Answer
$(\frac{1}{a(a-b)}, -\frac{1}{b(a-b)})$
Work Step by Step
Given system is-
$ax +by$ = $0$
i.e. $ax $ = $-by$
i.e. $x $ = $\frac{-by}{a}$ __ eq.1
$a^{2}x +b^{2}y$ = $1$ __ eq.2
Substituting for $x$ in eq.2 from eq.1
$a^{2}\frac{-by}{a}+b^{2}y$ = $1$
i.e. $-aby+b^{2}y$ = $1$
i.e. $y(b^{2}-ab)$ = $1$
i.e. $by(b-a)$ = $1$
i.e. $y$ = $\frac{1}{b(b-a)}$, Given $b\ne 0$ , $a\ne b$
Substituting for $y$ in eq.1
$x $ = $\frac{-by}{a}$
i.e. $x $ = $\frac{-b}{a} \frac{1}{b(b-a)}$, Given $a\ne 0$ , $a\ne b$
i.e. $x $ = $-\frac{1}{a(b-a)}$
i.e. $x $ = $\frac{1}{a(a-b)}$
Thus $(\frac{1}{a(a-b)}, \frac{1}{b(b-a)})$ is the solution of given system.
i.e. $(\frac{1}{a(a-b)}, -\frac{1}{b(a-b)})$ is the solution of given system.