Question

# If $\alpha$ and $\beta$ are the roots of the equation $a{{x}^{2}}+bx+c=0$, then the value of the determinant$\left| \begin{matrix} 1 & \cos \left( \beta -\alpha \right) & \cos \alpha \\ \cos \left( \beta -\alpha \right) & 1 & \cos \beta \\ \cos \alpha & \cos \beta & 1 \\\end{matrix} \right|$, isA. $\sin \left( \alpha +\beta \right)$B. $\sin \alpha \sin \beta$C. $1+\cos \left( \alpha +\beta \right)$D. $0$

Hint: Here, you have to find the determinant of $\left| \begin{matrix} 1 & \cos \left( \beta -\alpha \right) & \cos \alpha \\ \cos \left( \beta -\alpha \right) & 1 & \cos \beta \\ \cos \alpha & \cos \beta & 1 \\ \end{matrix} \right|$ by applying the formulas: $\left| \begin{matrix} {{a}_{11}} & {{a}_{12}} & {{a}_{13}} \\ {{a}_{21}} & {{a}_{22}} & {{a}_{23}} \\ {{a}_{31}} & {{a}_{32}} & {{a}_{33}} \\ \end{matrix} \right|={{a}_{11}}\left| \begin{matrix} {{a}_{22}} & {{a}_{23}} \\ {{a}_{32}} & {{a}_{33}} \\ \end{matrix} \right|-{{a}_{12}}\left| \begin{matrix} {{a}_{21}} & {{a}_{23}} \\ {{a}_{31}} & {{a}_{33}} \\ \end{matrix} \right|+{{a}_{13}}\left| \begin{matrix} {{a}_{21}} & {{a}_{22}} \\ {{a}_{31}} & {{a}_{32}} \\ \end{matrix} \right|$ and $\left| \begin{matrix} {{a}_{11}} & {{a}_{12}} \\ {{a}_{21}} & {{a}_{22}} \\ \end{matrix} \right|={{a}_{11}}{{a}_{22}}-{{a}_{12}}{{a}_{21}}$. Then, you have to reduce the terms by applying various trigonometric identities and cancel as many terms possible.

Complete Step-by-Step solution:
Here, we are given that $\alpha$ and $\beta$ are the roots of the equation $a{{x}^{2}}+bx+c=0$.
Now, we have to find the value of the determinant $\left| \begin{matrix} 1 & \cos \left( \beta -\alpha \right) & \cos \alpha \\ \cos \left( \beta -\alpha \right) & 1 & \cos \beta \\ \cos \alpha & \cos \beta & 1 \\ \end{matrix} \right|$.
We know that for a $3\times 3$ matrix the determinant is obtained by the formulas:
$\left| \begin{matrix} {{a}_{11}} & {{a}_{12}} & {{a}_{13}} \\ {{a}_{21}} & {{a}_{22}} & {{a}_{23}} \\ {{a}_{31}} & {{a}_{32}} & {{a}_{33}} \\ \end{matrix} \right|={{a}_{11}}\left| \begin{matrix} {{a}_{22}} & {{a}_{23}} \\ {{a}_{32}} & {{a}_{33}} \\ \end{matrix} \right|-{{a}_{12}}\left| \begin{matrix} {{a}_{21}} & {{a}_{23}} \\ {{a}_{31}} & {{a}_{33}} \\ \end{matrix} \right|+{{a}_{13}}\left| \begin{matrix} {{a}_{21}} & {{a}_{22}} \\ {{a}_{31}} & {{a}_{32}} \\ \end{matrix} \right|$
We also know that for a $2\times 2$ matrix we have the determinant formula:
$\left| \begin{matrix} {{a}_{11}} & {{a}_{12}} \\ {{a}_{21}} & {{a}_{22}} \\ \end{matrix} \right|={{a}_{11}}{{a}_{22}}-{{a}_{12}}{{a}_{21}}$
Now, let us denote:
$\Delta =\left| \begin{matrix} 1 & \cos \left( \beta -\alpha \right) & \cos \alpha \\ \cos \left( \beta -\alpha \right) & 1 & \cos \beta \\ \cos \alpha & \cos \beta & 1 \\ \end{matrix} \right|$
Now, we can say that:
$\Delta =1\left| \begin{matrix} 1 & \cos \beta \\ \cos \beta & 1 \\ \end{matrix} \right|-\cos \left( \beta -\alpha \right)\left| \begin{matrix} \cos \left( \beta -\alpha \right) & \cos \beta \\ \cos \alpha & 1 \\ \end{matrix} \right|+\cos \alpha \left| \begin{matrix} \cos \left( \beta -\alpha \right) & 1 \\ \cos \alpha & \cos \beta \\ \end{matrix} \right|$ … (1)
Now let us calculate the determinants of all terms separately.
First consider:
\begin{align} & \left| \begin{matrix} 1 & \cos \beta \\ \cos \beta & 1 \\ \end{matrix} \right|=\left( 1\times 1-\cos \beta \times \cos \beta \right) \\ & \left| \begin{matrix} 1 & \cos \beta \\ \cos \beta & 1 \\ \end{matrix} \right|=1-{{\cos }^{2}}\beta \\ \end{align}
We know that $1-{{\cos }^{2}}\beta ={{\sin }^{2}}\beta$. Hence, we will get:
$\left| \begin{matrix} 1 & \cos \beta \\ \cos \beta & 1 \\ \end{matrix} \right|={{\sin }^{2}}\beta$ ….. (2)

Next, let us consider $\left| \begin{matrix} \cos \left( \beta -\alpha \right) & \cos \beta \\ \cos \alpha & 1 \\ \end{matrix} \right|$.
$\left| \begin{matrix} \cos \left( \beta -\alpha \right) & \cos \beta \\ \cos \alpha & 1 \\ \end{matrix} \right|=\cos \left( \beta -\alpha \right)\times 1-\cos \beta \times \cos \alpha$
$\left| \begin{matrix} \cos \left( \beta -\alpha \right) & \cos \beta \\ \cos \alpha & 1 \\ \end{matrix} \right|=\cos \left( \beta -\alpha \right)-\cos \beta \cos \alpha$
We know that:
$\cos \left( \beta -\alpha \right)=\cos \beta \cos \alpha +\sin \beta \sin \alpha$
By applying this in the above equation we obtain:
$\left| \begin{matrix} \cos \left( \beta -\alpha \right) & \cos \beta \\ \cos \alpha & 1 \\ \end{matrix} \right|=\cos \beta \cos \alpha +\sin \beta \sin \alpha -\cos \beta \cos \alpha$
$\left| \begin{matrix} \cos \left( \beta -\alpha \right) & \cos \beta \\ \cos \alpha & 1 \\ \end{matrix} \right|=\sin \beta \sin \alpha$ ….. (3)
Now, we, have to find the determinant of $\left| \begin{matrix} \cos \left( \beta -\alpha \right) & 1 \\ \cos \alpha & \cos \beta \\ \end{matrix} \right|$.
$\left| \begin{matrix} \cos \left( \beta -\alpha \right) & 1 \\ \cos \alpha & \cos \beta \\ \end{matrix} \right|=\cos \left( \beta -\alpha \right)\times \cos \beta -1\times \cos \alpha$
$\left| \begin{matrix} \cos \left( \beta -\alpha \right) & 1 \\ \cos \alpha & \cos \beta \\ \end{matrix} \right|=\cos \left( \beta -\alpha \right)\cos \beta -\cos \alpha$ …. (4)

Now, substituting the equations (2), (3) and (4) in equation (1), we obtain:
$\Delta =1\times {{\sin }^{2}}\beta -\cos \left( \beta -\alpha \right)\times \sin \beta \sin \alpha +\cos \alpha \times \left( \cos \left( \beta -\alpha \right)\cos \beta -\cos \alpha \right)$

$\Delta ={{\sin }^{2}}\beta -\cos \left( \beta -\alpha \right)\sin \beta \sin \alpha +\cos \alpha \cos \beta \cos \left( \beta -\alpha \right)-{{\cos }^{2}}\alpha$
Now, for the second term and third term $\cos \left( \beta -\alpha \right)$ is the common factor. Therefore, we can take it outside, we will obtain:
$\Delta ={{\sin }^{2}}\beta -\cos \left( \beta -\alpha \right)\left( \sin \beta \sin \alpha \text{ }+\cos \alpha \cos \beta \right)-{{\cos }^{2}}\alpha$
We know the formula that:
$\cos (\beta -\alpha )=\cos \beta \cos \alpha +\sin \beta \sin \alpha$
Therefore our equation becomes:
$\Delta ={{\sin }^{2}}\beta -\cos \left( \beta -\alpha \right)\cos \left( \beta +\alpha \right)-{{\cos }^{2}}\alpha$ ….. (5)
We also know the formula that:
$\cos A\cos B=\dfrac{1}{2}\left( \cos (A+B)+\cos (A-B) \right)$
Hence, in equation (5) for the second term we have, $A=\beta +\alpha$ and $B=\beta -\alpha$. Therefore, we will get:
$\cos \left( \beta -\alpha \right)\cos \left( \beta +\alpha \right)=\dfrac{1}{2}\left[ \cos \left( \beta +\alpha +\beta +\alpha \right)+\cos \left( \beta +\alpha -\left( \beta +\alpha \right) \right) \right]$
$\cos \left( \beta -\alpha \right)\cos \left( \beta +\alpha \right)=\dfrac{1}{2}\left[ \cos \left( \beta +\alpha +\beta -\alpha \right)+\cos \left( \beta +\alpha -\beta +\alpha \right) \right]$
By addition and subtraction we obtain:
$\cos \left( \beta -\alpha \right)\cos \left( \beta +\alpha \right)=\dfrac{1}{2}\left[ \cos 2\beta +\cos 2\alpha \right]$ ….. (6)
We also know that:
${{\sin }^{2}}\beta =\dfrac{1-\cos 2\beta }{2}$ …… (7)
${{\cos }^{2}}\beta =\dfrac{1+\cos 2\beta }{2}$ ……. (8)
Now, by substituting equations (6), (7) and (8) in equation (5) we obtain:
$\Delta =\dfrac{1-\cos 2\beta }{2}-\dfrac{1}{2}\left[ \cos 2\beta +\cos 2\alpha \right]-\dfrac{1+\cos 2\beta }{2}$
Next, by taking the LCM we get:
$\Delta =\dfrac{1-\cos 2\beta -\left[ \cos 2\beta +\cos 2\alpha \right]-1+\cos 2\beta }{2}$
Now, by taking the terms outside of the bracket we get:
$\Delta =\dfrac{1-\cos 2\beta -\cos 2\beta -\cos 2\alpha -1+\cos 2\beta }{2}$
Next, by subtracting the similar terms we obtain:
\begin{align} & \Delta =\dfrac{0}{2} \\ & \Delta =0 \\ \end{align}
Hence, we can say that:
$\Delta =\left| \begin{matrix} 1 & \cos \left( \beta -\alpha \right) & \cos \alpha \\ \cos \left( \beta -\alpha \right) & 1 & \cos \beta \\ \cos \alpha & \cos \beta & 1 \\ \end{matrix} \right|=0$
Therefore, the correct answer for this question is option (d).

Note: Here, you can also find the determinant by making any two rows or columns similar. If any two rows or columns are similar then we can say that the determinant is zero.