Topic three is systems of linear
equations.
So what I'm building up to, is I want to
be able to solve problems like this.
So I have three equations and three
unknowns.
So the unknowns are x, y and z.
And then each row here is one equation and
I want to find a, I want to find
x y and z that simultaneously satisfy all
three of these equations.
So there are two ways I can look at this
if I, if I try and translate this into
vectors.
So I can look at something called the row
picture.
So that's trying to describe each one of
these rows, and if I,
if I do that, what I end up with is the
dot product.
So I'd have the dot product of the
coefficient, so there's an invisible
1 here, so if I just say X, I really mean
1 times X.
That's where this 1 comes from.
So, the first row is 1, 2, 3 dot X, Y, Z
is equal to 6.
Second row is 2, 5, 2 dot X, Y, Z is equal
to 4 and so on.
And it turns out I can also do this using
a column picture.
So I can also think of this as X times 1,
2, 6, y as 2 5 minus 3 and z
times 3, 2, 1, sounds just the same
numbers are the coefficients here.
So, I can think of this either as dot
products, they give me the
rows or I can think of this as a linear
combination of the columns of
my coefficients.
So the problem this is trying to describe,
let's see, so
you can see there is sort of this dark
point here.
And each one of these equations in the
row picture, it's describing one of these
planes.
And so it's, it's really difficult to try
and put three planes onto one picture.
Even, even this one, I failed miserably
trying to
do my own artwork, so I just copied this
out
of the book.
And even there, I still can't really
convince myself
what direction this plane is supposed to
be tilting.
you know sometimes it goes from looking
like this to looking like this to me.
But the point of this is I'm supposed to
have these two planes.
And the space where they intersect is
going to be this black line, here.
So if I have one point that's kind of like
this and one point
that's like this, where they intersect is
going to be described by a line.
So the intersection of the first two
planes will give me a line.
So I've gone from having sort of
two-dimensional objects.
So these two planes, they're
two-dimensional objects that exist
in three dimensions.
When I intersect them, I now have a
one-dimensional object, so this line that
exists in three dimensions, and then to
solve my problem.
So to finish solving my system of
equations, I
need to intersect that line with the third
plane.
And so, if I have a plane and one line,
it's going to hit at just a single point.
And that point is going to be the value of
x, y, and z that solves my system of
equations.
So, I want to now try so, so that's sort
of the row picture.
So remember the rows were giving me
equations of planes and those planes,
then, were
giving me this picture and that was giving
me one approach for solving this problem.
And now I'm also going to try and consider
solving it using the, the column approach.
So to do that I'm going to start by
stacking rows, or binding columns.
So
I can either think of the, the rows as
being the coefficients
from each equation or the columns of being
the coefficients for each variable.
So if I stack the rows, I put the
coefficients for the
third equation on bottom, second equation
in the middle, and the third
equation on top, or I can take all of the
x coor-,
x coeffcients and put them in the first
column, y coefficients go in
the second column, z coefficients go in
the third column.
So, the, the nice thing about this is we
get a duality.
So it doesn't matter which approach I
take, I'm going to get the same answer.
So, quite often, like in, in statistics,
it turns out that looking at the columns
of a matrix makes calculating statistical
properties of estimators very easy.
And looking at the rows of the data in a
matrix makes
plotting the data.
So, actually looking at it you know, with
your eye, very easy.
So, you can use the rows to make plots,
and you can use the columns to do
mathematics.
But, because you're really doing the same
thing, regardless of whether
you try to use the row problem, or the
column problem.
both of those are going to be descriptions
of the same actual problem.
So if A is my coefficient matrix, then I
can write my system of three
equations and three unknowns like this.
So I have the coefficient matrix.
Times a, a vector of unknowns and that's
going
to equal a vector of the, the right hand
sine.
So the, you know these were equations like
somethi- you know,
a x plus b y plus c z equals d so these
are these values d.
They go on the right hand side here in a
vector called b.
And so you can recover pretty easily the
row picture here, so I just have 1,
2, 3 dot x, y ,z equals 6, that's going to
give me the first equation.
And I can repeat that for the second or
the third row, or I can say x.
Times 1, 2,6 Plus y times 2, 5, -3 Plus z
times 3,2,1
Equals 6, 4, 2 and that gives me the
column picture of the problem.
And so we can think of this as the left
hand side multiplies a times the unknowns
v to get v.
And so what we want to move towards now is
having some sort of
multiplication rule for matrices and
vectors that's
going to make sense out of this.
So the, the properties we want it to have
are it has
to give this linear combination for the
column picture, and it needs to
Give me the row picture, so I need the dot
product here,uh you
know that this row of this matrix times
this column vector of unknowns,
that needs to be a don't product.
So I can represent this as Av being just
row 1 dot v row 2 dot v or row 3 dot v.
Or the column picture, a times v, that's
just going to be, so x here was the first
component of my vector v, y was the second
component, and z was the third component.
And so just some examples here, so if I
have 1, so 1
0 0.4 5 6 that's going to equal
4.
1 0 0.4 5 6
again is going to equal 4 and 1 0 0.4 5 6
is going to equal 4.
So multiplying by this matrix A Just
plucks
this first value out and puts that in the
answer.
Another matrix- this matrix has a special
name.
It's called the identity matrix.
If I have ones down the diagonal so the
diagonal is where the row is equal to the
column.
So here I'm in the first row and first
column, here I'm in the second row and
second column.
Here, I'm in the first row but the second
column, so it's not on the diagonal.
So if I have ones down the diagonal and
zeroes everywhere else,
then that's going to give me the same
vector back.
So I have one zero zero times four five
six, that's going to just pluck out this
four.
Then 010 times 456.
That's going to take the 5.
Then 001, that's
going to take this 6.
So suppose I have this system
of equations.
So 3x minus Y equals 3.
X plus Y equals 5.
I can write that in the matrix form as a
coefficient
matrix times an unknown vector XY is equal
to the vector 35.
And
I can draw two pictures of this now.
So as long as these are, you know,
two equations and 2 unknowns, it's easy to
draw.
So the row picture, this is probably how
you want to think about this.
So each one of these rows determines a
line in the X-Y plane,
and then the answer is just going to be
where the two lines intersect.
So I drew these two lines and I
intersected at the point 2,3,
and so you can see, if I put in 2,3 here,
I get 6 minus 3 is equal to 3, and
2 plus 3 is equal to 5 so that is actually
the solution to this system of equations.
I can also draw the column picture.
So this one's a little bit trickier to
see, but the columns are 3 1 so that gives
me my first column.
And so then this arrow here, labeled
column one.
I go over three and up one.
So this is column one.
Column 2 is -1, 1 so that's this vector
here so -1 1 and
it's labeled column two.
And then I'm looking for a linear
combination, so I'm looking for something
times column 1 plus something times column
2 that's going to equal 3,5.
And so you can visualize that, if I
want to get,if I have to follow these two
directions, so either this direction or
this direction,
and I want to get to the point 3,5.
Then what
I need to do is go 2 times column 1, and
that gets me to this point.
And then I'm going to add 3 times column
2, that gets me up to this point.
So this point up here, which is 3,5, I can
describe that as 2 times column 1.
Plus 3 times column 2.
So 2 times column 1.
Plus 3 times column 2.
So that gives me the vector 6, 2.
Plus the vector -3, 3.
So six minus 3 gives me 3.
And 2, 3 gives me 5 so that gives me the,
the same answer
whether I try and do this using the row
view or the column view.
And then both of those are representations
of this matrix
form of the equation.
So some properties of systems of equations
so most of the time If you
have the same number of equations and
unknowns, there's going to be one
solution..
So, that means, you know, you have a plane
like this, a plane like this.
That determines a line and then you maybe
have a plane like
this and wherever that line hits the third
plane, that's your solution.
But, of course, things can go wrong, so
you
could have two planes that are actually
the same plane.
So Imagine if I took the equation of a
plane
and just multiplied both sides by 2,
every, every point
would still be, the, the points that
satisfy that are
going to be the points in the same actual
plane.
Or I could have two planes that were
parallel
and so there would be no intersection
between those.
So, there may be no solution.
So, if two lines are parallel and
basically the, my interpretation
of a system of equations is, where do
these two lines intersect.
If the two lines are parallel, we will get
no solution.
Or if you have two equations that
look different but are actually describing
the same
line in the plane, then you have
infinitely
many solutions, so any point on that plane
will satisfy the equation of both lines.
Sorry, any point on that line satisfies
the equation of each line.