Find $ax^5 + by^5$ by matrix from $ \left\{ \begin{aligned} ax+by&=3\\ ax^2+by^2&=7\\ ax^3+by^3&=16\\ ax^4+by^4&=42 \end{aligned} \right.$ 
Find $ax^5 + by^5$ if the real numbers $a$, $b$, $x$, and $y$ satisfy the equations
  $$
\left\{
\begin{aligned} 
ax+by&=3 \\
ax^{2}+by^{2}&=7 \\ 
ax^{3}+by^{3}&=16 \\
ax^{4}+by^{4}&=42
\end{aligned} 
\right.
$$
  with matrix form!

Interestingly, when I googled the equations, the problem comes from 1990 AIME Problems/Problem 15. The link gives two methods for the solution.
Is there a fashioned way to find the solution in matrix form?

The secuence $c_n=ax^n+by^n$ satisfies a linear two step recurrence relation of the form
$$c_{n+2}=Ac_{n+1}+Bc_n$$
for some constants $A,B$. This means that if we perform the row operations coming from that recurrence relation (subtract the second row multiplied by $A$ and the first row multiplied by $B$ from the third row) to the matrix
$$
\left(\begin{array}{ccc}c_1&c_2&c_3\\c_2&c_3&c_4\\c_3&c_4&c_5\end{array}\right)
$$
its last row becomes all zeros. This means that its determinant vanishes, i.e.
$$
0=\left\vert\begin{array}{ccc}c_1&c_2&c_3\\c_2&c_3&c_4\\c_3&c_4&c_5\end{array}\right\vert=\left\vert\begin{array}{ccc}3&7&16\\7&16&42\\16&42&c_5\end{array}\right\vert=20-c_5.
$$
Therefore $c_5=20$.
