Hey guys, in this video I'm going to go over this linear algebra problem. It says,
to use elementary row operations to reduce this augmented matrix to reduced
row echelon form and then the second part says to use the reduced row echelon
form to solve this system of linear equations. This is one of the first
type of problems you're going to see in a linear algebra class. To solve a system
of linear equations you usually either use row echelon form or reduced row
echelon form. In this video we're going to do the latter. Just as a recap
of what reduced row echelon form is, any matrix that is in reduced row echelon
form is in row echelon form as part of the definition and, in addition to that,
the leading coefficient of each nonzero row is equal to one which is called the
pivot or the leading one and each column containing a leading one must have zero
everywhere else. I'm going to do an example of that in the first part. We
have the matrix 1 -1 1 -1 is equal to 5 and then 1 3 5 7 is
equal to -3. We want to have--the 1 here is fine--and we want to get
the second row to be 0 as the leading number. The first operation that I'm
going to do is I'm going to subtract R2 from R1, so the second row--the first row
from the second row and that's going to give me 1 -1 1 -1 5.
So the same thing at the top. And then 0 4 4 8 -8 on the second row. That
first operation already gives me row echelon form because we have the leading
coefficient being 1--well the leading coefficient being farther to the left on
the first row then the second row--and now we just have to make this a 1 and
get this to be 0. To do that I did one quarter or 1/4 times R2 and that gives 1
-1 1 -1 5 0 1 1 2 -2. Now we have leading ones
on both rows and now we just have to get rid of the -1 and I'm going to
do that by adding row 1 and row 2 to get 1 0 2 1 is equal to 3 and 0 1 1 2 is
equal to -2. We're done with the first part and as you can see the
first part helps us solve the system of linear equations in the second part
because if you look at the augmented matrix, it is the augmented matrix for
this system. If you look at the coefficients 1 -1 1 -1
that's the first row equal to 5 and then for the second row it's the same thing.
For systems of linear equations you can either have one solution, infinite
solutions, or no solution. In this problem we have two equations and four
unknowns and we have the row reduced echelon form of the system of equations.
This is equivalent to saying x1 plus 0 x2 plus 2 x3 plus 1 x4 is equal to
3 for the first row and then 0 x1 plus x2 plus x3 plus 2 x4 is equal to
-2 and the solution is just going to be written out what x1 is
equal to. X1 is going to be equal to 3 minus 2 x3 minus x4 and we're going
to do the same thing for x2. That's going to be equal to -2 minus x3
minus 2 x4, so just getting x2 and x1 on one side. And so we have two
equations and four unknowns. We're going to say that x3 can be anything.
X3 is arbitrary and x4 is arbitrary because any values will satisfy that
equation for x1 and x2 and so this problem has infinite solutions and we
are done with this problem.
*chiptune music*