Question

The value of the determinant $∆=\left|\begin{array}{ccc}{\sin }^{2}33°& {\sin }^{2}57°& \cos 180°\\ -{\sin }^{2}57°& -{\sin }^{2}33°& {\cos }^{2}180°\\ \cos 180°& {\sin }^{2}33°& {\sin }^{2}57°\end{array}\right|$ is equal to _______________

Answer 1

Let $∆=\left|\begin{array}{ccc}{\sin }^{2}33°& {\sin }^{2}57°& \cos 180°\\ -{\sin }^{2}57°& -{\sin }^{2}33°& {\cos }^{2}180°\\ \cos 180°& {\sin }^{2}33°& {\sin }^{2}57°\end{array}\right|$


$$\begin{gathered} ∆=\left|\begin{array}{ccc}{\sin }^{2}33°& {\sin }^{2}57°& \cos 180°\\ -{\sin }^{2}57°& -{\sin }^{2}33°& {\cos }^{2}180°\\ \cos 180°& {\sin }^{2}33°& {\sin }^{2}57°\end{array}\right| \\ =\left|\begin{array}{ccc}{\sin }^{2}33°& {\sin }^{2}57°& -1\\ -{\sin }^{2}57°& -{\sin }^{2}33°& {\left(-1\right)}^2\\ -1& {\sin }^{2}33°& {\sin }^{2}57°\end{array}\right| \\ =\left|\begin{array}{ccc}{\sin }^{2}33°& {\sin }^{2}57°& -1\\ -{\sin }^{2}57°& -{\sin }^{2}33°& 1\\ -1& {\sin }^{2}33°& {\sin }^{2}57°\end{array}\right| \\ \mathrm{Applying}{R}_1\to R_1+R_2 \\ =\left|\begin{array}{ccc}{\sin }^{2}33°-{\sin }^{2}57°& {\sin }^{2}57°-{\sin }^{2}33°& -1+1\\ -{\sin }^{2}57°& -{\sin }^{2}33°& 1\\ -1& {\sin }^{2}33°& {\sin }^{2}57°\end{array}\right| \\ =\left|\begin{array}{ccc}{\sin }^{2}33°-{\sin }^{2}57°& {\sin }^{2}57°-{\sin }^{2}33°& 0\\ -{\sin }^{2}57°& -{\sin }^{2}33°& 1\\ -1& {\sin }^{2}33°& {\sin }^{2}57°\end{array}\right| \\ \mathrm{Taking}\mathrm{out}\left({\sin }^{2}33°-{\sin }^{2}57°\right)\mathrm{common}\mathrm{from}{R}_1 \\ =\left({\sin }^{2}33°-{\sin }^{2}57°\right)\left|\begin{array}{ccc}1& -1& 0\\ -{\sin }^{2}57°& -{\sin }^{2}33°& 1\\ -1& {\sin }^{2}33°& {\sin }^{2}57°\end{array}\right| \\ \mathrm{Applying}{C}_2\to C_2+C_1 \\ =\left({\sin }^{2}33°-{\sin }^{2}57°\right)\left|\begin{array}{ccc}1& -1+1& 0\\ -{\sin }^{2}57°& -{\sin }^{2}33°-{\sin }^{2}57°& 1\\ -1& {\sin }^{2}33°-1& {\sin }^{2}57°\end{array}\right| \\ =\left({\sin }^{2}33°-{\sin }^{2}57°\right)\left|\begin{array}{ccc}1& 0& 0\\ -{\sin }^{2}57°& {\cos }^{2}33°-1-{\sin }^{2}57°& 1\\ -1& -\left(1-{\sin }^{2}33°\right)& {\sin }^{2}57°\end{array}\right| \\ =\left({\sin }^{2}33°-{\sin }^{2}57°\right)\left|\begin{array}{ccc}1& 0& 0\\ -{\sin }^{2}57°& {\cos }^{2}33°-1-{\sin }^{2}57°& 1\\ -1& -{\cos }^{2}33°& {\sin }^{2}57°\end{array}\right| \\ =\left({\sin }^{2}33°-{\sin }^{2}57°\right)\left|\begin{array}{ccc}1& 0& 0\\ -{\sin }^{2}57°& {\cos }^2\left(90°-57°\right)-1-{\sin }^{2}57°& 1\\ -1& -{\cos }^2\left(90°-57°\right)& {\sin }^{2}57°\end{array}\right| \\ =\left({\sin }^{2}33°-{\sin }^{2}57°\right)\left|\begin{array}{ccc}1& 0& 0\\ -{\sin }^{2}57°& {\sin }^{2}57°-1-{\sin }^{2}57°& 1\\ -1& -{\sin }^{2}57°& {\sin }^{2}57°\end{array}\right| \\ =\left({\sin }^{2}33°-{\sin }^{2}57°\right)\left|\begin{array}{ccc}1& 0& 0\\ -{\sin }^{2}57°& -1& 1\\ -1& -{\sin }^{2}57°& {\sin }^{2}57°\end{array}\right| \\ \mathrm{Taking}\mathrm{out}\left(-1\right)\mathrm{common}\mathrm{from}{C}_2 \\ =\left(-1\right)\left({\sin }^{2}33°-{\sin }^{2}57°\right)\left|\begin{array}{ccc}1& 0& 0\\ -{\sin }^{2}57°& 1& 1\\ -1& {\sin }^{2}57°& {\sin }^{2}57°\end{array}\right| \\ =\left(-1\right)\left({\sin }^{2}33°-{\sin }^{2}57°\right)\left(0\right)\left(\because C_2\mathrm{and}{C}_3\mathrm{are}\mathrm{identical},\therefore \mathrm{the}\mathrm{value}\mathrm{of}\mathrm{determinant}\mathrm{is}\mathrm{zero}\right) \\ =0 \end{gathered}$$

Hence, the value of the determinant $∆=\left|\begin{array}{ccc}{\sin }^{2}33°& {\sin }^{2}57°& \cos 180°\\ -{\sin }^{2}57°& -{\sin }^{2}33°& {\cos }^{2}180°\\ \cos 180°& {\sin }^{2}33°& {\sin }^{2}57°\end{array}\right|$ is equal to 0.