Question

$ax+by+cz=k$
$a^{2}x+b^{2}y+c^{2}z=k^2$
$a^{3}x+b^{3}y+c^{3}z=k^3$
Solve by Crammer's rule

Answer 1

$ax+by+cz=k$

$a^{2}x+b^{2}y+c^z=k^2$

$a^{3}x+b^{3}y+c^z=k^3$

using cramer's rule,

$x=\frac{D_1}{D},y=\frac{D_2}{D},Z=\frac{D_3}{D}$

where,

$D=\left[\begin{array}{ccc}a& b& c\\ a^2& b^2& c^2\\ a^3& b^3& c^3\end{array}\right]=abc\left[\begin{array}{ccc}1& 1& 1\\ a& b& c\\ a^2& b^2& c^2\end{array}\right]$

$=abc\left[\begin{array}{ccc}0& 1& 0\\ a-b& b& c-b\\ a^2-b^2& b^2& c^2-b^2\end{array}\right]=abc(a-b)(c-b)\left[\begin{array}{ccc}0& 1& 0\\ 1& b& 1\\ a+b& b^2& c+b\end{array}\right]$

$=(a-b)(c-b)(a+b-b-c)abc$

$=(a-b)(b-c)(c-a)abc$

$D_1=\left[\begin{array}{ccc}k& b& c\\ k^2& b^2& c^2\\ k^3& b^3& c^3\end{array}\right]=kbc\left[\begin{array}{ccc}1& 1& 1\\ k& b& c\\ k^2& b^2& c^2\end{array}\right]=kbc\left[\begin{array}{ccc}0& 1& 0\\ k-b& b& c-b\\ k^2-b^2& b^2& c^2-b^2\end{array}\right]$

$-kbc(k-b)(c-b)\left[\begin{array}{ccc}0& 1& 0\\ 1& b& 1\\ k+b& b^2& c+b\end{array}\right]$

$=kbc(k-b)(c-b)(k+b-c-b)$

$=kbc(k-b)(b-c)(c-k)$

$\therefore x=\frac{D_1}{D}=\frac{kbc(k-b)(b-c)(c-k)}{kbc(a-b)(b-c)(c-a)}$

$=\frac{k(k-b)(b-c)(c-k)}{a(a-b)(b-c)(c-a)}$

$D_2=\left[\begin{array}{ccc}a& k& c\\ a^2& k^2& c^2\\ a^3& k^3& c^3\end{array}\right]=akc\left[\begin{array}{ccc}1& 1& 1\\ a& k& c\\ a^2& k^2& c^2\end{array}\right]=akc\left[\begin{array}{ccc}0& 1& 0\\ a-b& k& c-k\\ a^2-k^2& k^2& c^2-k^2\end{array}\right]$

$-akc(a-k)(c-k)\left[\begin{array}{ccc}0& 1& 0\\ 1& k& 1\\ a+k& k^2& c+k\end{array}\right]$

$=akc(a-k)(c-k)(a+k-c-k)$

$=akc(a-k)(k-c)(c-a)$

$\therefore y=\frac{D_2}{D}=\frac{akc(a-k)(k-c)(c-a)}{abc(a-b)(b-c)(c-a)}$

$=\frac{k(a-k)(k-c)(c-a)}{b(a-b)(b-c)(c-a)}$

$D_3=\left[\begin{array}{ccc}a& b& k\\ a^2& b^2& k^2\\ a^3& b^3& k^3\end{array}\right]=abk\left[\begin{array}{ccc}1& 1& 1\\ a& b& k\\ a^2& b^2& k^2\end{array}\right]=abk\left[\begin{array}{ccc}0& 1& 0\\ a-b& b& k-b\\ a^2-b^2& b^2& k^2-b^2\end{array}\right]$

$-abk(a-b)(k-b)\left[\begin{array}{ccc}0& 1& 0\\ 1& b& 1\\ a+b& b^2& k+b\end{array}\right]$

$=abk(a-b)(k-b)(a+b-k-b)$

$=abk(a-b)(b-k)(k-a)$

$\therefore z=\frac{D_3}{D}=\frac{abk(a-b)(b-k)(k-a)}{abc(a-b)(b-c)(c-a)}$

$=\frac{k(a-b)(b-k)(k-a)}{c(a-b)(b-c)(c-a)}$