Question

If $\sin 2x = 1$, then \left( {\begin{array}{*{20}{c}} 0&{\cos x}&{ - \sin x} \\\ {\sin x}&0&{\cos x} \\\ {\cos x}&{\sin x}&0 \end{array}} \right) equals

Answer 1

Hint : As we know that the above question is related to trigonometric function and matrices. We know that a matrix is a collection of numbers, which are organised in rows and columns. Or we can say that it is an array of numbers arranged in a rectangular way and they are divided between rows and columns. In this question we will use the trigonometric formula and with the help of that we will solve the determinant of the given matrix.

Complete step by step solution:
As per the question we have $\sin 2x = 1$. We know that the value of $\sin 90$ or we can write it as $\sin \dfrac{\pi }{2}$ is $1$.
So we can write s$\sin 2x = \sin \dfrac{\pi }{2} \Rightarrow 2x = \dfrac{\pi }{2}$. It gives us the value $x = \dfrac{\pi }{{2 \cdot 2}} = \dfrac{\pi }{4}$.
We have the matrix \left( {\begin{array}{*{20}{c}} 0&{\cos x}&{ - \sin x} \\\ {\sin x}&0&{\cos x} \\\ {\cos x}&{\sin x}&0 \end{array}} \right). We will calculate the determinant of the matrix now.
We can calculate the determinant now: 0\left( {\begin{array}{*{20}{c}} 0&{\cos x} \\\ {\sin x}&0 \end{array}} \right) - \cos x\left( {\begin{array}{*{20}{c}} {\sin x}&{\cos x} \\\ {\cos x}&0 \end{array}} \right) + ( - \sin x)\left( {\begin{array}{*{20}{c}} {\sin x}&0 \\\ {\cos x}&{\sin x} \end{array}} \right).
So the above can be written as $0(0 \times 0 - \sin x\cos x) - \cos x(\sin x \times 0 - \cos x \times \cos x) + ( - \sin x)(\sin x \times \sin x - 0 \times \cos x)$
We know that anything multiplied by zero gives the value zero, so we can write the above as $- \cos x(0 - {\cos ^2}x) - \sin x({\sin ^2}x - 0)$.
Therefore the expression is ${\cos ^3}x - {\sin ^3}x$. From the above we put the value $x = \dfrac{\pi }{4}$ ; ${\cos ^3}\dfrac{\pi }{4} - {\sin ^3}\dfrac{\pi }{4}$ and also we know that $\cos \dfrac{\pi }{4} = \sin \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }}$.
So by putting the values we can write the above as${\left( {\dfrac{1}{{\sqrt 2 }}} \right)^3} - {\left( {\dfrac{1}{{\sqrt 2 }}} \right)^3} = \dfrac{1}{{2\sqrt 2 }} - \dfrac{1}{{2\sqrt 2 }}$.
It gives us the value $0$.
Hence the required answer is \left( {\begin{array}{*{20}{c}} 0&{\cos x}&{ - \sin x} \\\ {\sin x}&0&{\cos x} \\\ {\cos x}&{\sin x}&0 \end{array}} \right) equals $0$.
So, the correct answer is “0”.

Note : We should note that determinant of \left( {\begin{array}{*{20}{c}} {{a_{11}}}&{{a_{12}}}&{{a_{13}}} \\\ {{a_{21}}}&{{a_{22}}}&{{a_{23}}} \\\ {{a_{31}}}&{{a_{32}}}&{{a_{33}}} \end{array}} \right) can be written as {a_{11}}\left( {\begin{array}{*{20}{c}} {{a_{22}}}&{{a_{23}}} \\\ {{a_{32}}}&{{a_{33}}} \end{array}} \right) - {a_{12}}\left( {\begin{array}{*{20}{c}} {{a_{21}}}&{{a_{23}}} \\\ {{a_{31}}}&{{a_{33}}} \end{array}} \right) + {a_{13}}\left( {\begin{array}{*{20}{c}} {{a_{21}}}&{{a_{22}}} \\\ {{a_{31}}}&{{a_{32}}} \end{array}} \right). Before solving this kind of question we should be fully aware of the functions and the formulas of determinants and how to solve them. The above given matrix is $3 \times 3$ matrix.