PROFESSOR: Hello, I'm Linan.
Welcome to the recitation
of Linear Algebra.
It's my great
pleasure to guide you
through the first recitation.
In the first lecture, we
learned some important concepts.
We discussed how to view a
linear system of equations
from different points.
And we discussed the concepts
such as row picture, column
picture, and the form in matrix.
Some of them may be new to you.
So today, we're going to
use this simple example
to review those concepts.
We're going to work with
this simple system of two
equations with two unknowns.
So I would like you
to first solve it,
and then to find out the
associated row picture
and column picture
with this system.
After you're done,
we're also going
to discuss the matrix form
of this linear system.
Why don't you pause
the video now and try
to work them out on your own.
A good suggestion
would be, you can
try to sketch your answer in
an xy coordinate like this.
OK, I'll see you in a while.
I hope you have just
had some fun with it.
Let's work on it together.
Well, we're going to
solve these equations.
As you can see, we have
two unknowns, x and y.
And they have to satisfy these
two equations at the same time.
How would you solve it?
A very simple way would
be you substitute x by y,
in terms of y.
So let's do it this way.
So we use the second equation
to rewrite x as 2y minus 1.
So 2y minus 1.
Then you plug this into
the first equation.
This implies twice-- x would
be replaced by this-- 2 times
2y minus 1 plus y is equal to 3.
When you simplify
this equation here,
you will arrive at that
5y minus 2 is equal to 3.
That simply tells
you y is equal to 1.
If y is equal to 1, then
we go back to this formula.
We see that x is
also equal to 1.
That's it.
This is the answer to
this linear system.
And it's easy enough.
Both x and y are 1.
Now, let's try to find out its
row picture and column picture.
So I'm going to work on
this xy coordinate here.
First, let's look
at row picture.
Please review what
a row picture is.
So by row picture,
I mean you have
to look at this linear
system according to each row.
So what is the first row?
Well, the first row is an
equation with two unknowns.
So 2 times x plus
y is equal to 3.
What is this equation, exactly?
As you may remember,
this equation
actually gives you a
straight line in xy-plane.
So now let's put the line here.
I want a line that
satisfies 2x plus y is 3.
Let's first set x to be 0.
If x is 0, by the
first equation,
y should be 3, which is here.
Now, let's set x to be 1.
If x is 1, then y is also 1.
So what I have now are
two points on the line.
And this is enough,
because on the xy-plane,
two points will uniquely
determine a straight line.
And that's the line
we're looking for.
So all we need to do is to
connect these two points.
Let me try to draw
this line straight.
So this is the
line that is given
by 2x plus y is equal to 3.
This is the easiest way
to determine a line.
You just need to pick two
points that are on the line.
Then you connect them, and here
you have the straight line.
So that's the first row.
Now let's look at
the second row.
The second row is x
minus 2y is equal to -1.
Same thing.
We're going to locate two
points on the second line.
So again, let's set
x to be 0 first.
If x is 0, then y
should be 1/2, right?
So this line has to
cross this point.
That's x, 0; y, 1/2.
Then let's put x to be 1 again.
So if x is 1, then y is also 1.
So we're going to use
the same point here.
Now we have two points in the
second line to connect them.
So by connecting them, I will
have my second straight line.
So this is the line
corresponding to x minus 2y
is -1.
This is the row
picture, because we
have separated
the two equations,
and we look at
them respectively.
The first equation
gives me this line,
and the second equation
gives me this line.
Then what do I mean by
solving this linear system?
Well, we are putting
these two lines together,
and we are looking
at a point which
is on the first line and
second line at the same time.
Then clearly, that's the point
where they intersect, right?
We can see from the
answer over there,
the coordinate of this
point should be (1, 1).
This is also clear from the
construction of the two lines.
We have noticed
that this point is
on the first line and the
second line at the same time.
Two lines meeting
at the point (1, 1).
That's the row picture
of this linear system.
Now, let's move on to
the column picture.
Again, I will need
an xy coordinate.
So, where can I find my columns?
If you look at
the two equations,
then you focus on the
coefficient in front
of x in both equations.
What would that be?
Well, in the first equation,
I have a 2 in front of x.
In the second equation,
I have a 1 in front of x.
I want to put them together
as a column vector.
Let me call it v_1.
And I'm going to do the same
thing to the coefficients
in front of y.
In the first equation, I
have a 1 in front of y.
In the second equation, I
have a -2 in front of y.
Put them together.
Call it a column vector v_2.
These are the columns
I'm looking for.
Then what does that
linear system say?
Well, I have extracted the
coefficients in front of x.
Now I can consider x to be the
coefficient of this vector.
And same thing,
I'm going to view y
as the coefficient
of this vector.
Then what that linear system is
doing is just to sum them up.
That gives you the left-hand
side of the linear system.
What is the right-hand side?
Again, you put the two
constants as a column
vector, which is [3, -1].
That is the right-hand side.
So what I'm doing here is I'm
taking the linear combination
of v_1 and v_2.
And the coefficients are given
by x and y, respectively.
And I want the result of this
combination to be [3, -1].
Now let's incorporate
this into this picture.
I have a v_1, so I'm going
to draw a vector v_1.
x is 2, y is 1, so v_1 is here.
That's v_1.
And I need a v_2.
x is 1.
y is -2.
So that's my v_2.
I want to take the sum
of x multiple of v_1
and y multiple of v_2.
And I want the result
to be [3,  -1].
Well, taking a hint from
the previous consideration,
we know that both x
and y should be 1.
So I'm actually summing one
copy of v_1 and one copy of v_2.
So how do you indicate the
sum of these two vectors?
You complete the parallelogram
spanned by these two vectors.
Then the vector
given by the diagonal
is the sum of v_1 and v_2.
Is this vector going
to be [3,  -1]?
Well, we can check.
The x-coordinate will
be 2 plus 1, which is 3.
That's 2 plus 1, 3.
And the y-coordinate
will be 1 minus 2.
So 1 minus 2,
which should be -1.
That's it.
That's one multiple of v_1
and one multiple of v_2.
The sum will be [3, -1].
And that's the row picture.
Where does that that "x is equal
to y is equal to 1" come from?
It comes from solving
the linear system.
It comes from the row picture.
So here, we have found out
the row picture and the column
picture of this linear system.
What I would like to
mention is the matrix form
of this linear system.
So what is the matrix form?
What if I put these two
column vectors together?
So I want to put them
back to back, v_1 and v_2.
And I call this matrix to be
A. So if you write it out,
A should be given
by 2, 1; 1, -2.
Matrix A has v_1 and v_2
as its column vectors,
and it's a 2-by-2 matrix.
If I consider-- if I take
this into account, then
what will be the left-hand
side of the linear system?
In other words, what would
be the left-hand side
of this equation?
This is actually matrix
A multiplying a vector
given by x, y.
So that's [2, 1; 1, -2]
multiplying x and y.
So you put both unknowns
together as a column vector.
That's the left-hand
side of the equation.
And again, the
right-hand side is given
by this column vector, [3; -1].
This is the matrix form
of this linear system.
We can actually
solve this directly.
In other words, we can get
this unknown vector at once,
both x-coordinate
and y-coordinate.
Let's recall, if you have a
scalar equation like this--
let's say a is some
constant times x is unknown
is equal to b.
If a is non-zero,
what would be x?
So clearly, x
should be b over a.
I can also write it
as a inverse times b.
That's what we do
when we have numbers.
So here, what we
want to apply is
a similar idea but to matrix.
What you want to
find is a matrix
A inverse such that
A inverse times A
is equal to an identity
matrix, which is [1, 0; 0, 1].
This may be new to you,
but as you go further
into this course, this idea will
become more and more natural.
If such an inverse
matrix exists, then
what would be this vector here?
Then [x; y] will simply be
A inverse times [3; -1].
That will give you the answer.
Here, I'm not going
to go into detail,
but we will return to
this later in this course.
I hope this simple
example is helpful to you
in reviewing what you've
learned in the lecture.
Thank you for watching,
and I'm looking forward
to seeing you again.
