are expressed in grams:
20x1 + 25x2 = 50
200x1 + 150x2 = 300
20x1 + 60x2 = 100
(the total amount of sugar)
(the total amount of protein)
(the total amount of fat)
She now does a Gaussian elimination for this system:
Ohio University – Since 1804
Winfried Just, Ohio University
Department of Mathematics
MATH3200, Lecture 15: Linear Combinations
Gaussian elimination for Homework 29

20
200
20
25
150
60


50 subtract 10(row 1) from row 2 20
0
300
−→
100
20
25
−100
60

50
−200
100




20
25
50
20
25
50
1 from row 3 
 0 −100 −200 subtract row
0 −100 −200
−→
20
60
100
0
35
50

20
25
 0 −100
0
35


50
1
divide row 1 by 20 
−200
0
−→
50
0
Ohio University – Since 1804
Winfried Just, Ohio University
1.25
−100
35

2.5
−200
50
Department of Mathematics
MATH3200, Lecture 15: Linear Combinations
Gaussian elimination for Homework 29, continued

1
0
0

1
0
0
1.25
−100
35
1.25
1
35

1
0
0


2.5
1
divide row 2 by -100 
−200
0
−→
50
0
1.25
1
35

2.5
2
50


2.5 subtract 35(row 2) from row 3 1 1.25
0
2
1
−→
50
0
0
1.25
1
0


2.5
1
divide
row
3
by
-20
0
2 
−→
−20
0
Ohio University – Since 1804
Winfried Just, Ohio University
1.25
1
0

2.5
2 
−20

2.5
2
1
Department of Mathematics
MATH3200, Lecture 15: Linear Combinations
Marvin is out of luck
Marilyn has found the augmented matrix


1 1.25 2.5
~
1
2
Ab = 0
0
0
1
of the equivalent system
x1 + 1.25x2 = 2.5
x2 = 2
0=1
This system is inconsistent. Marvin cannot compose his meal
without adding some Losit-Fast.
Or so Marilyn says.
Ohio University – Since 1804
Winfried Just, Ohio University
Department of Mathematics
MATH3200, Lecture 15: Linear Combinations
Marvin is not conceding yet
Marvin: (looking dejected) ”Ok, we can’t figure it out with your
Gaussian elimination. But what if we approach the problem from
my angle and look for coefficients of a linear combination instead?”
Marilyn: ”Er, well, ... I don’t really understand your linear
combination stuff. Can you explain it to me?”
Marvin: (becomes animated again) “This is really quite simple. It
works like this: If you have any vectors ~v1 ,~v2 , . . . ,~vn and another
~ , then w
~ is a linear combination of ~v1 ,~v2 , . . . ,~vn if, and
vector w
only if, there exist numbers (or as the textbook says, scalars)
d1 , d2 , . . . , dn such that
~ .”
d1~v1 + d2~v2 + · · · + dn~vn = w
Marilyn: “Can you give me an example?”
Marvin: Sure. Let ~vQ ,~vF ,~vE ,~vM be the vectors of amounts of
sugar, protein, and fat in one serving of Losit-Quick, Losit-Fast,
Losit-Easy, and in the meal that each of us needs to compose.
Ohio University – Since 1804
Winfried Just, Ohio University
Department of Mathematics
MATH3200, Lecture 15: Linear Combinations
Marvin explains it really nicely
Marvin: “Then ~vQ + 2~vF = ~vM . So if we let d1 = 1, d2 = 2, then
we see that ~vM is a linear combination of ~vQ and ~vF .”
Marvin continues: “We can look at this coordinate by coordinate.


20
~vQ = 200
20
 
15
~vF = 50
40


25
~vE = 150
60

~vM

50
= 300
100
So we have


  
  
20
15
d1 20 + d2 15
50
d1~vQ + d2~vF = d1 200 + d2 50 = d1 200 + d2 50 = 300 = ~vM
20
40
d1 20 + d2 40
100
When d1 = 1 and d2 = 2 it all adds up.”
Marilyn: “Wow! Can you write this for me as a system of linear
equations? You know, I find systems easier to understand.”
Ohio University – Since 1804
Winfried Just, Ohio University
Department of Mathematics
MATH3200, Lecture 15: Linear Combinations
Marvin translates it into systems
Marvin: “When we look at the coordinates that give us the
amounts of sugar, protein, and fat one by one, we get the following
system of linear equations:”
20d1 + 15d2 = 50
200d1 + 50d2 = 300
20d1 + 40d2 = 100
(the total amount of sugar)
(the total amount of protein)
(the total amount of fat)
Marilyn: “Could I use x1 , x2 instead of d1 , d2 here?”
Marvin: “Absolutely! You see, it doesn’t matter which letters you
use for your variables. These are just names.”
Marilyn: “So the coefficients d1 = 1, d2 = 2 that you showed me
here are precisely the numbers that I got from my solution of the
system by Gaussian elimination?”
Marvin: “Right!”
Ohio University – Since 1804
Winfried Just, Ohio University
Department of Mathematics
MATH3200, Lecture 15: Linear Combinations
Marvin’s meal
Marilyn: (with a sly smile) “So how about your meal?”
Marvin: “I need coefficients d1 , d2 such that d1~vQ + d2~vE = ~vM .
We can look at this coordinate by coordinate.”



 
 

20
25
d1 20 + d2 25
50
d1~vQ +d2~vE = d1 200 +d2 150 = d1 200 + d2 150 = 300 = ~vM
20
60
d1 20 + d2 60
100
Marvin: ”Since you like systems better ... ”
Marilyn: “And I like x1 , x2 better than d1 , d2 .”
Marvin: (ever the gentleman) “We will write it as”
20x1 + 25x2 = 50
200x1 + 150x2 = 300
20x1 + 60x2 = 100
(the total amount of sugar)
(the total amount of protein)
(the total amount of fat)
(A moment of profound silence ... )
Ohio University – Since 1804
Winfried Just, Ohio University
Department of Mathematics
MATH3200, Lecture 15: Linear Combinations
Marvin’s meal, Plan B. Solution of Homework 30
Marvin: (Deep breath. Takes it like a man.) “Now you can see
that my approach is really the same as yours, only using a more
condensed notation and different terminology. Since the system is
inconsistent, we will need to use ingredient Losit-Fast.
We can find the serving sizes x1 , x2 , x3 by solving the following
system:”
20x1 + 15x2 + 25x3 = 50
200x1 + 50x2 + 150x3 = 300
20x1 + 40x2 + 60x3 = 100
(the total amount of sugar)
(the total amount of protein)
(the total amount of fat)
Marilyn interjects: ”By using Gaussian elimination.”
Marvin: ”We can leave this an an exercise for the students. I can
see right off the bat that the solution is x1 = 1, x2 = 2, x3 = 0.”
(It dawns upon Marvin what x3 = 0 means.
No beer-flavored Losit-Easy. Long and profound silence ... )
Ohio University – Since 1804
Winfried Just, Ohio University
Department of Mathematics
MATH3200, Lecture 15: Linear Combinations
Marvin’s meal, Plan C. Maybe.
Marilyn: (soothingly) “Maybe there is a different solution?”
Homework 31: Solve the system on the previous slide by using
Gaussian elimination. Determine all solutions.
Marilyn: (gently) “Should we try your approach?”
Marvin: (with a glimmer of hope returning to his face) “Good
idea. Let’s assume that [y1 , y2 , y3 ]T is a different solution.
Then ~vM can be expressed as a linear combination of ~vQ ,~vF ,~vE in
at least two different ways:”
x1~vQ + x2~vF + x3~vE = ~vM = y1~vQ + y2~vF + y3~vE .
Marilyn: ”When we subtract the right-hand side from the left-hand
side, ~vM cancels and we get
(x1 − y2 )~vQ + (x2 − y1 )~vF + (x3 − y3 )~vE = ~0,
where ~0 is the zero vector that has entry 0 in all coordinates.”
Marvin and Marilyn: “What does this tells us about the meal?”
Ohio University – Since 1804
Winfried Just, Ohio University
Department of Mathematics
MATH3200, Lecture 15: Linear Combinations
Marvin’s meal, Plan D.
Marvin: ”We can simplify the notation by writing the equation
(x1 − y2 )~vQ + (x2 − y1 )~vF + (x3 − y3 )~vE = ~0, in the form
c1~vQ + c2~vF + c3~vE = ~0,
where c1 = x1 − y2 , c2 = x2 − y2 , c3 = x3 − y3 .”
Marilyn: ”If [x1 , x2 , x3 ]T and [y1 , y2 , y3 ]T are really different
solutions, then not all of the constants c1 , c2 , c3 can be zero.”
Marvin: ”So this means a second solution can exist only if the
vectors ~vQ ,~vF ,~vE are what is called linearly dependent.
But how can we find out whether or not this is the case?”
Marilyn: ”We could use Gauss ...” (senses a looming relationship
crisis and bites her tongue) “You know what, Prof. Just is going to
teach us some great methods for determining linear dependence in
a few days. So how about postponing the diet and having some
burgers with fries”
Marvin: “and a couple of beers!”
Just, Ohio University
Ohio University – Since Winfried
1804
MATH3200, Lecture 15: Linear
Combinations
Department
of Mathematics
Homework problems
Our protagonists enjoy the restored harmony of their relationship
in a friendly neighborhood pub. Let’s not intrude and get back to
work. Perhaps we will meet them again some other day.
Homework 32: What does the solution of Homework 31 tell you
about linear dependence of the vectors ~vQ ,~vF , and ~vE ?
Homework 33: To spare Marvin the ordeal of ingesting Losit-Fast
or doing without Losit-Easy, our couple will buy another product
from the company. Let’s call it Losit-Other. There are many
options, and our protagonists will look at the corresponding
vector ~vO first. By now they know that in order for Marvin to be
able to combine Losit-Other with Losit-Quick and Losit-Easy for
meals to his liking, this vector must satisfy several conditions. How
can these conditions be expressed in the language of linear
combinations?
Ohio University – Since 1804
Winfried Just, Ohio University
Department of Mathematics
MATH3200, Lecture 15: Linear Combinations
Linear combinations: The definition
Definition
~ is a linear combination of vectors ~v1 ,~v2 , . . . ,~vn if there
A vector w
exist scalars d1 , d2 , . . . , dn such that
~ = d1~v1 + d2~v2 + · · · + dn~vn .
w
The vectors in the above definition must all be of the same order to
make addition meaningful, but they could be either row vectors (of
order 1 × m) or column vectors (of order m × 1). Their common
dimension m can be smaller than, equal to, or larger than n.
The zero vector ~0 of a given order is always a linear combination of
any nonempty set of vectors {~v1 ,~v2 , . . . ,~vn } of the same order, as
we can choose the coefficients dj in the above definition to be all 0.
Ohio University – Since 1804
Winfried Just, Ohio University
Department of Mathematics
MATH3200, Lecture 15: Linear Combinations
An example of linear combinations: A bee
Assume a bee colony nests in a hollow tree.
Let [0, 0, 0] denote the position of the hive.
A foraging bee travels t1 = 3 time units in the direction of vector
~v1 = [2, 3, 0.5] and then t2 = 2 time units in the direction of vector
~v2 = [0, −2, −0.5] and discovers some tasty blossoms on another
tree.
We can express the position of these blossoms as the linear
combination
~ = t1~v1 + t2~v2 = 3[2, 3, 0.5] + 2[0, −2, −0.5] = [6, 5, 0.5].
w
On the way back to the hive, our foraging bee discovers a better
route to the blossoms, and upon return to the hive, she shares this
info with her sisters by performing a wiggle dance.
Ohio University – Since 1804
Winfried Just, Ohio University
Department of Mathematics
MATH3200, Lecture 15: Linear Combinations
Two sisters
After the dance, sister Beezee of the foraging bee travels first t3
time units in the direction of vector ~v3 = [1, 1, 0.2] and then t4
time units in the direction of vector ~v4 = [2, 1, −0.3],
while sister Buzzy travels first t5 time units in the direction of
vector ~v5 = [3, 3, 0.2] and then t6 time units in the direction of
vector ~v6 = [0, −1, −0.1].
Homework 34: (a) One of the two sisters didn’t pay attention
during the wiggle dance and never found the blossoms. Which one
was it, and how can you tell?
(b) How long did it take the other sister to arrive at the blossoms?
Ohio University – Since 1804
Winfried Just, Ohio University
Department of Mathematics
MATH3200, Lecture 15: Linear Combinations
Another example: A chemical reaction network
Assume a vat contains a mixture of chemical compounds (or, more
generally, species) A, B, C , D.
Chemists know that the following reactions could in principle occur:
−→
A + 2B ←− 2C
−→
A + 2C ←− 2D
−→
A + B ←− D
−→
B + D ←− 2C
Homework 35: If the compounds in this system were oxygen O2 ,
carbon monoxide CO, carbon dioxide CO2 and carbon C, how
would these match to the letter symbols A, B, C , D above? Which
directions of the above reactions would be ergetically implausible
under normal conditions?
We will from now on just work with letters A, B, C , D that could
represent various chemical species and assume that each reaction
could proceed in either direction.
Just, Ohio University
Ohio University – Since Winfried
1804
MATH3200, Lecture 15: Linear
Combinations
Department
of Mathematics
Hint for Homework 35
If the chemical species in this system are oxygen O2 , carbon
monoxide CO, carbon dioxide CO2 and carbon C, the only possible
reactions are
−→
O2 + C ←− CO2
−→
O2 + 2C ←− 2CO
−→
O2 + 2CO ←− 2CO2
−→
C + CO2 ←− 2CO
Ohio University – Since 1804
Winfried Just, Ohio University
Department of Mathematics
MATH3200, Lecture 15: Linear Combinations
Acknowledgement
This example is loosely adapted from
Gerhard Just et al. (1988) Mathematik für Chemiker.
(Mathematics for Chemists.) 3rd edition. VEB Deutscher Verlag
für Grundstoffindustrie.
It was developed by our family business in discussions with the first
author of this book.
Ohio University – Since 1804
Winfried Just, Ohio University
Department of Mathematics
MATH3200, Lecture 15: Linear Combinations
A chemical reaction network, continued
Chemists know that the following reactions could in principle occur:
−→
A + 2B ←− 2C
−→
A + B ←− D
−→
A + 2C ←− 2D
−→
B + D ←− 2C
However, they do not know which of these reactions actually occur
and want to figure it out by observing how the the concentrations
change over time.
The initial concentrations (in moles per volume of the vat) will be
denoted by [A]0 , [B]0 , [C ]0 , [D]0 .
Now suppose concentrations are measured again after one hour
and denoted by [A]1 , [B]1 , [C ]1 , [D]1 .
~ = [[A]1 − [A]0 , [B]1 − [B]0 , [C ]1 − [C ]0 , [D]1 − [D]0 ].
Let w
~ represents the net change in concentrations. If
The vector w
some coordinate [X ]1 − [X ]0 is positive, then a net production of
compound X was observed, if some coordinate [X ]1 − [X ]0 is
negative, then a net consumption of compound X was observed.
Ohio University – Since Winfried
1804
Just, Ohio University
Department
of Mathematics
MATH3200, Lecture 15: Linear
Combinations
Which reactions do occur?
−→
A + 2B ←− 2C
−→
A + 2C ←− 2D
−→
A + B ←− D
−→
B + D ←− 2C
Chemists know that only the above reactions could occur, but
suspect that not all of them in fact do.
(When) can one deduce, based exclusively on observing the
~ of net change of concentrations, that at least one,
vector w
at least two, or at least three of these reactions do occur?
(When) can we further narrow down the possibilities based
on energetic constraints that rule out one direction for each
of the four reactions?
Ohio University – Since 1804
Winfried Just, Ohio University
Department of Mathematics
MATH3200, Lecture 15: Linear Combinations
Vectors for each reaction
1
2
3
4
−→
A + 2B ←− 2C
−→
A + 2C ←− 2D
−→
A + B ←− D
−→
B + D ←− 2C
Consider the vector ~v1 = [−1, −2, 2, 0].
This vector represents the net change in concentrations if only the
first reaction occurs and consumes one mole of the first reactant,
which is compound (or species) A in this case.
Homework 36: (a) Find analogues ~v2 ,~v3 ,~v4 of the vector ~v1 for
reactions 2, 3, and 4.
(b) How is the vector
~ = [[A]1 − [A]0 , [B]1 − [B]0 , [C ]1 − [C ]0 , [D]1 − [D]0 ]
w
related to the vectors ~v1 ,~v2 ,~v3 ,~v4 ?
Ohio University – Since 1804
Winfried Just, Ohio University
Department of Mathematics
MATH3200, Lecture 15: Linear Combinations
Systems of linear equations as linear combinations
Consider a system of linear equations
a11 x1 + a12 x2 + · · · + a1n xn = b1
...
am1 x1 + am2 x2 + · · · + amn xn = bm
Let ~a1 , . . . , ~an be the column vectors of the coefficient matrix A,
and let ~b be the column vector that represents the right-hand side:


a11


~a1 =  ... 
am1




a12
a1n




~a2 =  ...  . . . ~an =  ... 
am2
amn


b1
. 
~b = 
 .. 
bm
Then the above system can be written as
x1~a1 + x2~a2 + · · · + xn~an = ~b.
Ohio University – Since 1804
Winfried Just, Ohio University
Department of Mathematics
MATH3200, Lecture 15: Linear Combinations
Systems of linear equations as linear combinations
Consider a system of linear equations
6x1 − x2 + x3 = 5
x2 − 7x3 = 0
−x1 + 2x2 − 3x3 = 9
Homework 37: Find vectors ~a1 , . . . , ~an such that the above
system can be written in the form
x1~a1 + x2~a2 + · · · + xn~an = ~b.
Ohio University – Since 1804
Winfried Just, Ohio University
Department of Mathematics
MATH3200, Lecture 15: Linear Combinations
Consistency of systems and linear combinations
a11 x1 + a12 x2 + · · · + a1n xn = b1
...
am1 x1 + am2 x2 + · · · + amn xn = bm
Let ~a1 , . . . , ~an be the column vectors of the coefficient matrix A,
and let ~b be the column vector that represents the right-hand side.
By writing the system as x1~a1 + x2~a2 + · · · + xn~an = ~b we
essentially get a one-line proof of the following result:
Theorem
In the notation introduced above, the system of linear equations is
consistent if, and only if, the vector ~b is a linear combination of
the vecors ~a1 , ~a2 , . . . , ~an .
Ohio University – Since 1804
Winfried Just, Ohio University
Department of Mathematics
MATH3200, Lecture 15: Linear Combinations
Solutions of systems and linear combinations
a11 x1 + a12 x2 + · · · + a1n xn = b1
...
am1 x1 + am2 x2 + · · · + amn xn = bm
Let ~a1 , . . . , ~an , ~b be as on the previous three slides.
By writing the system as x1~a1 + x2~a2 + · · · + xn~an = ~b,
we also see that solving the system is the same as finding all
vectors of coefficients x1 , x2 , . . . , xn that work for the linear
combination.
Homework 38: Reread the story of Marilyn and Marvin and
convince yourself that Marilyn had known this all along.
Ohio University – Since 1804
Winfried Just, Ohio University
Department of Mathematics
MATH3200, Lecture 15: Linear Combinations