WEBVTT

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MARTINA BALAGOVIC: Hi.

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Welcome.

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Today's problem is
about finding solutions

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of this non-homogeneous linear
system: x minus 2y minus 2z

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equals b_1, 2x minus
5y minus 4z equals b_2,

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and 4x minus 9y
minus 8z equals b_3.

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And as you can see,
the system doesn't only

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have numbers and
unknowns, it also

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has parameters,
b_1, b_2, and b_3,

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and the solution will
depend on these parameters,

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but also the existence
of the solution

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will depend on these parameters.

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And we're asked
to find a solution

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and find when it exists,
depending on the values

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of b_1, b_2, and b_3.

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So now you should pause the
video, solve the problem,

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and come back and compare
your solution with mine.

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And we're back.

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Let's try it.

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Let's start by solving this
system as though b_1, b_2,

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and b_3 were numbers.

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So we write the matrix of the
system, which is 1, minus 2,

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minus 2, b_1; and then 2,
minus 5, minus 4, b_2; and 4,

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minus 9 minus 8, b_3.

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And we do elimination.

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So we multiply the
first row by minus 2

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and add it to the second row.

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And we multiply it by minus 4
and add it to the third row.

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And we get 1, minus 2, minus 2,
b_1; 0, 4 minus 5 is minus 1,

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4 minus 4 is 0, and minus
2 times b_1 plus b_2.

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And here we get 0, 8 minus 9
is minus 1, and 8 minus 8 is 0.

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Finally, on the right-hand
side, minus 4*b_1 plus b_3.

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And you can already
see that something's

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going to happen here.

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But let's do one more step.

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So eliminating further, we get
1, minus 2, minus 2, b_1; 0,

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minus 1, 0, minus
2*b_1 plus b_2.

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And in the last row we replace
it with the last row minus

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the second row, and we get
0, 0, 0, minus 4*b_1 plus 2--

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so minus minus 2*b_2 is minus
2*b_1 minus b_2 and plus b_3.

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I hope I did this right.

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So now let's think of
it as a system again.

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The last equation says 0 equals
this expression in b_1, b_2,

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and b_3.

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So this is something
to note down.

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If minus 2*b_1 minus b_2 plus
b_3 is some number that's not

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0, then the last equation is
going to say 0 equals nonzero.

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It's never going
to be satisfied,

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and the entire system is never
going to have a solution.

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So in this case, we
have no solutions.

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If this is equal to 0, so minus
2*b_1 minus b_2 plus b_3 is

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equal to 0, then let's do one
more step on this matrix here.

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Let's turn this number into
1 by multiplying this row

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by negative 1.

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And let's use it to eliminate
this number here as well.

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So in this case, we
get-- let me write it

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from the last row,
which now becomes 0,

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0, 0, equals 0, which is fine.

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The second row becomes
0, 1, 0, 2*b_1 minus b_2.

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And the first one, to
get rid of this minus 2,

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we multiply this row by negative
2 and add it to the first one.

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We get 1, 0, negative 2, and
here we get b_1 plus 4*b_1

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which is 5*b_1, minus 2*b_2.

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The reason why we did it was to
get the identity matrix here.

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And now let's solve this.

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These two columns, corresponding
to variables x and y,

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have pivots in them.

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So these are the
pivot variables.

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This column here has no pivot
in it, so it's a free variable.

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And now we're going to
calculate the solutions,

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but by picking
particular values for z,

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and then calculating
the values for x and y.

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We have two kinds of solution.

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One kind is the
particular solution.

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So this one solves A*x equals b.

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There's only one of them.

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And we get it by setting the
free variable equal to 0.

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Setting the free
variable equal to 0,

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we get, well this is equal to 0.

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The second equation says
y equals this thing here,

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so 2*b_1 minus b_2.

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And the first equation
says x minus 2 times 0

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equals this expression here.

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So 5*b_1 minus 2*b_2.

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That's our particular solution.

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The next kind is the
special solution.

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So remember, those
solve A*x equals 0.

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There's as many of them as
there are free variables.

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In our case, there's only one.

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And we get it by setting all
free variables equal to 0,

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except one equal to 1.

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And do it for every
free variable.

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So in our case there's
only one free variable

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and we set z equal to 1.

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The solution that
we get in this case,

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and remember we're
solving Ax equals

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0-- we don't care about the
right-hand side anymore-- so z

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is 1.

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This second equation
says y equals 0,

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and the first equation says
x minus 2 times 1 equals 0.

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In other words, x equals 2.

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So the special
solution is [2,  0, 1].

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And now all solutions
are of the form

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x equals the particular
solution plus any multiple

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of the special solution.

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Let me recap.

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In case this particular
combination of parameters

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is not 0, there's no solutions.

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In case this particular
combination of parameters

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is equal to 0,
there are solutions,

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there are as many of them
as there are real numbers c,

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and they're all of this
form for these two vectors.

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And that's all I
wanted to say today.