Matrices: Reduced Row Echelon Form 3 And another example of solving a system of linear equations by putting an augmented matrix into reduced row echelon form
Matrices: Reduced Row Echelon Form 3
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- I have here three linear equations of four unknowns.
- And like the first video, where I talked about reduced
- row echelon form, and solving systems of linear equations
- using augmented matrices, at least my gut feeling says,
- look, I have fewer equations than variables, so I probably
- won't be able to constrain this enough.
- Or maybe I'll have an infinite number of solutions.
- But let's see if I'm right.
- So let's construct the augmented matrix for this
- system of equations.
- My coefficients on the x1 terms are 1, 1, and 2.
- Coefficients on the x2 are 2, 2, and 4.
- Coefficients on x3 are 1, 2, and 0.
- There's of course no x3 term, so we can view it as a 0
- Coefficients on the x4 are 1, minus 1, and 6.
- And then on the right-hand side of the equals sign, I
- have 8, 12, and 4.
- There's my augmented matrix, now let's put this guy into
- reduced row echelon form.
- The first thing I want to do is, I want to zero out these
- two rows right here.
- So what can we do?
- I'm going to keep my first row the same for now, so it's 1,
- 2, 1, 1, 8.
- That line essentially represents my equals sign.
- What I can do is, let me subtract-- let me replace the
- second row with the second row minus the first row.
- So 1 minus 1 is 0, 2 minus 2 is 0, 2 minus 1 is 1, 1--
- negative 1 minus 1 is minus 2, and then 12 minus 8 is 4.
- There you go, that looks good so far, it looks like column,
- or x2 which is represented by column two, looks like it
- might be a free variable, but we're not 100% sure yet.
- Let's do all of our rows.
- So let's take-- to get rid of this guy right here, let's
- replace our third equation with the third equation minus
- two times our first equation.
- So we get 2 minus 2 times 1 is 0, 4 minus 2 times 2, well
- that's 0, 0 minus 2 times 1, that's minus 2.
- 6 minus 2 times 1, well that's 4, right?
- 6 minus 2.
- And then 4 minus 2 times 8, is minus 16, 4
- minus 16 is minus 12.
- Now what can we do?
- Well, let's see if we can get rid of this minus
- 2 term right there.
- So let me rewrite my augmented matrix.
- I'm going to keep row two the same this time, so I get a 0,
- 0, 1, minus 2, and essentially my equals sign, or the
- augmented part of the matrix.
- And now let's see what I can do.
- Let me get rid of this 0 up here, because I want to get
- into reduced row echelon form.
- So any of my pivot entries, which are always going to have
- the coefficient 1, or the entry 1, it should be the only
- non-zero term in my row.
- How do I get rid of this one here?
- Well I can subtract-- I can replace row one with row 1
- minus row 2.
- So 1 minus 0 is 1, 2 minus 0 is 2, 1 minus 1 is 0, 1 minus
- minus 2, that's 1 plus 2, which is 3.
- And then 8 minus 4 is 4.
- And now how can I get rid of this guy?
- Well let me replace row 3 with row 3 plus 2 times row 1.
- Sorry, with row 3 plus 2 times row 2.
- Because then you'd have minus 2, plus 2 times this, and
- they'd cancel out.
- So let's see the zeros.
- 0 plus 2 times 0, that's 0.
- 0 plus 2 times 0, that's 0, minus 2 plus 2 times 1 is 0.
- 4 plus 2 times minus 2, that's 4 minus 4, that's 0.
- And then I have minus 12, plus 2 times 4.
- That's minus 12 plus 8, that's minus 4.
- Now, this is interesting right now-- this is interesting.
- I have essentially put this in reduced row echelon form.
- I have two pivot entries, that's a pivot entry right
- there, and that's a pivot entry right there.
- They're the only non-zero term in their respective columns.
- And this is just kind of a style issue, but this pivot
- entry is in a lower row than that one.
- So it's in a column to the right of this one right there.
- And I just inspected, this looks like a-- this column two
- looks kind of like a free variable-- there's no pivot
- entry there, no pivot entry there.
- But let's see, let's map this back to
- our system of equations.
- These are just numbers to me and I just kind of
- mechanically, almost like a computer, put this in reduced
- row echelon form.
- Actually, almost exactly like a computer.
- But let me put it back to my system of linear equations, to
- see what our result is.
- So we get 1 times x1, let me write it in yellow.
- So I get 1 times x1, plus 2 times x2, plus 0 times x3,
- plus 3 times x4 is equal to 4.
- Obviously I could ignore this term right there, I didn't
- even have to write it.
- I'm not going to write that.
- Then I get 0 times x1, plus 0 times x2, plus 1 times x3, so
- I can just write that.
- I'll just write the one.
- 1 times x3, minus 2 times x4, is equal to 4.
- And then this last term, what do I get?
- I get 0 x1, plus 0 x2 plus 0 x3 plus 0 x4, well, all of
- that's equal to 0, and I've got to write something on the
- left-hand side.
- So let me just write a 0, and that's got to be
- equal to minus 4.
- Well this doesn't make any sense whatsoever.
- 0 equals minus 4.
- This is this is a nonsensical constraint, this is
- 0 can never equal minus 4.
- This is impossible.
- Which means that it is essentially impossible to find
- an intersection of these three systems of equations, or a
- solution set that satisfies all of them.
- When we looked at this initially, at the beginning of
- the of the video, we said there are only three
- equations, we have four unknowns, maybe there's going
- to be an infinite set of solutions.
- But turns out that these three-- I guess you can call
- them these three surfaces-- don't intersect in R4.
- These are all four dimensional, we're dealing in
- R4 right here, because we have-- I guess each vector has
- four components, or we have four variables, is the way you
- could think about it.
- And it's always hard to visualize things in R4.
- But if we were doing things in R3, we can imagine the
- situation where, let's say we had two planes in R3.
- So that's one plane right there, and then I had another
- completely parallel plane to that one.
- So I had another completely parallel plane
- to that first one.
- Even though these would be two planes in R3, so let me give
- an example.
- So let's say that this first plane was represented by the
- equation 3x plus 6y plus 9z is equal to 5, and the second
- plane was represented by the equation 3x plus 6y plus 9z is
- equal to 2.
- These two planes in R3-- this is the case of R3, so this is
- R3 right here.
- These two planes, clearly they'll never intersect.
- Because obviously this one has the same coefficients adding
- up to 5, this one has the same coefficient adding up to 2.
- And when, if we just looked at this initially, if it wasn't
- so obvious, we would have said, we have only two
- equations with three unknowns, maybe this has an infinite set
- of solutions.
- But it won't be the case, because you can actually just
- subtract this equation, from the bottom equation, from the
- top equation.
- And what do you get?
- You would get a very familiar-- so if you just
- subtract the bottom equation from the top equation, and you
- get 3x minus 3x, 6y minus 6y, 9z minus 9z-- actually, let me
- do it right here.
- So for that minus that, you get zero is equal to 5 minus
- 2, which is 3.
- Which is a very similar result that we got up there.
- So when you have two parallel planes, in this case in R3, or
- really to any kind of two parallel equations, or a set
- of parallel equations, they won't intersect.
- And you're going to get, when you put it in reduced row
- echelon form, or you just do basic elimination, or you
- solve the systems, you're going to get a statement that
- zero is equal to something, and that means that there are
- no solutions.
- So the general take-away, if you have zero equals
- something, no solutions.
- If you have the same number of pivot variables, the same
- number of pivot entries as you do columns, so if you get the
- situations-- let me write this down, this is good to know.
- if you have zero is equal to anything, then
- that means no solution.
- If you're dealing with R3, then you probably have
- parallel planes, in R2 you're dealing with parallel lines.
- If you have the situation where you have the same number
- of pivot entries as columns, so it's just 1, 1, 1, 1, this
- is the case of R4 right there.
- I think you get the idea.
- That equals a, b, c, d.
- Then you have a unique solution.
- Now if, you have any free variables-- so free variables
- look like this, so let's say we have 1, 0, 1, 0, and then I
- have the entry 1, 1, let me be careful.
- 0, let me do it like this.
- 1, 0, 0, and then I have the entry 1, 2, and then I have a
- bunch of zeroes over here.
- And then this has to equal zero-- remember, if this was a
- bunch of zeroes equaling some variable, then I would have no
- solution, or equalling some constant, let's say this is
- equal to 5, this is equal to 2.
- If this is our reduced row echelon form that we
- eventually get to, then we have a few free variables.
- This is a free, or I guess we could call this column a free
- column, to some degree this one would be too.
- Because it has no pivot entries.
- These are the pivot entries.
- So this is variable x2 and that's variable x4.
- Then these would be free, we can set
- them equal to anything.
- So then here we have unlimited solutions,
- or no unique solutions.
- And that was actually the first example we saw.
- And these are really the three cases that you're going to see
- every time, and it's good to get familiar with them so
- you're never going to get stumped up when you have
- something like 0 equals minus 4, or 0 equals 3.
- Or if you have just a bunch of zeros and a bunch of rows.
- I want to make that very clear.
- Sometimes, you see a bunch of zeroes here, on the left-hand
- side of the augmented divide, and you might say, oh maybe I
- have no unique solutions, I have an
- infinite number of solutions.
- But you have to look at this entry right here.
- Only if this whole thing is zero and you have free
- variables, then you have an infinite number of solutions.
- If you have a statement like, 0 is equal to a, if this is
- equal to 7 right here, then all of the sudden you would
- have no solution to this.
- That you're dealing with parallel surfaces.
- Anyway, hopefully you found that helpful.
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