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# Solving linear systems with matrices

Sal solves a linear system with 3 variables by representing it with an augmented matrix and bringing the matrix to

**reduced row-echelon form**. Created by Sal Khan.Video transcript

I figure it never hurts getting
as much practice as possible solving systems of
linear equations, so let's solve this one. What I'm going to do is I'm
going to solve it using an augmented matrix, and I'm going
to put it in reduced row echelon form. So what's the augmented matrix
for this system of equations? Three unknowns with
three equations. I just have to do
the coefficents. So the coefficients of x
terms are just 1, 1, 1. Coefficients of the y terms
are 1, 2, and 3. Coefficients of the z terms
are 1, 3, and 4. And let me show that
it's augmented. And then they equal
3, 0, and minus 2. Now, I want to get this
augmented matrix into reduced row echelon form. So the first thing, I have
a leading 1 here that's a pivot entry. Let me make everything else in
that column equal to a 0. So I'm not going to change
my first row. So it will just be a 1, a 1, a
1, and then my dividing line, and then I have a 3. Now, to zero this out, let me
just replace the second row with the first row minus
the second row. So 1 minus 1 is 0. 1 minus 2 is-- actually, a
better thing to do, because I eventually want this to be 1
anyway, let me replace this row with this row, with the
second row minus the first row instead of the first row
minus the second row. I can do it either way. So the second row minus
the first row. So 1 minus 1 is 0. 2 minus 1 is 1. 3 minus 1 is 2. And then 0 minus 3 is minus 3. Now I want to also
zero this out. So let me replace this
guy with this equation minus that equation. So 1 minus 1 is 0. 3 minus 1 is 2. 4 minus 1 is 3. Minus 2 minus 3 is minus 5. Fair enough. So I got my pivot entry here. I have another pivot
entry here. It's to the right of this one,
which is what I want for reduced row echelon form. Now, I need to target this
entry and that entry. I need to zero them out. So let's do it. So I'm going to keep my
second row the same. My second row is 0, 1, 2, and
then I have a minus 3, the augmented part of it. And to zero this guy out, what
I can do is I can replace the first row with the first row
minus the second row. So I get 1 minus 0 is 1. 1 minus 1-- there's
a bird outside. Let me close my window. So where was I? I'm replacing the first row with
the first row minus the second row. So 1 minus 0 is 1. 1 minus 1 is 0. 1 minus 2 is minus 1. And then 3 minus minus 3, so
that's equal to 3 plus 3, so that's equal to 6. 1 minus 0 is 1. 1 minus 1 is 0, negative 1. And then 3 minus negative
3, that's 6. I always want to make sure I
don't make a careless mistake. Now, let me get rid of this
entry right here. Let me zero that out. So let me replace the third row
with the third row minus 2 times the second row. So we have 0 minus--
well, 2 times 0. That's just going to be 0. 2 minus 2 times 1, that's
2 minus 2, that's 0. 3 minus 2 times 2, that's
3 minus 4, or minus 1. And then finally, minus 5
minus 2 times minus 3. Let me write that down. That's minus 5, minus minus 6. That's minus 5 plus
6 is equal to 1. I really wanted to make
sure I didn't make a careless mistake there. So that is equal to 1. So I'm almost done, but I'm
still not in reduced row echelon form. This has to be a positive
1 in order to get there. It can't be anything
other than a 1. That's just the style of reduced
row echelon form. And then these guys up here
have to be zeroed out. Well, the easy thing to do,
let me just multiply this equation by minus 1. So then this becomes a plus 1
and that becomes a minus 1. And then I just need to zero
out these two guys up here. So let's do it. So my equation, I'm going to
keep my third row the same. My third row is now
0, 0, 1, minus 1. And now I want to zero
this guy out. So what I can do is I could set
my first row equal to my first row plus my last row,
because if these two add up, they're going to
be equal to 0. So what do I get? 1 plus 0 is 1. 0 plus 0 is 0. Minus 1 plus 1 is 0. 6 plus minus 1 is 5. Now, I want to zero
this row out. And to zero this row out, what
I can do is I'll replace it with the second row minus
2 times the first row. So 0 minus 2 times
0 is just 0. 1 minus 2 times 0 is just 1. 2 minus 2 times 1 is 0. Minus 3 minus 2 times
negative 1. Let me write that down. Minus 3 minus 2 times minus 1. I don't want to make
a careless mistake. So what is that equal to? This is equal to minus 3 minus
minus 2, or minus 3 plus 2, which is equal to minus 1. So that's equal to minus 1. And now I have my augmented
matrix in reduced row echelon form. My pivot entries are the only
entries in their columns. Each pivot entry in each
successive row is to the right of the pivot entry before it. And actually, I have
no free variables. Every column has
a pivot entry. So let's go back from the
augmented matrix world and kind of put back our
variables there. So what do we get? We get x plus 0y plus
0z is equal to 5. That's that row right there. We get 0x plus 1y plus 0z
is equal to minus 1. That's that row right there. And then finally, we have
0x plus 0y plus 1z is equal to minus 1. That's that row right there. And just like that, we've
actually solved our system of three equations with
three unknowns. That's the solution
right there. I just wrote it in this way
just so you can see the corresponding, but obviously,
I could have written them closer to their equal sign. So hopefully, you found
that vaguely useful.