MARTINA BALAGOVIC: Hi.
Welcome.
Today's problem is
about finding solutions
of this non-homogeneous linear
system: x minus 2y minus 2z
equals b_1, 2x minus
5y minus 4z equals b_2,
and 4x minus 9y
minus 8z equals b_3.
And as you can see,
the system doesn't only
have numbers and
unknowns, it also
has parameters,
b_1, b_2, and b_3,
and the solution will
depend on these parameters,
but also the existence
of the solution
will depend on these parameters.
And we're asked
to find a solution
and find when it exists,
depending on the values
of b_1, b_2, and b_3.
So now you should pause the
video, solve the problem,
and come back and compare
your solution with mine.
And we're back.
Let's try it.
Let's start by solving this
system as though b_1, b_2,
and b_3 were numbers.
So we write the matrix of the
system, which is 1, minus 2,
minus 2, b_1; and then 2,
minus 5, minus 4, b_2; and 4,
minus 9 minus 8, b_3.
And we do elimination.
So we multiply the
first row by minus 2
and add it to the second row.
And we multiply it by minus 4
and add it to the third row.
And we get 1, minus 2, minus 2,
b_1; 0, 4 minus 5 is minus 1,
4 minus 4 is 0, and minus
2 times b_1 plus b_2.
And here we get 0, 8 minus 9
is minus 1, and 8 minus 8 is 0.
Finally, on the right-hand
side, minus 4*b_1 plus b_3.
And you can already
see that something's
going to happen here.
But let's do one more step.
So eliminating further, we get
1, minus 2, minus 2, b_1; 0,
minus 1, 0, minus
2*b_1 plus b_2.
And in the last row we replace
it with the last row minus
the second row, and we get
0, 0, 0, minus 4*b_1 plus 2--
so minus minus 2*b_2 is minus
2*b_1 minus b_2 and plus b_3.
I hope I did this right.
So now let's think of
it as a system again.
The last equation says 0 equals
this expression in b_1, b_2,
and b_3.
So this is something
to note down.
If minus 2*b_1 minus b_2 plus
b_3 is some number that's not
0, then the last equation is
going to say 0 equals nonzero.
It's never going
to be satisfied,
and the entire system is never
going to have a solution.
So in this case, we
have no solutions.
If this is equal to 0, so minus
2*b_1 minus b_2 plus b_3 is
equal to 0, then let's do one
more step on this matrix here.
Let's turn this number into
1 by multiplying this row
by negative 1.
And let's use it to eliminate
this number here as well.
So in this case, we
get-- let me write it
from the last row,
which now becomes 0,
0, 0, equals 0, which is fine.
The second row becomes
0, 1, 0, 2*b_1 minus b_2.
And the first one, to
get rid of this minus 2,
we multiply this row by negative
2 and add it to the first one.
We get 1, 0, negative 2, and
here we get b_1 plus 4*b_1
which is 5*b_1, minus 2*b_2.
The reason why we did it was to
get the identity matrix here.
And now let's solve this.
These two columns, corresponding
to variables x and y,
have pivots in them.
So these are the
pivot variables.
This column here has no pivot
in it, so it's a free variable.
And now we're going to
calculate the solutions,
but by picking
particular values for z,
and then calculating
the values for x and y.
We have two kinds of solution.
One kind is the
particular solution.
So this one solves A*x equals b.
There's only one of them.
And we get it by setting the
free variable equal to 0.
Setting the free
variable equal to 0,
we get, well this is equal to 0.
The second equation says
y equals this thing here,
so 2*b_1 minus b_2.
And the first equation
says x minus 2 times 0
equals this expression here.
So 5*b_1 minus 2*b_2.
That's our particular solution.
The next kind is the
special solution.
So remember, those
solve A*x equals 0.
There's as many of them as
there are free variables.
In our case, there's only one.
And we get it by setting all
free variables equal to 0,
except one equal to 1.
And do it for every
free variable.
So in our case there's
only one free variable
and we set z equal to 1.
The solution that
we get in this case,
and remember we're
solving Ax equals
0-- we don't care about the
right-hand side anymore-- so z
is 1.
This second equation
says y equals 0,
and the first equation says
x minus 2 times 1 equals 0.
In other words, x equals 2.
So the special
solution is [2,  0, 1].
And now all solutions
are of the form
x equals the particular
solution plus any multiple
of the special solution.
Let me recap.
In case this particular
combination of parameters
is not 0, there's no solutions.
In case this particular
combination of parameters
is equal to 0,
there are solutions,
there are as many of them
as there are real numbers c,
and they're all of this
form for these two vectors.
And that's all I
wanted to say today.
