So we've been looking at different examples of solving systems of linear equations.
This video is similar in that we're going to be looking at a system of linear equations,
but it's a little bit different because the system of equations actually has a variable
in it.
So here are the equations we're going to look at.
So the first one is x1 plus 3x2 equals 4, and the second equation is 2x1 plus ax2 equals
a negative 3.
So right here, we have this variable a that is a variable or an unknown-- it's not an
unknown.
It's a variable system the system of equations.
And what we're going to do is we're going to figure out for which values of a does this
system of equations have "a" solution.
So for what value of a is the system consistent?
So that's the question that we're going to try to solve here.
So the way we're going to do this is we're in still form an augmented matrix.
So here are the coefficients of x1's and x2's.
And then the right side of the equation-- on the equal side, 4, negative 3, they go
right here.
This is our standard augmented matrix.
This looks a little different from the previous videos though because the number right here
for the x2 factor is not a number, but it's this variable a.
We can still manipulate these equations and do the same type of operations that we're
used to doing though.
We can still manipulate this into a form that we can try to solve the equations for.
So let's go ahead and do that.
Let's go ahead and let the second equation equal itself minus 2 times the first.
So I'm going to try to get a 0 right here just like I always do when working with systems
of equations in this augmented matrix.
If I do that, then I end up with a 0 there, and I have a minus 6 here, and then I'll have
a negative 11 right here.
So if I look at this equation now if we write out what this means, what do I have?
This equation says that 0x1 plus the quantity a minus 6 times x2 is equal to a negative
11.
Well, 0 times x1 is just 0, so this just simplifies to the quantity a minus 6 times x2 is equal
to a negative 11.
So the way I like to think about this is what would happen if this factor a minus 6 was
equal to zero?
So if this term right here was 0 in its totality, I'd have 0 times x2 equals negative 11 or
0 equals a negative 11.
So I'd end up with a contradiction.
So I can't have a minus 6 equals 0.
Another way to think of it is if I was to solve for x2 by dividing this side of the
equation by a minus 6 and dividing the other side of the equation by a minus 6, I'd end
up with this expression right here for x2. x2 is equal to a negative 11 divided by the
quantity a minus 6.
Well, obviously, I can only do that division as long as a minus 6 is not 0.
So same condition pops up, I don't want a minus 6 to be equal to 0.
In the first case, I would have an inconsistent equation, 0 equals negative 11.
In the other case, I've violated a fundamental law of divided by a 0.
Either case, that's a bad thing.
Both of those conditions tell me that as long as a is not 6, I'm OK.
As long as a isn't 6, so factor won't be a 0 and I won't be dividing by 0 here.
So this system will actually have a solution for all a, so this is the symbol for all,
except when a is equal to 6.
When a equals 6, things fall apart.
I have a contradiction.
But for any other value, I am OK.
So we've answered the question.
This system of equations is consistent.
It has a solution for any value of a except for the value a equals six 6.
So just to wrap things up, as a specific example, if I happened to have a equals 5, I can actually
go through and solve for the values of x1 and x2, and I would get x1 as a negative 29
and x2 is equal to 11.
This would be a unique solution to the system of equations.
Or if for instance, a was equal to negative 4.
If I solve for x1 and x2, I would get the unique solution x1 equals 0.7, and x2 equals
1.1.
I could keep doing this for forever.
I could take-- there's an infinite number of a's I could pick, and for each value of
a I pick, there is a unique solution as long as I don't pick a equals 6.
So that's this problem.
It's really similar to what we've done before in terms of solving systems of equations,
but it was a little bit different because we had a variable that was a coefficient of
one of our unknowns.
In this case, it was the coefficient of the x2 unknown in the second equation.
And we found out that we can solve this system of equations for all values of a except for
one value and that value was a equal to 6.