Matrices and Determinants 2 Question 41
45. If $\alpha$ be a repeated root of a quadratic equation $f(x)=0$ and $A(x), B(x)$ and $C(x)$ be polynomials of degree 3,4 and 5 respectively, then show that
$ \left|\begin{array}{lll} A(x) & B(x) & C(x) \\ A(\alpha) & B(\alpha) & C(\alpha) \\ A^{\prime}(\alpha) & B(\alpha) & C^{\prime}(\alpha) \end{array}\right| $
is divisible by $f(x)$, where prime denotes the derivatives.
$(1984,3$ M)
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Answer:
Solution:
- Since, $\alpha$ is repeated root of $f(x)=0$.
$\therefore f(x)=a(x-\alpha)^{2}, a \in$ constant $(\neq 0)$
Let $\quad \varphi(x)=\left|\begin{array}{ccc}A(x) & B(x) & C(x) \\ A(\alpha) & B(\alpha) & C(\alpha) \\ A^{\prime}(\alpha) & B^{\prime}(\alpha) & C^{\prime}(\alpha)\end{array}\right|$
To show $\varphi(x)$ is divisible by $(x-\alpha)^{2}$, it is sufficient to show that $\varphi(\alpha)$ and $\varphi^{\prime}(\alpha)=0$.
$ \begin{aligned} & \therefore \quad \varphi(\alpha)=\left|\begin{array}{ccc} A(\alpha) & B(\alpha) & C(\alpha) \\ A(\alpha) & B(\alpha) & C(\alpha) \\ A^{\prime}(\alpha) & B^{\prime}(\alpha) & C^{\prime}(\alpha) \end{array}\right| \\ & =0 \quad\left[\because R_{1} \text { and } R_{2} \text { are identical }\right] \\ & \text { Again, } \quad \varphi^{\prime}(x)=\left|\begin{array}{ccc} A^{\prime}(x) & B^{\prime}(x) & C^{\prime}(x) \\ A(\alpha) & B(\alpha) & C(\alpha) \\ A^{\prime}(\alpha) & B^{\prime}(\alpha) & C^{\prime}(\alpha) \end{array}\right| \\ \end{aligned} $
$ \begin{aligned} & =0 \quad\left[\because R_{1} \text { and } R_{3} \text { are identical }\right] \end{aligned} $
Thus, $\alpha$ is a repeated root of $\varphi(x)=0$.
Hence, $\varphi(x)$ is divisible by $f(x)$.
