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.