JEE Mains · Maths · STD 12 - 3 and 4 . metrices and determinant
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 \(\left[\begin{array}{ccc}\alpha^2 & \alpha & 1 \\1 & 1 & 1 \\a & b & c\end{array}\right]\) is singular. Then the value of \(\frac{(a-c)^2}{(b-a)(c-b)}+\frac{(b-a)^2}{(a-c)(c-b)}+\frac{(c-b)^2}{(a-c)(b-a)}\)
- A \(6\)
- B \(3\)
- C \(9\)
- D \(12\)
Answer & Solution
Correct Answer
(B) \(3\)
Step-by-step Solution
Detailed explanation
\(\begin{array}{l}\Delta=0=\left|\begin{array}{ccc}\alpha^2 & \alpha & 1 \\1 & 1 & 1 \\a & b & c\end{array}\right| \\\Rightarrow \alpha^2( c - b )-\alpha( c - a )+( b - a )=0\end{array}\) It is singular when \(\alpha=1\)…
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