Question

If $ \alpha $ is a repeated root of a quadratic equation $ f\left( x \right) = 0 $ and $ A\left( x \right) $ , $ B\left( x \right) $ and $ C\left( x \right) $ be polynomials of degree > 2 then the determinant $ \left( {\begin{array}{*{20}{c}}
{A\left( x \right)}&{B\left( x \right)}&{C\left( x \right)}\\
{A\left( \alpha \right)}&{B\left( \alpha \right)}&{C\left( \alpha \right)}\\
{{A'}\left( \alpha \right)}&{{B'}\left( \alpha \right)}&{{C'}\left( \alpha \right)}
\end{array}} \right) $ is divisible by
A. $ A\left( x \right) $
B. $ B\left( x \right) $
C. $ C\left( x \right) $
D. $ f\left( x \right) $

Answer 1

Hint: To find that which function divides the given matrix, we will find the factors of the matrix. Firstly, we will find the differential of the matrix and then we will substitute $ \alpha $ for $ x $ in the matrix. We will apply the properties of matrices to find the determinant. If the value of the determinant comes out to be zero then $ \alpha $ is the root. Then we will check for the factors of the determinant of the matrix and the given quadratic equation. If both of them have the same factors, then the determinant of the matrix is divisible by the given quadratic equation.

Complete step-by-step answer:
We assume the given matrix as $\left( {\begin{array}{*{20}{c}}
{A\left( x \right)}&{B\left( x \right)}&{C\left( x \right)}\\
{A\left( \alpha \right)}&{B\left( \alpha \right)}&{C\left( \alpha \right)}\\
{{A'}\left( \alpha \right)}&{{B'}\left( \alpha \right)}&{{C'}\left( \alpha \right)}
\end{array}} \right)$ is equal to $g\left( x \right)$.
Then the differential of $g\left( x \right)$ can be represented as:
${G'}\left( x \right) = \left( {\begin{array}{*{20}{c}}
{{A'}\left( x \right)}&{{B'}\left( x \right)}&{{C'}\left( x \right)}\\
{A\left( \alpha \right)}&{B\left( \alpha \right)}&{C\left( \alpha \right)}\\
{{A'}\left( \alpha \right)}&{{B'}\left( \alpha \right)}&{{C'}\left( \alpha \right)}
\end{array}} \right)$
If we are putting $\alpha $ for $x$ then $g\left( \alpha \right)$ can be represented as:
$g\left( \alpha \right) = \left( {\begin{array}{*{20}{c}}
{A\left( \alpha \right)}&{B\left( \alpha \right)}&{C\left( \alpha \right)}\\
{A\left( \alpha \right)}&{B\left( \alpha \right)}&{C\left( \alpha \right)}\\
{{A'}\left( \alpha \right)}&{{B'}\left( \alpha \right)}&{{C'}\left( \alpha \right)}
\end{array}} \right)$
From the properties of Matrices, we know that when two rows or columns are identical, then the determinant of the matrix becomes equal to zero. Hence, $g\left( \alpha \right) = 0$ . Therefore, we can say that $\left( {x - \alpha } \right)$ is the root of $g\left( \alpha \right) = 0$.
Then the differential of $g\left( \alpha \right)$ can be represented as:
${G'}\left( \alpha \right) = \left( {\begin{array}{*{20}{c}}
{A'\left( \alpha \right)}&{{B'}\left( \alpha \right)}&{{C'}\left( \alpha \right)}\\
{A\left( \alpha \right)}&{B\left( \alpha \right)}&{C\left( \alpha \right)}\\
{{A'}\left( \alpha \right)}&{{B'}\left( \alpha \right)}&{{C'}\left( \alpha \right)}
\end{array}} \right)$
Now again, since two rows are identical. Hence, the determinant of the matrix becomes equals to zero. Hence, ${G'}\left( \alpha \right) = 0$. Therefore, we can say that $\left( {x - \alpha } \right)$ is the root of ${G'}\left( \alpha \right) = 0$.
From the above expressions, we have $g\left( \alpha \right) = 0$ and ${G'}\left( \alpha \right) = 0$, $\alpha $ is the root and \[\left( {x - \alpha } \right)\] is the factor of the determinant of the $g\left( \alpha \right) = 0$ and ${G'}\left( \alpha \right) = 0$.

 Since, it is given in the question that $\alpha $ is the repeated root of the quadratic equation $f\left( x \right) = 0$. This can be expressed as ${\left( {x - \alpha } \right)^2}$ is the factor of $f\left( x \right) = 0$ .
Since, $g\left( \alpha \right) = 0$,${G'}\left( \alpha \right) = 0$and $f\left( x \right) = 0$ has common root ${\left( {x - \alpha } \right)^2}$. Hence, we can say that $g\left( x \right)$ is divisible by $f\left( x \right)$.
The determinant of $\left( {\begin{array}{*{20}{c}}
{A\left( x \right)}&{B\left( x \right)}&{C\left( x \right)}\\
{A\left( \alpha \right)}&{B\left( \alpha \right)}&{C\left( \alpha \right)}\\
{{A'}\left( \alpha \right)}&{{B'}\left( \alpha \right)}&{{C'}\left( \alpha \right)}
\end{array}} \right)$ is divisible by $f\left( x \right)$ .
So, the correct answer is “Option D”.

Note: We can define roots of a matrix as the value of $ x $ for which the value of determinant becomes zero. So, in this question we will find the roots of the determinant and then the factors. A factor is a term or a combination of the terms which when divides an equation or number, the remainder is always equals to zero.