A fellow user had an interesting question:
Less-tedious way of solving this system of linear equations?
Is there a shortcut to solving for $w_1, w_2, w_3$ in $$
\left[ \begin{array}{ccc} 1 & 1 & 1\\ \frac{3}{4}a+\frac{1}{4}b &
\frac{1}{2}a+\frac{1}{2}b & \frac{1}{4}a+\frac{3}{4}b\\
(\frac{3}{4}a+\frac{1}{4}b)^2 & (\frac{1}{2}a+\frac{1}{2}b)^2 &
(\frac{1}{4}a+\frac{3}{4}b)^2\\ \end{array} \right] \cdot
\left[ \begin{array}{c} w_1\\ w_2\\ w_3 \end{array} \right] =
\left[ \begin{array}{c} b-a\\ \frac{b^2-a^2}{2}\\
\frac{b^3-a^3}{3} \end{array} \right] $$ where $a, b$ are
constants? (I'm trying to derive the formula for a 3-point open
Newton-Cotes quadrature rule.) Thanks!
Unfortunately the question was deleted (link), but I think it might be interesting in general.
Best Answer
One way is to use the free Maxima computer algebra system.
You can enter your matrix like this:
$$ \begin{pmatrix}1 & 1 & 1\\ \frac{b}{4}+\frac{3a}{4} & \frac{b}{2}+\frac{a}{2} & \frac{3b}{4}+\frac{a}{4}\\ {{\left( \frac{b}{4}+\frac{3a}{4}\right) }^{2}} & {{\left( \frac{b}{2}+\frac{a}{2}\right) }^{2}} & {{\left( \frac{3b}{4}+\frac{a}{4}\right) }^{2}}\end{pmatrix} $$
and the result vector like that:
$$ \begin{pmatrix}b-a\\ \frac{{{b}^{2}}-{{a}^{2}}}{2}\\ \frac{{{b}^{3}}-{{a}^{3}}}{3}\end{pmatrix} $$ This command will solve your linear system by $LU$ decomposition:
You extract and simplify the solution vector like this:
$$ \begin{pmatrix}\frac{2b-2a}{3}\\ -\frac{b-a}{3}\\ \frac{2b-2a}{3}\end{pmatrix} $$ where
first
picks the first element of the result andratsimp
is one of Maxima's simplification functions.With this you can verify the result:
$$ \begin{pmatrix}b-a\\ \frac{{{b}^{2}}-{{a}^{2}}}{2}\\ \frac{{{b}^{3}}-{{a}^{3}}}{3}\end{pmatrix} $$ Note that Maxima uses a dot for the matrix multiplication operator.