Linear system of equations that arises from recurrence relation.

Question

So when solving a third order homogenous recurrence relation I end upp with the task of determening the constants $a,b$ and $c$. The initial conditions give rise to the following system of equations:

$$\left\{ \begin{array}{rcr} 2a+(1+\sqrt{6})b+(1-\sqrt{6})c & = & 0\\ 4a+(1+\sqrt{6})^2b+(1-\sqrt{6})^2c & = & 0\\ 8a+(1+\sqrt{6})^3b+(1-\sqrt{6})^3c & = & 1 \end{array} \right.$$

I've verified that this system is correct but now I need to solve it. I find brute force Gaussian elimination quite tedious to do (or maybe there is some trick to make it simple?). Is there any other easier method to solve this without having to spend 3-4 pages of hand written scribble?

The answer is

$$a=-\frac{1}{10}, \ b=\frac{1}{10\sqrt{6}}, \ c=-\frac{1}{10\sqrt{6}}.$$

Answer 1

I would suggest that you use Cramer's method, i.e. using determinants.

One thing that will help you is to notice the following: Let $\phi=1+\sqrt6$. Then you have $$1-\sqrt6 = \frac{(1-\sqrt6)(1+\sqrt6)}{1+\sqrt6}=\frac{-5}{\phi}\ \ \ (*)$$ Put this into the equations, get rid of the fractions and find the determinants. One property that will be useful is that if a row or column of a determinant has a common factor, you can take the factor out in front of the determinant. After solving the system you will get expressions that contain $\phi$ and you will have to calculate some powers of $\phi$ which is not too hard to do.

Edit: More steps.

By using (*), you can eventually get to the system: \begin{eqnarray*} 2\phi a+\phi^2 b-5c &=& 0 \\ 4\phi^2 a+\phi^4 b+25c &=& 0 \\ 2\phi^3 a+\phi^6 b-125c &=& \phi^3 \end{eqnarray*}

Now, the determinants are: $$D = \left|\begin{array}{ccc} 2\phi & \phi^2 & -5 \\ 4\phi^2 & \phi^4 & 25 \\ 8\phi^3 & \phi^6 & -125 \\ \end{array}\right| = 2\phi\cdot\phi^2\cdot(-5) \left|\begin{array}{ccc} 1 & 1 & 1 \\ 2\phi & \phi^2 & -5 \\ 4\phi^2 & \phi^4 & 25 \\ \end{array}\right| $$ This determinant can be calculated using Sarrus' rule. $$ D_a = -5\phi^2 \left|\begin{array}{ccc} 0 & 1 & 1 \\ 0 & \phi^2 & -5 \\ \phi^3 & \phi^4 & 25 \\ \end{array}\right| =-5\phi^5 \left|\begin{array}{cc} 1 & 1 \\ \phi^2 & -5 \\ \end{array}\right| $$ $$ D_b = -10\phi \left|\begin{array}{ccc} 1 & 0 & 1 \\ 2\phi & 0 & -5 \\ 4\phi^2 & \phi^3 & 25 \\ \end{array}\right| = 10\phi^4 \left|\begin{array}{cc} 1 & 1 \\ 2\phi & -5 \\ \end{array}\right| $$ $$ D_c = 2\phi^3 \left|\begin{array}{ccc} 1 & 1 & 0 \\ 2\phi & \phi^2 & 0 \\ 4\phi^2 & \phi^4 & \phi^3 \\ \end{array}\right| = 2\phi^6 \left|\begin{array}{cc} 1 & 1 \\ 2\phi & \phi^2 \\ \end{array}\right| $$ You should get $$D=-10\phi^4(2\phi^4+\phi^3+5\phi-50)$$ $$D_a = 5\phi^5(\phi^2+5)$$ $$D_b=-10\phi^4(2\phi+5)$$ $$D_c=2\phi^7(\phi-2)$$

The solutions are then $a=D_a/D$, $b=D_b/D$, $c=D_c/D$. Plugging in the numbers I get the same solution that you gave.

Please let me know if I have made any errors along the way.

Answer 2

Very basic (done by hand in less than a page with pen and paper).

Consider $$2 a+A b+B c = 0\tag 1$$ $$ 4 a+A^2 b+B^2 c =0\tag 2$$ $$ 8 a+A^3 b+B^3 c =1\tag 3$$

Use $(1)$ and $(2)$ and solve for $(a,b)$ to get $$a=\frac{B (B-A)}{2 (A-2)} c\qquad \text{and} \qquad b=\frac{(2-B) B }{(A-2) A}c$$ Plug in $(3)$ and solve for $c$ to get $$c=\frac{1}{(2-B) B (A-B)}\implies a=\frac{1}{2 (A-2) (B-2)}\qquad \text{and} \qquad b=\frac{1}{(A-2) A (A-B)}$$

Now, make $A=1+\sqrt 6$ and $B=1-\sqrt 6$.