Question

(i) Solve the equation
$\left|\begin{array}{ccc}x-1& 2& 3\\ 0& x-2& 4\\ 0& 0& x-3\end{array}\right|$
(ii) Show that
$\left|\begin{array}{ccc}1& a& bc\\ 1& b& ca\\ 1& c& ab\end{array}\right|=\left|\begin{array}{ccc}1& a& a^2\\ 1& b& b^2\\ 1& c& c^2\end{array}\right|$

Answer 1

$\left(i\right)\left|\begin{array}{ccc}x-1& 2& 3\\ 0& x-2& 4\\ 0& 0& x-3\end{array}\right|=0$

$\Rightarrow (x-1)(x-2)(x-3)-2\left(0\right)+3\left(0\right)=0$

$\Rightarrow (x-1)(x-2)(x-3)=0$

$\Rightarrow x=1,x=2,x=3$

$\left(ii\right)$

LHS =

$\left|\begin{array}{ccc}1& a& bc\\ 1& b& ca\\ 1& c& ab\end{array}\right|=1(a{b}^2-c^{2}a)-a(ab-ca)+bc(c-b)$

$=a{b}^2-c^{2}a-a^{2}b+c{a}^2+b{c}^2-b^{2}c$

$=a{b}^2+c{a}^2+b{c}^2-c^{2}a-a^{2}b-b^{2}c$

RHS

$\left|\begin{array}{ccc}1& a& a^2\\ 1& b& b^2\\ 1& c& c^2\end{array}\right|$

$=1(b{c}^2-c{b}^2)-a(c^2-b^2)+a^2(c-b)$

$=b{c}^2-c{b}^2-a{c}^2+a{b}^2+a^{2}c-a^{2}b$

$=a{b}^2+a^{2}c+b{c}^2-c{b}^2-a{c}^2-a^{2}b$

$\Rightarrow LHS=RHS$