- #1
BiGyElLoWhAt
Gold Member
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I'm working with matrices, so maybe this should be in the calc and beyond, but my problem is with the algebra (-.-)
I've reduced my matrix to this:
##(b-a)(c-a)\left | \begin{array}{cccc}
1 & (b-a)^{-1} & (c-a)^{-1} & 1 \\
0 & 1 & 1 & (d-a) \\
0 & 0 & (c+a) - (b+a) & (d^2- a^2) - (b + a)(d - a) \\
0 & 0 & (c^2 + ca +a^2) - (b^2 +ba +a^2) & (d^3 - a^3)- (b^2 + ba + a^2)(d - a) \\
\end{array} \right | ##
Now here's my dilemma, I need ##a_{4\ 3} = 0 ##, which means that I need to multiply ##[(c+a) - (b+a)]## by something to get that ugly term ##(c^2 + ca +a^2) - (b^2 +ba +a^2)##...
I'm not seeing anything, but I'm sure it can be done.
I've reduced my matrix to this:
##(b-a)(c-a)\left | \begin{array}{cccc}
1 & (b-a)^{-1} & (c-a)^{-1} & 1 \\
0 & 1 & 1 & (d-a) \\
0 & 0 & (c+a) - (b+a) & (d^2- a^2) - (b + a)(d - a) \\
0 & 0 & (c^2 + ca +a^2) - (b^2 +ba +a^2) & (d^3 - a^3)- (b^2 + ba + a^2)(d - a) \\
\end{array} \right | ##
Now here's my dilemma, I need ##a_{4\ 3} = 0 ##, which means that I need to multiply ##[(c+a) - (b+a)]## by something to get that ugly term ##(c^2 + ca +a^2) - (b^2 +ba +a^2)##...
I'm not seeing anything, but I'm sure it can be done.