Linear Algebra Problem #4: Infinitely Many Solutions and Parameter Expressions

[Saladsamurai]

Homework Statement

a) Show (using Gaussian Elimination) that the following sets of equations have infinitely many solutions and b) express the solution in terms of parameters.

x_1+2x_2+x_3=2
2x_1-x_2+7x_3=-6
-x_1+x_2-4x_3=4
x_1-2x_2+5x_3=-6

After G.E. I end up with the array

\left[\begin{array}{cccc}1&2&1&2\\ 0&1&-1&2\\0&0&0&0\\0&-4&4&-8\end{array}\right]

Row 3 implies infinite solutions.


It's part b I am stuck on.

For a solution the text gives:x_1=-2-3t,\ x_2=2+t,\ x_3=t\ \ \ (t\in R)

I know it must be real simple to arrive at the above line, but what is the procedure to do it?

 

[rootX]

I got:

1 2 1 2
2 -1 7 -6
-1 1 -4 4
1 -2 5 -6

rref:

1 0 3 -2
0 1 -1 2
0 0 0 0
0 0 0 0

Let z = t;
now,
see second equation:
y-z = 2
put t
y = 2+t

similarly find x

 

[Defennder]

As rootx has shown usually it's better if you row reduce the matrix to its reduced row echelon form so the solution can be extracted easily.

Once you get the RREF as rootX got, then you can let either x1, x2 or x3 be the free variable, meaning it can take on any real value and then express x1 and x2 in terms of that free variable The answer key uses x3 as the free variable.

Actually the fact that 3rd row is 0s doesn't necessarily imply infinite solutions. You will have infinite solutions if in your RREF matrix, you end up with more unknowns (x1...x4) than equations. In this case, you start off with 4 equations, but row-reduction shows that there are really only 2 linearly independent equations.

 

[Saladsamurai]

I got:

1 2 1 2
2 -1 7 -6
-1 1 -4 4
1 -2 5 -6

rref:

1 0 3 -2
0 1 -1 2
0 0 0 0
0 0 0 0

Let z = t;
now,
see second equation:
y-z = 2
put t
y = 2+t

similarly find x

What step did I miss? How did you get the 1st row to look like that?

As rootx has shown usually it's better if you row reduce the matrix to its reduced row echelon form so the solution can be extracted easily.

Once you get the RREF as rootX got, then you can let either x1, x2 or x3 be the free variable, meaning it can take on any real value and then express x1 and x2 in terms of that free variable The answer key uses x3 as the free variable.

Actually the fact that 3rd row is 0s doesn't necessarily imply infinite solutions. You will have infinite solutions if in your RREF matrix, you end up with more unknowns (x1...x4) than equations. In this case, you start off with 4 equations, but row-reduction shows that there are really only 2 linearly independent equations.

Sounds good. Thanks

 

[Saladsamurai]

What step did I miss? How did you get the 1st row to look like that?

No really. How? I thought in Gaussian Elimination, after changing the leading coefficient to 1, the first row doesn't change... ?

 

[Defennder]

He added a multiple of one of the other rows to the first one, to reduce it fully to reduced-row echelon form.

 

[Mindscrape]

Ask yourself whether the information contained in your R2 and R4 is the same.

 

[Saladsamurai]

He added a multiple of one of the other rows to the first one, to reduce it fully to reduced-row echelon form.

I guess I am a little confused as to the definition of rref. That is, what was it about row 1 when it was 1 2 1 2 that makes it NOT rref?

Silly question I'm sure, but my book is not that thorough.

 

[Defennder]

There is a pivot column for row 2 2nd column if you can reduce it to RREF. Put all the zero rows at the bottom.

 

[rootX]

What step did I miss? How did you get the 1st row to look like that?

I used MATLAB

But, my approach is to cancel everything out if possible; leaving with a simplest form of matrix (that cannot be simplified further)
I think doing things by algorithms make life hard (they are meant for computers)

 

[Saladsamurai]

I used MATLAB

But, my approach is to cancel everything out if possible; leaving with a simplest form of matrix (that cannot be simplified further)
I think doing things by algorithms make life hard (they are meant for computers)

Okay. But I am just wondering how the text got x_1=-2-3t, which clearly came from your simplified row 1. if all it has introduced is Gaussian elimination by hand?

It has not even introduced the concept of "reduced" row echelon form... just row echelon form.

How would you go about reducing R1?

I guess my question is WHEN do you reduce R1 to that stage?

 

[HallsofIvy]

He added a multiple of one of the other rows to the first one, to reduce it fully to reduced-row echelon form.

I guess I am a little confused as to the definition of rref. That is, what was it about row 1 when it was 1 2 1 2 that makes it NOT rref?

Silly question I'm sure, but my book is not that thorough.

I think Defennder meant "to the last one". There is nothing wrong with your first row- it is the fact that your last row is not all 0s that keeps it from being "rref". You cannot have a non-zero entry below a "pivot". If you add 4 times the second row to the fourth you get
\left[\begin{array}{cccc}1&2&1&2\\ 0&1&-1&2\\0&0&0&0\\0&0&0&0\end{array}\right]

THAT'S "rref".

What if it had been
\left[\begin{array}{cccc}1&2&1&2\\ 0&1&-1&2\\0&0&0&0\\0&-4&5&-8\end{array}\right]
so the last can not be made 0 like that? You would still get a 0 in the second column:
\left[\begin{array}{cccc}1&2&1&2\\ 0&1&-1&2\\0&0&0&0\\0&0&1&0\end{array}\right]
and now swap third and fourth rows:
\left[\begin{array}{cccc}1&2&1&2\\ 0&1&-1&2\\0&0&1&0\\0&0&0&0end{array}\right]
again "rref".

 

[Saladsamurai]

I think Defennder meant "to the last one". There is nothing wrong with your first row- it is the fact that your last row is not all 0s that keeps it from being "rref". You cannot have a non-zero entry below a "pivot". If you add 4 times the second row to the fourth you get
\left[\begin{array}{cccc}1&2&1&2\\ 0&1&-1&2\\0&0&0&0\\0&0&0&0\end{array}\right]

THAT'S "rref".

What if it had been
\left[\begin{array}{cccc}1&2&1&2\\ 0&1&-1&2\\0&0&0&0\\0&-4&5&-8\end{array}\right]
so the last can not be made 0 like that? You would still get a 0 in the second column:
\left[\begin{array}{cccc}1&2&1&2\\ 0&1&-1&2\\0&0&0&0\\0&0&1&0\end{array}\right]
and now swap third and fourth rows:
\left[\begin{array}{cccc}1&2&1&2\\ 0&1&-1&2\\0&0&1&0\\0&0&0&0\end{array}\right]
again "rref".

Thanks Halls I did not think there was a problem with R1, that's why I was so confused. And I figured out the X1 part... I just had to sub in the value of X2 that I got from R2 into R1 and solve for X1