Question

By using properties of determinants, show that:(i)\begin{bmatrix}1&a&a^2\\\ 1&b&b^2\\\ 1&c&c^2\end{bmatrix}=(a-b)(b-c)(c-a)$$(ii)\begin{bmatrix}1&1&1\\\ a&b&c\\\ a^3&b^3&c^3\end{bmatrix}=(a-b)(b-c)(c-a)(a+b+c)

Answer 1

(i) Let$\triangle=\begin{bmatrix}1&a&a^2\\\ 1&b&b^2\\\ 1&c&c^2\end{bmatrix}$
Applying $R_1 → R_1 − R_3$ and $R_2 → R_2 − R_3$, we have:
$△=\begin{bmatrix}0& a-c& a^2-c^2\\\ 0& b-c& b^2-c^2\\\ 1&c&c^2\end{bmatrix}$
$=(c-a)(b-c)\begin{bmatrix}0& -1& -a-c\\\ 0& 1& b+c\\\ 1&c&c^2\end{bmatrix}$
Applying $R_1 → R_1 + R_2$, we have:
$△=(b-c)(c-a)\begin{bmatrix}0&0&-a+b\\\ 0&1&b+c\\\ 1&c&c^2\end{bmatrix}$
$=(a-b)(b-c)(c-a)\begin{bmatrix}0&0&-1\\\ 0&1&b+c\\\ 1&c&c^2\end{bmatrix}$
Expanding along $C_1$, we have:
$△=(a-b)(b-c)(c-a)\begin{bmatrix}0&1\\\1&b+c\end{bmatrix}=(a-b)(b-c)(c-a)$
Hence, the given result is proved.
(ii)$Let \triangle=\begin{bmatrix}1&1&1\\\ a&b&c\\\ a^3&b^3&c^3\end{bmatrix}$
Applying $C_1 → C_1 − C_3$ and $C_2 → C_2 − C_3$, we have:
$△=\begin{bmatrix}0&0&1\\\ a-c& b-c& c\\\ a^3-c^3& b^3-c^3& c^3\end{bmatrix}$
$=\begin{bmatrix}0&0&1\\\ a-c& b-c& c\\\ (a-c)(a^2+ac+c^2)& (b-c)(b^2+bc+c^2)& c^3\end{bmatrix}$
$=(c-a)(b-c)\begin{bmatrix}0&0&1\\\ -1&1&c\\\ -(a^2+ac+c^2)& (b^2+bc+c^2)& c^3\end{bmatrix}$
Applying $C_1 → C_1 + C_2$, we have:
$△=(c-a)(b-c)\begin{bmatrix}0&0&1\\\ 0&1&c\\\ (b^2-a^2+(bc-ac)& (b^2+bc+c^2)& c^3\end{bmatrix}$
$△=(a-b)(c-a)(b-c)\begin{bmatrix}0&0&1\\\ 0&1&c\\\ (b^2-a^2+(bc-ac)& (b^2+bc+c^2)& c^3\end{bmatrix}$
Expanding along $C_1$, we have:
$△=(a-b)(c-a)(b-c)(a+b+c)(-1)\begin{bmatrix}0&1\\\1&c\end{bmatrix}$
$=(a-b)(b-c)(c-a)(a+b+c)$
Hence, the given result is proved.