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

Hint: 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 \[P \times P\]
\[{P^2} = \left[ {\begin{array}{*{20}{c}}
  {{{\left( { - z} \right)}^r}}&{{z^{2s}}} \\
  {{z^{2s}}}&{{z^r}}
\end{array}} \right] \times \left[ {\begin{array}{*{20}{c}}
  {{{\left( { - z} \right)}^r}}&{{z^{2s}}} \\
  {{z^{2s}}}&{{z^r}}
\end{array}} \right]\]
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 \[i,j\].