Question

Let $z = \dfrac{{ - 1 + \sqrt {3i} }}{2}$, where $i = \sqrt { - 1}$ and r,s \in \left\\{ {1,2,3} \right\\}. Let P = \left[ {\begin{array}{*{20}{c}} {{{\left( { - z} \right)}^r}}&{{z^{2s}}} \\\ {{z^{2s}}}&{{z^r}} \end{array}} \right] and $I$ be the identity matrix of order $2$. Then the total number of ordered pairs $\left( {r,s} \right)$ for which ${P^2} = - I$ is

Answer 1

Here we are asked to find the number of ordered pairs $\left( {r,s} \right)$ in total with the given data. For that, we will try to derive the value $z$ from the given data then we will substitute it in the given matrix $P$. The by using the given condition that ${P^2} = - I$ equating the matrix we got to the identity matrix we will find the possible values of $s$ & $r$.

Complete step-by-step solution:
It is given that $z = \dfrac{{ - 1 + \sqrt {3i} }}{2}$ where, $i = \sqrt { - 1}$ and also given that r,s \in \left\\{ {1,2,3} \right\\}. We aim to find the total number of ordered pairs $\left( {r,s} \right)$ when ${P^2} = - I$where P = \left[ {\begin{array}{*{20}{c}} {{{\left( { - z} \right)}^r}}&{{z^{2s}}} \\\ {{z^{2s}}}&{{z^r}} \end{array}} \right] and $I$ the identity matrix.
Consider the given complex number $z = \dfrac{{ - 1 + \sqrt {3i} }}{2}$ this can be written as $z = \cos \dfrac{{2\pi }}{3} + i\sin \dfrac{{2\pi }}{3} = {e^{i\dfrac{{2\pi }}{3}}} = \omega$ where $\omega$ is the cubic root of unity.
Thus, we got that $z = \omega$.
Now consider the given matrix P = \left[ {\begin{array}{*{20}{c}} {{{\left( { - z} \right)}^r}}&{{z^{2s}}} \\\ {{z^{2s}}}&{{z^r}} \end{array}} \right] and let us find the value of ${P^2}$ that is $P \times P$

{{{\left( { - z} \right)}^r}}&{{z^{2s}}} \\\ {{z^{2s}}}&{{z^r}} \end{array}} \right] \times \left[ {\begin{array}{*{20}{c}} {{{\left( { - z} \right)}^r}}&{{z^{2s}}} \\\ {{z^{2s}}}&{{z^r}} \end{array}} \right]$$ On multiplying these matrices, we get $${P^2} = \left[ {\begin{array}{*{20}{c}} {{{\left( { - z} \right)}^{2r}} + {z^{4s}}}&{{{\left( { - z} \right)}^r}{z^{2s}} + {z^r}{z^{2s}}} \\\ {{{\left( { - z} \right)}^r}{z^{2s}} + {z^r}{z^{2s}}}&{{z^{4s}} + {z^{2r}}} \end{array}} \right]$$ Here we are aiming to find the total number of ordered pairs $$\left( {r,s} \right)$$ when $${P^2} = - I$$. Substituting the values, we have in this condition we get $$\left[ {\begin{array}{*{20}{c}} {{{\left( { - z} \right)}^{2r}} + {z^{4s}}}&{{{\left( { - z} \right)}^r}{z^{2s}} + {z^r}{z^{2s}}} \\\ {{{\left( { - z} \right)}^r}{z^{2s}} + {z^r}{z^{2s}}}&{{z^{4s}} + {z^{2r}}} \end{array}} \right] = - \left[ {\begin{array}{*{20}{c}} 1&0 \\\ 0&1 \end{array}} \right]$$ $$\left[ {\begin{array}{*{20}{c}} {{{\left( { - z} \right)}^{2r}} + {z^{4s}}}&{{{\left( { - z} \right)}^r}{z^{2s}} + {z^r}{z^{2s}}} \\\ {{{\left( { - z} \right)}^r}{z^{2s}} + {z^r}{z^{2s}}}&{{z^{4s}} + {z^{2r}}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} { - 1}&0 \\\ 0&{ - 1} \end{array}} \right]$$ From the above expression, we get $${\left( { - z} \right)^{2r}} + {z^{4s}} = - 1$$ $${\left( { - z} \right)^r}{z^{2s}} + {z^r}{z^{2s}} = 0$$ $${\left( { - z} \right)^r}{z^{2s}} + {z^r}{z^{2s}} = 0$$ $${z^{4s}} + {z^{2r}} = - 1$$ From the above set of equations consider the third equation $${\left( { - z} \right)^r}{z^{2s}} + {z^r}{z^{2s}} = 0$$ On simplifying this equation, we get $$\left( {{{\left( { - z} \right)}^r} + {z^r}} \right){z^{2s}} = 0$$ We already derived that $$z = \omega $$ substituting this in the above equation we get $$\left( {{{\left( { - \omega } \right)}^r} + {\omega ^r}} \right){\omega ^{2s}} = 0$$ $$ \Rightarrow {\omega ^{2s}} = 0\& {\left( { - \omega } \right)^r} + {\omega ^r} = 0$$ But $${\omega ^{2s}} \ne 0$$ (Since $$\omega $$- cubic root of unity). Therefore, $${\left( { - \omega } \right)^r} + {\omega ^r} = 0$$ when $$r$$ is odd thus, $$r = \left\\{ {1,3} \right\\}$$ Thus, we have found the values of $$r$$ now we will find the values of $$s$$. Substituting $$r = \left\\{ {1,3} \right\\}$$ in the equation $$ \Rightarrow {\omega ^{4s}} = - 1 - {\omega ^2}$$ we get: When $$r = 1$$, $$i,j$$ $$ \Rightarrow {\omega ^{4s}} = - 1 - {\left( { - \omega } \right)^2}$$ $$ \Rightarrow {\omega ^{4s}} = - 1 - {\omega ^2}$$ $$ \Rightarrow {\omega ^{4s}} = \omega $$ $$ \Rightarrow s = 1$$ When $$r = 3$$, $${\left( { - \omega } \right)^6} + {\omega ^{4s}} = - 1$$ $${\omega ^{4s}} = - 1 - 1$$ (Since $$\omega $$ is cubic root of unity) $${\omega ^{4s}} = - 2$$ This implies that no value $$s$$ is possible. Therefore, $$s = 1$$ thus we get, $$\left( {r,s} \right) = \left( {1,1} \right)$$ Therefore, the total number of ordered pairs $$\left( {r,s} \right)$$ is only one that is $$\left( {1,1} \right)$$.** **Note:** Points we need to remember that $$\omega $$ is the cubic root of unity and we have that $${\omega ^2} + \omega + 1 = 0$$. When two matrices are equal then their terms will be equal concerning their positions. That is if two matrices are equal $$\left[ A \right] = \left[ B \right]$$ then $${a_{ij}} = {b_{ij}}$$ for all $$i,j$$.