RREF for 3-Variable Equations with Geometric Ratios

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May 26th 2010, 06:28 AM
chaosier

RREF for 3-Variable Equations with Geometric Ratios

Hello everyone, I'll start off my first post by explaining a few things.

The equations used in this matrix all have their coefficients related by a common ratio, for example:

8x+4y+2z=1 all have a ratio of 2 and so forth.

When I have three of such equations I find that the there is always a definite solution to these equations. With its appearance being that of an identity matrix I was wondering if there were any pattern related to the solutions as I found one for a 2x2 matrix.

Here are my algebraic expressions:

ax+(ar)y+(ar^2)z=ar^3
bx+(bn)y+(bn^2)z=bn^3
jx+(jk)y+(jk^2)z=jk^3

where a,b and j are not equal, and r/n/k are ratios respectively.

I put this into a matrix in its condensed form

Could anybody help me with using the Guassian method of elimination of help me obtain a resulting solution where it will look like a 3x3 identity matrix.

Thanks. (Cool)
May 26th 2010, 08:38 AM
Soroban

Hello, chaosier!

Quote:

$\begin{array}{cccc}ax+(ap)y+(ap^2)z &=& ap^3 & [1] \\ bx+(bq)y+(bq^2)z &=& bq^3 & [2] \\ cx+(cr)y+(cr^2)z &=& cr^3 & [3] \end{array}$

$\begin{array}{cccccc}\text{Divide [1] by }a\!: & x + py + p^2z &=& p^3 \\ \text{Divide [2] by }b\!: & x + qy +q^2z &=& q^2 \\ \text{Divide [3] by }c\!: & x + ry + r^2z &=& r^3 \end{array}$

$\text{We have: }\;\left[\begin{array}{ccc|c} 1 & p & p^2 & p^3 \\ 1 & q & q^2 & q^3 \\ 1 & r & r^2 & r^3 \end{array}\right]$

$\begin{array}{c} \\ R_2-R_1 \\ R_3-R_1\end{array} \left[\begin{array}{ccc|c}1 & p & p^2 & p^3 \\ 0 & q-p & q^2-p^2 & q^3-p^3 \\ 0 & r-p & r^2-o^2 & r^3-p^3 \end{array}\right]$

. . $\begin{array}{ccc} . \\ . \\ \frac{1}{q-p}R_2 \\ \frac{1}{r-p}R_3 \end{array}$ . $\left[ \begin{array}{ccc|c} 1 & p & p^2 & p^3 \\ 0 & 1 & q+r & q^2+qp + p^2 \\ 0 & 1 & r+p & r^2+rp + p^2 \end{array} \right]$

$\begin{array}{c} R_1 - p\!\cdot\!R_2 \\ \\ R_3 - R_2 \end{array} \left[\begin{array}{ccc|c}1 & 0 & -pq & -pq(p+q) \\ 0 & 1 & p+q & p^2 + pq + q^2 \\ 0 & 0 & r-q & (r^2-q^2) + p(r-q) \end{array}\right]$

. . $\begin{array}{ccc} \\ \\ \frac{1}{r-q}R_3 \end{array} \left[\begin{array}{ccc|c} 1 & 0 & -pq & -pq(p+q) \\ 0 & 1 & p+q & p^2 + pq + q^2 \\ 0 & 0 & 1 & p+q+r \end{array}\right]$

$\begin{array}{c}R_1 + pq\!\cdot\!R_3 \\ R_2 - (p+q)R_3 \\ \end{array} \left[\begin{array}{ccc|c}1 & 0 & 0 & pqr \\ 0 & 1 & 0 & -(pq+qr+pr) \\ 0 & 0 & 1 & p+q+r \end{array}\right]$
May 26th 2010, 03:50 PM
chaosier

Thanks a lot, by the way how do you get that straight line separating the 3x3 from the 3x1 in MathType