Let $\alpha$ be a root of the equation $(a - c){x^2} + (b - a)x + (c - b) = 0$ where a, b, c are distinct real numbers such that the matrix $$\left[ {\matrix{ {{\alpha ^2}} & \alpha & 1 \cr 1 & 1 & 1 \cr a & b & c \cr } } \right]$$ is singular. Then, the value of $${{{{(a - c)}^2}} \over {(b - a)(c - b)}} + {{{{(b - a)}^2}} \over {(a - c)(c - b)}} + {{{{(c - b)}^2}} \over {(a - c)(b - a)}}$$ is
Solution
$$
\begin{aligned}
& \Delta=0=\left|\begin{array}{ccc}
\alpha^2 & \alpha & 1 \\\\
1 & 1 & 1 \\\\
\mathrm{a} & \mathrm{b} & \mathrm{c}
\end{array}\right| \\\\
& \Rightarrow \alpha^2(\mathrm{c}-\mathrm{b})-\alpha(\mathrm{c}-\mathrm{a})+(\mathrm{b}-\mathrm{a})=0
\end{aligned}
$$<br/><br/>
It is singular when $\alpha=1$<br/><br/>
$$
\begin{aligned}
& \frac{(a-c)^2}{(b-a)(c-b)}+\frac{(b-a)^2}{(a-c)(c-b)}+\frac{(c-b)^2}{(a-c)(b-a)} \\\\
& \frac{(a-b)^3+(b-c)^3+(c-a)^3}{(a-b)(b-c)(c-a)} \\\\
& =3 \frac{(a-b)(b-c)(c-a)}{(a-b)(b-c)(c-a)}=3
\end{aligned}
$$
About this question
Subject: Mathematics · Chapter: Matrices and Determinants · Topic: Types of Matrices and Operations
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