So in this video, we're going to look at a few more examples of how we would go about
solving systems of equations using augmented matrices, except we're going to skip some
of the details about constructing the augmented matrix and performing row operations to get
to the row reduced echelon form. I have some other videos on the details of that. We want
to just jump to the final form and make sure we know how to interpret the final form in
terms of what the solution is. So we're going to look at a few different
matrices and row reduce them and then take a look at that final form of the augmented
matrix. So let's take a look at this augmented matrix. So this, remember, this augmented
matrix represents three equations with three unknowns and the final column of this matrix
represents the equals sign part of the equation. So this first row represents 1x 1 plus 0x
2 plus 2x 3 equals 5, that's what an augmented matrix means. When we solve an augmented matrix
to find the solution to this system of linear equations, we perform row reduction, right.
So we do these row operations and we get it into a reduced row echelon form. That's something
that we do and we've done that in previous videos.
If we do that here, we'll end up with a matrix that I call A sub rr, meaning it's been row
reduced and it looks like this. So again, we're skipping the details here. We just want
to get to this row reduced form and interpret this matrix to find the solution.
So in this row reduced matrix we look at this final equation and we have all zeros except
for a number there in the final column. So 0, 0, 0, 1. Anytime you see all zeros and
then a number, what this means is that you have zero x1 plus 0 x2 plus 0 x3 equals a
number. Which is the same thing as saying 0 equals 1.
So this is actually what we call a contradiction. 0 is not equal to 1. That is completely false.
0 is not 1. So we have encountered a contradiction which means this system of equations does
not have a solution. So anytime you see all zeros and then a number in that final column,
whether the number is 1 or 7 or pi or square root of 4, it doesn't matter what the number
is, we end up with the contradiction of the form 0 equals a number which is not correct.
So be on the lookout for row reduced augmented matrices that have this form.
All right, let's look at another example. So in this example, we'll look at a slightly
different matrix. We're going to look at this matrix. 1, 2, 3 4, 5, 6, 7, 8, 9, 10, 11,
12. To solve this system of equations, again, we perform row operations to get it into a
row reduced form. And in this case, the row reduced echelon form of the matrix looks like
this. So you might say, oh, wow, looks like we in
have the same problem here. But the big difference between this example and the first one we
looked at is this is 0 this time. So we don't have a problem here, this is not a contradiction.
This is really just saying that 0 times x1 plus 0 times x2 plus 0 times x3 equals 0.
And 0 does equals 0 so that's not a contradiction. So it's not quite the same thing. So we don't
have a contradiction here but we do have a solution to this system of equations. And
to find out the solution, we use some terminology that we call basic variables and free variables.
So the basic variables are easy to identify. We just look for each column that has a pivot.
So this column right here has a pivot in that first position. Since it's in the first position
that means that x1 is what we call a basic variable. Also column 2 has a 1 in the second
position has a pivot. So x2 is what we call a basic variable.
The other columns do not have pivots so no other basic variables, we found all of them.
So anything that's left over is what we call a free variable. So the variable x3 wasn't
a basic variable so it must be a free variable. So x3 is a free variable.
So what this means is we write our system of equations solution, we're going to write
equations for x1 and x2, we're going to write an equation for all the basic variables. And
then those equations will contain the free variables as essentially a variable. So let's
go to this first row of the equation and write down what we have in terms of the matrix.
What we have is x1 plus 0 x2 minus x3 is equal to negative 2.
If I move the x3 to the other side of this equation, I can write x1 as a negative 2 plus
x3. And I've now written x1 in terms of numbers and the three variables. So this is actually
the solution for x1. Let's go to the second row. The second row
says that 0 x1 plus x2 plus 2x3 is equal to 3. Again, I can move the x3 term to the other
side and I have x2 equals 3 minus 2x3. So again, I've been able to write the basic variable
x2 in terms of numbers and free variables. So I now have an equation for x2.
So the total solution to this system of equations are these two equations, x1 equals a negative
2 plus x3. And x2 equals 3 minus 2x3. And x3 is a free variable. So this is the solution
to this system of equations. There's actually an infinite number of solutions. For any number
x3 you give me, I can plug into these equations compute x1 and compute x2 and give you a solution
in the form of x1, x2, and x3. Let's do another example. So a final example
here. We'll have a new matrix. Here's our new matrix A. In general, to solve this augmented
matrix, I would do row reduction techniques to get it down to a reduced row echelon form.
We're going to skip those details but you can go to one of my other videos to watch
how to go about doing that. If I do row reduction, I get down to this form of a matrix. And our
goal now is to look at this row reduced augmented matrix, and figure out what the solution is
of this system of equations. So our strategy is going to be the same as
the last example. Let's identify the basic variables and the free variables. The basic
variables I get from the pivot columns. The first one is x1. So x1 is a basic variable.
I have a pivot in this column, the second column, so x2 is a basic variable. And I also
have a pivot in this column. I have all zeros with a one in that final column, this is the
fourth column so that means that x4 is a basic variable.
This system of equations has five unknowns, x1, 2x, 3x, x4, and x5. I've figured out that
x1, x2, and x4 are basic variables. So that means the leftover ones are the free variables.
So the free variables in this case are x3 and x5. So now I can go ahead and write equations
for each of the basic variables. From the first row of my row reduced augmented matrix,
I have that x1 plus 5x3 plus 9x5 is equal to 1.
If I move the x3 and x5 terms to the other side, I can rewrite x1 in terms of numbers
and free variables. So I now have an equation for x1. From the second row of my row reduce
matrix, I have the equation x2 plus 7x3 plus 4x5 is equal to negative 9. If I move the
terms x3 terms and x5 terms to the side, I can write an equation for x2. So I've written
x2 in terms of numbers and the free variables x3 and x5. And then finally, from the third
row I have the equation x4 plus x5 is equal to 3. Which if I move the x5 to the other
side I have an equation for x4. x4 has been written in terms of numbers and free variables.
So this is my system of equations solution. I have an equation for each one of my basic
variables, x1, x2, and x4. And I know that x3 in x5 are free variables. So if you tell
me any value for x3 and x5, I can plug into the equation for x1, I compute x1. I can compute
x2 and I can compute x4 and I have a solution to this system of equations.
Since x3 and x5 are both free, there is an infinite number of ways to choose them, so
I actually have an infinite number of solutions to this system of equations. But that's OK,
they're all valid solutions. So that's the final example I wanted to work
through. We didn't do any details of how to do row reduction in this video. We basically
just started with an augmented matrix, showed the row reduced version of it, and then learned
how to identify the general solution of that row reduced augmented matrix.
In the first example there was no solution, we encountered a contradiction. In the last
two examples we had both basic variables and free variables. And the strategy is to write
an equation for each basic variable in terms of numbers and the other free variables.