How to symbolically solve a system of linear equations 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.
 A: One way is to use the free Maxima computer algebra system.
You can enter your matrix like this:A: matrix(
 [1,1,1], 
 [(3/4)*a+(1/4)*b,(1/2)*a+(1/2)*b,(1/4)*a+(3/4)*b], 
 [((3/4)*a+(1/4)*b)^2,((1/2)*a+(1/2)*b)^2,((1/4)*a+(3/4)*b)^2]
 );

$$
\begin{pmatrix}1 & 1 & 1\\
\frac{b}{4}+\frac{3a}{4} & \frac{b}{2}+\frac{a}{2} & \frac{3b}{4}+\frac{a}{4}\\
{{\left( \frac{b}{4}+\frac{3a}{4}\right) }^{2}} & {{\left( \frac{b}{2}+\frac{a}{2}\right) }^{2}} & {{\left( \frac{3b}{4}+\frac{a}{4}\right) }^{2}}\end{pmatrix}
$$
and the result vector like that:
bcol: matrix(
 [b-a], 
 [(b^2-a^2)/2], 
 [(b^3-a^3)/3]
);

$$
\begin{pmatrix}b-a\\
\frac{{{b}^{2}}-{{a}^{2}}}{2}\\
\frac{{{b}^{3}}-{{a}^{3}}}{3}\end{pmatrix}
$$
This command will solve your linear system by $LU$ decomposition:
ls : linsolve_by_lu(A,bcol);

You extract and simplify the solution vector like this:
xycol: ratsimp(first(ls));

$$
\begin{pmatrix}\frac{2b-2a}{3}\\
-\frac{b-a}{3}\\
\frac{2b-2a}{3}\end{pmatrix}
$$
where first picks the first element of the result and ratsimp is one of Maxima's simplification functions.
With this you can verify the result:
ratsimp(A . xycol);

$$
\begin{pmatrix}b-a\\
\frac{{{b}^{2}}-{{a}^{2}}}{2}\\
\frac{{{b}^{3}}-{{a}^{3}}}{3}\end{pmatrix}
$$
Note that Maxima uses a dot for the matrix multiplication operator.
A: One can also make use of the Mathematica programming language and proceed as follows:

where the lines with MatrixForm[$\cdots$] were just to check if the matrices were fine.
The command LinearSolve[matrix, result] finds the solution and the added // MatrixForm is just for pretty printing.
