Buszmenan

2022-09-01

Huge linear system of equations with powers of $2$:
$\begin{array}{rl}\left({2}^{0}{\right)}^{n}{a}_{n}+\left({2}^{0}{\right)}^{n-1}{a}_{n-1}+\cdots +\left({2}^{0}{\right)}^{1}{a}_{1}& ={4}^{0}\\ \left({2}^{1}{\right)}^{n}{a}_{n}+\left({2}^{1}{\right)}^{n-1}{a}_{n-1}+\cdots +\left({2}^{1}{\right)}^{1}{a}_{1}& ={4}^{1}\\ ⋮\\ \left({2}^{n-1}{\right)}^{n}{a}_{n}+\left({2}^{n-1}{\right)}^{n-1}{a}_{n-1}+\cdots +\left({2}^{n-1}{\right)}^{1}{a}_{1}& ={4}^{n-1}\end{array}$
for $n\ge 2$. Show that the (unique) solution of this system is when ${a}_{2}=1$ and all other variables are zero.

ordonansexa

Suppose that $\left({a}_{1},\dots ,{a}_{n}\right)$ is a solution of the proposed system. Consider $P\left(X\right)=-{X}^{2}+\sum _{1}^{n}{a}_{j}{X}^{j}$. This is a polynomial of degree smaller or equal to $n$, with $P\left(0\right)=0$. Moreover, by assumption $P\left({2}^{i}\right)=0$ for every$i=0,\dots n-1$, so $P$ has at $n+1$ zeros, and it must be identically zero. So $\left({a}_{1},{a}_{2},\dots ,{a}_{n}\right)=\left(0,1,\dots ,0\right)$. The converse is trivially true.

hikstac0

Use Cramer's rule remembering the ${4}^{k}=\left({2}^{k}{\right)}^{2}$ which makes the right hand side the same as the second to last column. This means that two columns will be the same (and so the determinant will be zero) for all variables except for ${a}_{2}$ where the determinant in the numerator will be the same as the determinant in the denominator.