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I have a set of equations representing a linear transformation from the variables $A, B, C$ to $X, Y, Z$. I know that this can be solved by substitution, but I can't figure out how to carry out that substitution. How can this be done? Here are my equations:

$A = \frac{1}{4}X + \frac{1}{2}Y + \frac{1}{4}Z$

$B = X - Y$

$C = Z - Y$

Now I want to retrieve the opposite transformation. Given $A, B, C$, what are the values for $X, Y, Z$? This can be solved by inverting a matrix about them, but I am hoping for a simple approach by substitution.

For full disclosure, this is exam review for me. I know the question and answer, but how the substitution is carried out doesn't make sense to me :)

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up vote 2 down vote accepted

From the second and third equation, you get $X = B + Y$ and $Z = C + Y$. Substitute in (four times) the first to get $4A = (B + Y) + 2Y + (C + Y) = B + C + 4Y$. So $Y = (4A - B - C)/4$ and so on ($X = (4A + 3B - C)/4$ and $Z = (4A - B + 3C)/4$).

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So, you have the following matrix equation $A=PX$:

$$\begin{bmatrix}A\\B\\C \end{bmatrix}=\begin{bmatrix}\frac 1 4 & \frac 1 4 &\frac 1 4\\ 1&0&-1&\\0&-1&1\end{bmatrix}\begin{bmatrix}X\\Y\\Z\end{bmatrix}$$

Yo're looking for a $Q$ such that $X=QA$. Working from what you have, you must see that $Q=P^{-1}$.

Hope you can evaluate the inverse...


Invert the co-efficient matrix...

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Sorry. I pointed out in the question that I know this can be solved by inverting a matrix, but I'm looking for the substitution approach :) – Deod Apr 17 '12 at 17:37
I am sorry, I did not read the question properly. :-( – user21436 Apr 17 '12 at 17:44

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