WEBVTT

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PROFESSOR: Hi, everyone.

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So for the first
part of this course,

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we've learned basically
the ins and outs

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of solving linear
systems of equations.

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Today we're going to do a little
review of the basic concepts.

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Hopefully we'll see a few of
them in the following problem.

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We're given a square
matrix A-- a three

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by three square matrix A-- where
the last entry is a parameter

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k.

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And this parameter will vary.

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And we'll see what happens to
the system of equations A*x

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equal to [2, 3, 7], for which
k it has a unique solution,

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for which k it has
infinitely many solutions.

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Then we'll find the
LU decomposition.

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And finally, we'll write
down the complete solution,

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the system.

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So I'll give you a
few minutes to work

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this problem on your own.

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And then please come
back and see how I do it.

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All right, welcome back.

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So let's start with part A.
For which k does this system,

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A*x equal [2, 3, 7],
have a unique solution?

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So what do we know about square
systems of linear equations?

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They have a unique solution
when the matrix A is invertible.

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So now, when is A invertible?

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It is invertible when
it is of full rank.

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And how do we figure this out?

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We do it by performing
row operations.

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We do it by doing
eliminations on the matrix.

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But since we want to
simulate an exam setting,

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it always pays off to see
what tasks lie ahead of us.

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So in part C, we're asked to
find the LU decomposition.

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This means that when
we do row operations,

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we'd better keep track what row
operations we're doing exactly.

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In particular, we'll write them
down as elementary matrices.

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And in part D, we'll be asked to
compute the complete solution.

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And therefore it's good
to do row operations

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on the augmented matrix
A. So let's do this.

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I'm going to write this.

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The augmented matrix A is the
following beast, 1, 1, 1; 1 2,

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3; 3, 4, k; and then 2, 3, 7.

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So first thing, we subtract
a multiple of row one

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from row two.

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And it's exactly
negative 1 times

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the first row plus the second.

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Let me write down the
corresponding elementary matrix

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that does this.

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It's E_(2,1).

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And it's lower diagonal
with 1's on the diagonal.

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And it's going to
be exactly minus 1

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in the first entry
of the second row.

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So we get 1, 1,
1, 2; 0, 1, 2, 1.

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And we copy down the third row.

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Now we subtract a
multiple of the first row

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from the third one.

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And let me write this here.

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Yeah, we'll multiply the
first row by negative 3,

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and add it to the third one.

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This is accomplished by the
elementary matrix E_(3,1) which

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is 1, 1, 1, negative
three, and then 0, 0.

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OK, 1, 1, 1, 2; 0, 1, 2, 1.

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We copy the first two rows.

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And then the third one
will be 0, 1, k minus 3,

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and 7 minus 3 times
2, 7 minus 6, 1.

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We have essentially one last
row operation to perform.

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Let me do it here.

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So we'll subtract the second
row from the third one.

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And we'll get 1, 1, 1, 2;
0, 1, 2, 1; 0, 0, k minus 5,

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and then 0.

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And this was achieved by the
elementary matrix E_(3,2),

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which was 1, 1, 1,
and then negative 1.

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Because we multiplied the
second row by negative 1

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and added it to the third one.

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So we got to a matrix,
which is upper triangular.

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And we want to figure out: for
which value of the parameter k

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is this matrix of full rank.

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This is a pivot.

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This is a pivot.

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And we want this one
to be a pivot as well.

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And that happens when
k minus 5 is not 0.

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So when k is different from 5,
the matrix A is of full rank.

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And therefore the system
A*x equals to [2, 3,

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7] has a unique solution.

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Now part B. For which k do we
have infinitely many solutions?

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So when are we in
such a situation?

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We are in such a situation when
the null space of the matrix A

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is nontrivial.

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So the null space
will be nontrivial

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when this k minus 5
number here, which

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is what's the pivot in
the first case, is 0.

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So k minus 5 equals to 0.

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You see there's a
little caveat here.

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When k is equal to 5,
we get the third row

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of the augmented matrix,
0, 0, 0, equal to 0.

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This means that the matrix
is actually consistent,

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and we indeed have a solution
But if this entry were nonzero,

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then we would get no solutions.

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Now off to part C. We want to
compute the LU decomposition.

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Well, we already got
what the matrix U

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is through performing
row operations.

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It's this guy here.

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Let me write it down-- 1, 1,
1; 0 1, 2; 0, 0, k minus 5.

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And when k is equal to 4,
this entry's negative 1.

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And now what about the matrix L?

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Well, how did we get to this U?

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We had the matrix A.
We got the matrix U.

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And what we did was
first we applied E_(2,1).

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Then we applied E_(3,1)
and then E_(3,2).

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We get A by inverting
this equation.

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So it's going to be E_(2,1)
inverse E_(3,1) inverse E_(3,2)

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inverse times U. And
this is our matrix L.

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And we know it's fairly easy
to invert these elementary

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matrices E. We flip the signs
of the off-diagonal entries.

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I'm not going to
write down again

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the inverses of these guys.

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I'm just going to write the
product of the inverses.

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And that's also very
easy to compute.

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Because one, we
invert the signs.

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We just send the numbers in
their corresponding entry of L.

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I mean the following thing.

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So minus 1 becomes a 1.

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And it comes here, in
its respective entry.

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For E_(3,1), we flip
the sign, first 3,

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and we plug it in here.

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And for E_(3,2), we flip the
sign of this guy, becomes 1,

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and we plug it in here.

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So give me a few moments
to erase the board,

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and then I'll do part four.

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We're back, and we're
going to do part D now.

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So we need to find
the complete solution

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of the system for all k.

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And we saw that
for k equal to 5,

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the system had many,
many solutions.

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And for k not equal
to 5, it had only one.

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So first let's look
at the case k not

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equal to 5, when the
matrix A was invertible.

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It's not hard to see what the
solution of the system then is.

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It's going to be-- let
me just write it down.

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So when k is not equal
to 5, this was nonzero.

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Therefore, x_3 needs to be 0.

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When x_3 is 0, we have
x_2 plus 2 times x_3.

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So x_2 plus 0 equals 1.

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Therefore x_2 is 1.

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And then we go back
to the first row.

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We have x_1 plus x_2
plus x_3 equals 2.

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So x_1 plus 1 plus 0 equals 2.

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So x_1 plus 1 equals 2.

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And therefore x_1 is 1.

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Good.

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Now what about k equal to 5?

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Then we see that x_3
is a free variable.

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So the solution will
be [x 1,  x 2, x 3].

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x_3 can be any number c.

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From the second row, we'll
get the value of x_2.

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It's 1 minus 2 times x_3.

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So 1 minus 2 times c.

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And x_1 is 2 minus
x_2 minus x_3.

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So let me rewrite this.

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It's 2 minus x_2
is 1 minus-- aah,

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chalk-- 2c, minus
c, 1 minus 2c, c.

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So we'll decompose
this vector here

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into a component which
is independent of c

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and a component which
is a multiple of c.

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So this is 2 minus 1.

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It should be [1, 1, 0]
plus c times--

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we'll have two c minus
c, c, so c times 1.

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Here we'll have minus
2c so negative 2.

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And here we'll have 1.

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And thus we get the particular
solution for the system

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and the special
solution for the system.

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We're kind of done here.

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If you're at an exam,
you should immediately

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start the next problem.

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Good luck, and
Ill see you later.