Question

If {\left| {\begin{array}{*{20}{c}} {\cos \dfrac{{2\pi }}{7}}&{ - \sin \dfrac{{2\pi }}{7}} \\\ {\sin \dfrac{{2\pi }}{7}}&{\cos \dfrac{{2\pi }}{7}} \end{array}} \right|^k} = \left| {\begin{array}{*{20}{c}} 1&0 \\\ 0&1 \end{array}} \right|, then the least positive integral value of k is:
A. $3$
B. $4$
C. $6$
D. $7$

Answer 1

According to the question, first we will assign $\dfrac{{2\pi }}{7} = \theta$. Then, we will square the given matrix. This means that in place of ‘k’, we will take the power to be 2. Then we will solve the matrix and again replace 2 from ‘k’. Now, we will try to solve for ‘k’ and get its value

Formulae Used:
$\cos 2\theta = {\cos ^2}\theta - {\sin ^2}\theta$ and $\sin 2\theta = 2\cos \theta \sin \theta$

Complete step by step answer:
The given matrix is:

{\cos \dfrac{{2\pi }}{7}}&{ - \sin \dfrac{{2\pi }}{7}} \\\ {\sin \dfrac{{2\pi }}{7}}&{\cos \dfrac{{2\pi }}{7}} \end{array}} \right|^k} = \left| {\begin{array}{*{20}{c}} 1&0 \\\ 0&1 \end{array}} \right|$$ Now let us assume $$\dfrac{{2\pi }}{7} = \theta $$, hence we can write the LHS of the given matrix as $${\left| {\begin{array}{*{20}{c}} {\cos \theta }&{ - \sin \theta } \\\ {\sin \theta }&{\cos \theta } \end{array}} \right|^k} - - (i)$$ Let us assume$$k = 2$$, hence we can rewrite the above matrix as $$ \Rightarrow {\left| {\begin{array}{*{20}{c}} {\cos \theta }&{ - \sin \theta } \\\ {\sin \theta }&{\cos \theta } \end{array}} \right|^2}$$ Now we will square the above obtained matrix, so we get

{\left| {\begin{array}{{20}{c}}
{\cos \theta }&{ - \sin \theta } \\
{\sin \theta }&{\cos \theta }
\end{array}} \right|^2} = \left| {\begin{array}{{20}{c}}
{\cos \theta }&{ - \sin \theta } \\
{\sin \theta }&{\cos \theta }
\end{array}} \right|\left| {\begin{array}{{20}{c}}
{\cos \theta }&{ - \sin \theta } \\
{\sin \theta }&{\cos \theta }
\end{array}} \right| \\
\Rightarrow {\left| {\begin{array}{{20}{c}}
{\cos \theta }&{ - \sin \theta } \\
{\sin \theta }&{\cos \theta }
\end{array}} \right|^2} = \left| {\begin{array}{*{20}{c}}
{{{\cos }^2}\theta - {{\sin }^2}\theta }&{ - 2\cos \theta \sin \theta } \\
{2\sin \theta \cos \theta }&{ - {{\sin }^2}\theta + {{\cos }^2}\theta }
\end{array}} \right| \\

From the trigonometric formulas, we know that: $$\cos 2\theta = {\cos ^2}\theta - {\sin ^2}\theta $$ $$\sin 2\theta = 2\cos \theta \sin \theta $$ Now, we will put the values from the formulas in the obtained equation, we get: $$\left| {\begin{array}{*{20}{c}} {\cos \theta }&{ - \sin \theta } \\\ {\sin \theta }&{\cos \theta } \end{array}} \right|\left| {\begin{array}{*{20}{c}} {\cos \theta }&{ - \sin \theta } \\\ {\sin \theta }&{\cos \theta } \end{array}} \right| = \left| {\begin{array}{*{20}{c}} {\cos 2\theta }&{ - \sin 2\theta } \\\ {\sin 2\theta }&{\cos 2\theta } \end{array}} \right|$$ We can say that in place of the power ‘k’, we applied the power 2 and so we squared it. Now, if we will replace it with ‘k’, then we get: $$ \Rightarrow \left| {\begin{array}{*{20}{c}} {\cos \theta }&{ - \sin \theta } \\\ {\sin \theta }&{\cos \theta } \end{array}} \right|\left| {\begin{array}{*{20}{c}} {\cos \theta }&{ - \sin \theta } \\\ {\sin \theta }&{\cos \theta } \end{array}} \right| = \left| {\begin{array}{*{20}{c}} {\cos k\theta }&{ - \sin k\theta } \\\ {\sin k\theta }&{\cos k\theta } \end{array}} \right|$$ So, according to the question, when we make it an equation, we get it as: $$ \Rightarrow \left| {\begin{array}{*{20}{c}} {\cos k\theta }&{ - \sin k\theta } \\\ {\sin k\theta }&{\cos k\theta } \end{array}} \right| = \left| {\begin{array}{*{20}{c}} 1&0 \\\ 0&1 \end{array}} \right|$$ From the rules of Matrices, we get $$\cos k\theta = 1$$or $$ - \sin k\theta = 0$$ So, in case of $$ - \sin k\theta = 0$$, then $$k\theta = n\pi $$. We have the value as $$\dfrac{{2\pi }}{7} = \theta $$. So, when we put this value, we get: $$ \Rightarrow k\dfrac{{2\pi }}{7} = n\pi $$ The similar terms get cancelled out, and we get: $$ \Rightarrow k\dfrac{2}{7} = n$$ Now, to find ‘k’, we will try to make ‘k’ alone. We will shift the term $$\dfrac{2}{7}$$to the other side of the equation, and we get: $$ \Rightarrow k = \dfrac{{7n}}{2}$$ Similarly, in case of $$\cos k\theta = 1$$, then $$k\theta = n\pi $$. We have the value as $$\dfrac{{2\pi }}{7} = \theta $$. So, when we put this value, we get: $$ \Rightarrow k\dfrac{{2\pi }}{7} = n\pi $$ The similar terms get cancelled out, and we get: $$ \Rightarrow k\dfrac{2}{7} = n$$ Now, to find ‘k’, we will try to make ‘k’ alone. We will shift the term $$\dfrac{2}{7}$$ to the other side of the equation, and we get: $$ \Rightarrow k = \dfrac{{7n}}{2}$$ Now, in both cases we got the same value of ‘k’. Now, to make ‘k’ the positive integral, we have to remove the fraction part. For that we will take the value of ‘n’ as 2. Then the value of ‘k’ is 7. **So, the correct answer is Option D.** **Note:** The above method was easy to solve. But there is another method to solve this question. We will first assign a variable to the given matrix and then square it. This means that in place of ‘k’ we will put the power as 2. Now, we will put the power as 3 or 4 or 6 or 7 and check that in which value of power the result is the identity matrix.