Question

How to do row reduce echelon form for this matrix?
given \(p^3+q^3+r^3-3pqr \neq 0\)
\(A= \left[\begin{matrix}p&q&r|p+q+r\\q&r&p|p+q+r\\r&p&q|p+q+r\end{matrix}\right]\)
Please, help.

Answer 1

I tried many ways:).
1) Use manual method
2) find x, y, z by Ax = b --> \(x=A^{-1} b\) to get the solution first, then derive from it to get what term I can manipulate on row reduced echelon method. But it is really a mess.
I would like to know the simpler way.
Notice: it is a symmetric matrix.
and the given equation is -det A

Answer 2

.

Answer 3

What did you get by manual method?

Answer 4

I got nothing!! That is why I have to derive from method 2

Answer 5

Let T = p+q+r
and we have A is the 3x3 matrix from the LHS
\(det A = -p^3-q^3-r^3 +3pqr = (-1) (p^3+q^3+r^3-3pqr)\neq 0\)
Hence, A has inverse.
By using cofactor transpose of matrix A, I got
\(A^{-1}=\dfrac{-1}{det A}\left[\begin{matrix}rq -p^2&pr-q^2&pq-r^2\\rp-q^2&pq-r^2&qr-r^2\\qp-r^2&qr-p^2&pr-q^2\end{matrix}\right]\)

Answer 6

Hence \[\vec x=A^{-1}b =\dfrac{-T}{det A}\left[\begin{matrix}rq -p^2&pr-q^2&pq-r^2\\rp-q^2&pq-r^2&qr-r^2\\qp-r^2&qr-p^2&pr-q^2\end{matrix}\right]\]

Answer 7

After simplifying, I got \(x_1= 1+\dfrac{qr^2+q^2r-pr^2-p^2r}{p^3+q^3+r^3-3pqr}\)

Answer 8

From here, I argue that, to get rref , we need entry \(a_{11} \) must be the form of \(\dfrac{p^3+q^3+r^3-3pqr}{p^3+q^3-r^3-3pqr}=1\) and the rhs, as above,
So that we can have the required solution.
However, I have no way to get \(a_{11}\) as above. Or, I made mistake at somewhere.
That is what I tried.

Answer 9

\(A= \left[\begin{matrix}p&q&r|p+q+r\\q&r&p|p+q+r\\r&p&q|p+q+r\end{matrix}\right]\)

Answer 10

Indeed it's messy.
If you find the row echelon for of A (3x4), you will find that the last row is [0,0,1,1], which means that r=1.
By symmetry of the three variables, we can conclude that p=q=r.
However, by the same token, as long as p=q=r=k, k can take on any real value.
So we have an infinite number of solutions!

Answer 11

Oops, did not read the first line. This means that symmetric solutions are out.

Answer 12

nope, if p=q=r =1, then \(p^3+q^3+r^3-3pqr=0\) contradict

Answer 13

The row echelon form turns out to be
[1, q/p, r/p,(p+q+r)/p],
[0, 1, (p^2-qr)/(pr-q^2), ((p-q)r+p^2-q^2)/(pr-q^2)],
[0, 0, 1, 1]
But I have no idea where to go from here.
@ganeshie8

Answer 14

We cannot divide by any p, q, r since we don't know whether they are =0 or not.

Answer 15

@Loser66
To investigate further, we have to invert
a=matrix A less the augmented part
=[p,q,r]
[q,r,p]
[r,p,q]
and
the inverse of
\(\large a^{-1}=\left[\begin{matrix}\frac{qr-p^2}{r(pq-r^2)+p(qr-p^2)+q(pr-q^2)} & \frac{pr-q^2}{r(pq-r^2)+p(qr-p^2)+q(pr-q^2)} & \frac{pq-r^2}{r(pq-r^2)+p(qr-p^2)+q(pr-q^2)}\\ \frac{pr-q^2}{r(pq-r^2)+p(qr-p^2)+q(pr-q^2)} & \frac{pq-r^2}{r(pq-r^2)+p(qr-p^2)+q(pr-q^2)} & \frac{qr-p^2}{r(pq-r^2)+p(qr-p^2)+q(pr-q^2)}\\ \frac{pq-r^2}{r(pq-r^2)+p(qr-p^2)+q(pr-q^2)} & \frac{qr-p^2}{r(pq-r^2)+p(qr-p^2)+q(pr-q^2)} & \frac{pr-q^2}{r(pq-r^2)+p(qr-p^2)+q(pr-q^2)}\end{matrix}\right]\)
and
\(a^{-1}.\left[\begin{matrix}p+q+r\\p+q+r\\p+q+r\end{matrix}\right]\)
=\(\left[\begin{matrix}1\\1\\1\end{matrix}\right]\)
Which means that any combination of p,q,r such that
p+q+r=1, and \(p^3+q^3+r^3-3pqr \neq 0\) would fit the bill.
For example, (1/4,1/4,1/2), or even (-1/4,-1/4,1) are solutions.

Answer 16

I got (1,1,1) also for x, y , z, not for p, q, r
the third way I approach is
\(p^3+q^3+r^3-3pqr=(p+q+r)*(p^2+q^2+r^2-pr-pq-rq)\)
the LHS\(\neq 0\), hence \((p+q+r)\neq 0 ~~and~~~ (p^2+q^2+r^2)\neq 0\)

Answer 17

So that to A, to get rref, I do
\((R_1+R_2+R_3) \) for all Rows
that gives me all Rows are the same and each entries for the rows is
\((p+q+r) ~~~(p+q+r) ~~~ (p+q+r) | 3 (p+q+r)\)

Answer 18

Then multiple that row by \((p^2+q^2+r^2-pr-pq-rq)\) and simplify
we have all entries are \(p^3+q^3+r^3-3pqr \), the right hand side is \(3 (p^3+q^3+r^3-3pqr)\)

Answer 19

divide the rows by its common entry, we have (1 1 1 = 3)
And we get 1x + 1y+ 1z = 3
That is x=3-y-z

Answer 20

@loser66
What's the difference between x,y,z and p,q,r.
Do they solve the same problem? Sorry, I don't get what you mean by:
"I got (1,1,1) also for x, y , z, not for p, q, r"

Answer 21

The original problem :
Solve the following system of equation for x, y, and z in \(\mathbb R^3\), by row reduction and check by substitution that your solution is correct.
px + qy + rz = p+q+r
qx + ry+pz = p+q+r
rx + py + qz = p+q+r
It is given that \(p^3+q^3+r^3-3pqr \neq 0\)
Do not divide by p, q, or r, as you do not know if any of these equals 0
Write the answer in
1) vector-parametric form
2) flat form
3) functional form
4) What happens to the solution set if \(p^3+q^3+r^3-3pqr=0\)

Answer 22

@loser66
Thank you for posting the complete original question. That's a little more clear to me now. I was confused earlier, because the x,y,z were understood in the matrix A, and it did not occur to me.

I have posted the following in an earlier reply, but not in fancy formatting as the following, so it's worthwhile to repeat as follows:

the row echelon form of A is:
\[\left[\begin{matrix}1 & \frac{q}{p} & \frac{r}{p} & \frac{p+q+r}{p}\\ 0 & 1 & \frac{p^2-qr}{pr-q^2} & \frac{(p-q)r+p^2-q^2}{pr-q^2}\\ 0 & 0 & 1 & 1\end{matrix}\right]\]
which by normal backward pass gives (x,y,z)=(1,1,1).

As a check,
let a= square matrix formed by the first three columns of A, i.e.
a=\(\left[\begin{matrix}p & q & r\\ q & r & p\\ r & p & q\end{matrix}\right]\)
the inverse of a is
\( a^{-1}=\left[\begin{matrix}\frac{qr-p^2}{r(pq-r^2)+p(qr-p^2)+q(pr-q^2)} & \frac{pr-q^2}{r(pq-r^2)+p(qr-p^2)+q(pr-q^2)} & \frac{pq-r^2}{r(pq-r^2)+p(qr-p^2)+q(pr-q^2)}\\ \frac{pr-q^2}{r(pq-r^2)+p(qr-p^2)+q(pr-q^2)} & \frac{pq-r^2}{r(pq-r^2)+p(qr-p^2)+q(pr-q^2)} & \frac{qr-p^2}{r(pq-r^2)+p(qr-p^2)+q(pr-q^2)}\\ \frac{pq-r^2}{r(pq-r^2)+p(qr-p^2)+q(pr-q^2)} & \frac{qr-p^2}{r(pq-r^2)+p(qr-p^2)+q(pr-q^2)} & \frac{pr-q^2}{r(pq-r^2)+p(qr-p^2)+q(pr-q^2)}\end{matrix}\right]\)

and
\(\left[\begin{matrix}x\\y\\z\end{matrix}\right]=a^{-1}.\left[\begin{matrix}p+q+r\\p+q+r\\p+q+r\end{matrix}\right]=\left[\begin{matrix}1\\1\\1\end{matrix}\right]\)
exactly the same solution as by row echelon form.

You may notice that the denominator of the inverse is exactly \(p^3+q^3+r^3-3pqr\), which therefore calls for the restriction given in the original question.