Need help simplifying a set of equations (and understanding how to solve it)

Question

i have three algebraic expressions, each using the others. in these equations a, b, c and t are known and plugged in later:

$x = a^{-1}(t + y + z)$

$y = b^{-1}(t + x + z)$

$z = c^{-1}(t + x + y)$

i have managed to successfully solve the equations for two statements using substitution:

$x = a^{-1}(t + y)$

$y = b^{-1}(t + x)$

which substitutes and simplifies as:

$x = a^{-1}t + a^{-1}b^{-1}t + a^{-1}b^{-1}x$

$x - a^{-1}b^{-1}x = a^{-1}t + a^{-1}b^{-1}t$

$x(1 - a^{-1}b^{-1}) = a^{-1}t(b^{-1} + 1)$

therefore:

$x = a^{-1}t(1+b^{-1}) / (1 - a^{-1}b^{-1})$

$y = b^{-1}t(1+a^{-1}) / (1 - a^{-1}b^{-1})$

but introducing a third term has me stumped, as i keep getting a nonsense result when substituting leaving me with $x = x$ when i simplify the equation. (this is obviously correct, but also useless.) I'm hoping that someone would be so kind as to explain how i go about substituting more than one term and solving x, y and z in terms of t, a, b and c. does the same principle apply as with two terms?

for completeness, here is my (incorrect) work:

$x = a^{-1}t + a^{-1}y + a^{-1}z$

$a^{-1}z = x - (a^{-1}t + a^{-1}y)$

$a^{-1}z = x - a^{-1}(t + y)$

$z = ax - (t + y)$ (multiply both sides by a)

therefore, to find y i substitute z:

$x = a^{-1}t + a^{-1}y + a^{-1}(ax - (t + y))$

$x = a^{-1}t + a^{-1}y + x - a^{-1}t - a^{-1}y$

$x = (a^{-1}t - a^{-1}t) + (a^{-1}y - a^{-1}y) + x$

$x = 0 + 0 + x$

$x = x$

please forgive me if my terminology isn't correct. i don't exactly have a maths background and english isn't my first language.

Answer 1

You can write your system as

$$\left\{\begin{align} ax = (y+z) + t\\ by = (x+z) + t\\ cz = (x+y) + t \end{align}\right.$$

and then in matrix form you'd get

$$\begin{align} &\pmatrix{a&0&0\\0&b&0\\0&0&c}\pmatrix{x\\y\\z} = \pmatrix{0&1&1\\1&0&1\\1&1&0}\pmatrix{x\\y\\z} + t\pmatrix{1\\1\\1}\\ \implies &\underbrace{\pmatrix{a&-1&-1\\-1&b&-1\\-1&-1&c}}_{M}\pmatrix{x\\y\\z} = t\pmatrix{1\\1\\1}. \end{align}$$

Now, $M$ has determinant $\det(M) = abc - a -b -c -2$, so provided $\det(M)\neq 0$ your system admits a unique solution. In this case, we have

$$M^{-1} = \frac1{\det(M)}\pmatrix{bc-1&c+1&b+1\\c+1&ac-1&a+1\\b+1&a+1&ab-1}$$

and

$$\pmatrix{x\\y\\z} = tM^{-1}\pmatrix{1\\1\\1}.$$

We can explicitly compute and obtain

$$\begin{align} \pmatrix{x\\y\\z} &= \frac t{abc-a-b-c-2}\pmatrix{b+c+bc+1\\a+c+ac+1\\a+b+ab+1} \\ &= \frac t{abc-a-b-c-2}\pmatrix{(b+1)(c+1)\\(a+1)(c+1)\\(a+1)(b+1)} .\end{align}$$