How to solve for these simultaneous equations

Question

I have the following set of equations

$$\pi_1 = \pi_3 + [1 - \alpha(1 - p)]\pi_4$$ $$\pi_2 = \alpha(1 - p)\pi_4$$ $$\pi_3 = \alpha(1 - p)]\pi_1$$ $$\pi_4 = [1 - \alpha(1 - p)]\pi_1 + \pi_2$$ $$\pi_1 + \pi_2 + \pi_3 + \pi_4 = 1$$

Using this, I'm supposed to get the answer

$$\pi_1 = \pi_4 = \frac{1}{2 + 2\alpha(1 - p)} \hspace{1.5cm} \pi_2 = \pi_3 = \frac{\alpha(1 - p)}{2 + 2\alpha(1 - p)} $$

But I can't seem to do this. I always just end up with all the $\pi$'s equaling each other.

How do I do this question

Answer 1

In all honesty, "Just do it."

For example, if you substitute the third equation into the first, you get that

$$ [1-\alpha(1-p)](\pi_1-\pi_4)=0$$

For now, I'm going to assume that the coefficient is non-zero, which gives $\pi_1 = \pi_4$. The second and third equation then give that $\pi_2=\pi_3$. Substituting this into the last equation, we get that $\pi_1 = \frac {1}{ 2 +2\alpha(1-p)}$ (which is equal to $\pi_4$). And the value of $\pi_3 = \pi_2$ drops out immediately.

Now, if the coefficient is 0, then $[1-\alpha(1-p)]=0$, which gives $\pi_1=\pi_3$, $\pi_2=\pi_4$, and you have infinitely many solutions subject to $\pi_1+\pi_2 = \frac {1}{2}$.

Answer 2

We have to make this easier to read, so please forgive my rewriting of terms.

Let: $s = \pi_1, t = \pi_2, u = \pi_3, v = \pi_4, w = \alpha(1 - p)$.

Rewriting your system yields:

$\tag 1 s = u + (1 - w)v$

$\tag 2 t = wv$

$\tag 3 u = ws$

$\tag 4 v = (1 - w)s + wv$

$\tag 5 s + t + u + v = 1$

From $(4)$, we collect like terms on each side of the equation by subtracting $wv$ and get: $$\tag 6 v(1 - w) = (1 - w)s, \text{hence} ~v = s$$

Now, just substitute all the values you have into (5), yielding:

$u + (1-w)v + wv + ws + (1-w)s + wv = 1$, and collecting like terms and simplifying, we get:

$u + v + s + wv = 1$, and from $(3)$ we have $u = ws$, so we get:

$ws + s + v + wv = 1$

$s(1+w) + v(1+w) = 1$, or

$(s+v) = \frac{1}{1+w}$, but we know from (6) that $v = s$ and now have:

$2s = \frac{1}{1+w}$, or

$s = \frac{1}{2(1+w)}$

We now can easily substitute into $(2)$, yielding:

$t = w v = w s = w \frac{1}{2(1+w)}$, and

Substituting into $(3)$, yields:

$u = w s = w \frac{1}{2(1+w)}$

Regards