let `z= (-1+sqrt(3i))/2, where i=sqrt(-1) and r,s epsilon P1,2,3}. Let P= [((-z)^r, z^(2s)),(z^(2s), z^r)]` and I be the idenfity matrix or order 2. Then the total number of ordered pairs (r,s) or which `P^2=-I` is

Explanation

To solve the problem, we need to find the total number of ordered pairs \((r, s)\) such that \(P^2 = -I\), where \(I\) is the identity matrix of order 2. The matrix \(P\) is defined as:

\[ P = \begin{pmatrix} -z^r & z^{2s} \\ z^{2s} & z^r \end{pmatrix} \]

Given that \(z = \frac{-1 + \sqrt{3}i}{2}\), we can express \(z\) in terms of \( \omega \), where \( \omega = z \).

### Step 1: Calculate \(P^2\)

To find \(P^2\), we perform matrix multiplication:

\[ P^2 = P \cdot P = \begin{pmatrix} -z^r & z^{2s} \\ z^{2s} & z^r \end{pmatrix} \cdot \begin{pmatrix} -z^r & z^{2s} \\ z^{2s} & z^r \end{pmatrix} \]

Calculating the elements of \(P^2\):

1. First element:
\[ (-z^r)(-z^r) + (z^{2s})(z^{2s}) = z^{2r} + z^{4s} \]

2. Second element:
\[ (-z^r)(z^{2s}) + (z^{2s})(z^r) = -z^{r + 2s} + z^{r + 2s} = 0 \]

3. Third element:
\[ (z^{2s})(-z^r) + (z^r)(z^{2s}) = -z^{2s + r} + z^{2s + r} = 0 \]

4. Fourth element:
\[ (z^{2s})(z^{2s}) + (z^r)(z^r) = z^{4s} + z^{2r} \]

Thus, we have:

\[ P^2 = \begin{pmatrix} z^{2r} + z^{4s} & 0 \\ 0 & z^{2r} + z^{4s} \end{pmatrix} \]

### Step 2: Set \(P^2 = -I\)

We know that the identity matrix \(I\) is:

\[ I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \]

Thus, \(-I\) is:

\[ -I = \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix} \]

Setting \(P^2 = -I\) gives us:

\[ \begin{pmatrix} z^{2r} + z^{4s} & 0 \\ 0 & z^{2r} + z^{4s} \end{pmatrix} = \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix} \]

This leads to the equation:

\[ z^{2r} + z^{4s} = -1 \]

### Step 3: Analyze the equation

We need to find values of \(r\) and \(s\) such that:

\[ z^{2r} + z^{4s} = -1 \]

### Step 4: Substitute values for \(r\) and \(s\)

Given that \(r, s \in \{1, 2, 3\}\), we will substitute these values and check:

1. For \(r = 1\), \(s = 1\):
\[ z^{2(1)} + z^{4(1)} = z^2 + z^4 \]

2. For \(r = 1\), \(s = 2\):
\[ z^{2(1)} + z^{4(2)} = z^2 + z^8 \]

3. For \(r = 1\), \(s = 3\):
\[ z^{2(1)} + z^{4(3)} = z^2 + z^{12} \]

4. Repeat for \(r = 2\) and \(r = 3\).

### Step 5: Find valid pairs

After substituting all combinations, we find that the only solution that satisfies the equation is:

\((r, s) = (1, 1)\)

### Conclusion

The total number of ordered pairs \((r, s)\) for which \(P^2 = -I\) is:

\[ \text{Total pairs} = 1 \quad \text{(only } (1, 1) \text{)} \]