If `A^(-1)=[(sin^2alpha,0,0),(0,sin^2 beta,0), (0,0,sin^2gamma)]` and `B^(-1)=[(cos^2alpha,0,0),(0,cos^2 beta,0), (0,0,cos^2gamma)]` where `alpha, beta, gamma` are any real numbers and `C=A^(-5)+B^(-5)+5A^(-1)B^(-1)(A^(-3) +B^(-3))+10 A^(-2)B^(-2)(A^(-1)+B^(-1))` then find |C|

Options

A. A) 0
B. B) 1
C. C) 2
D. D) 3

Answer

To solve the problem, we need to find the determinant of the matrix \( C \) defined as:

\[ C = A^{-5} + B^{-5} + 5 A^{-1} B^{-1} (A^{-3} + B^{-3}) + 10 A^{-2} B^{-2} (A^{-1} + B^{-1}) \]

Given the matrices:

\[ A^{-1} = \begin{pmatrix} \sin^2 \alpha & 0 & 0 \\ 0 & \sin^2 \beta & 0 \\ 0 & 0 & \sin^2 \gamma \end{pmatrix} \]

\[ B^{-1} = \begin{pmatrix} \cos^2 \alpha & 0 & 0 \\ 0 & \cos^2 \beta & 0 \\ 0 & 0 & \cos^2 \gamma \end{pmatrix} \]

### Step 1: Calculate \( A^{-5} \) and \( B^{-5} \)

Using the property of matrices, \( A^{-n} = (A^{-1})^n \):

\[ A^{-5} = (A^{-1})^5 = \begin{pmatrix} \sin^{10} \alpha & 0 & 0 \\ 0 & \sin^{10} \beta & 0 \\ 0 & 0 & \sin^{10} \gamma \end{pmatrix} \]

\[ B^{-5} = (B^{-1})^5 = \begin{pmatrix} \cos^{10} \alpha & 0 & 0 \\ 0 & \cos^{10} \beta & 0 \\ 0 & 0 & \cos^{10} \gamma \end{pmatrix} \]

### Step 2: Calculate \( A^{-3} \) and \( B^{-3} \)

\[ A^{-3} = (A^{-1})^3 = \begin{pmatrix} \sin^6 \alpha & 0 & 0 \\ 0 & \sin^6 \beta & 0 \\ 0 & 0 & \sin^6 \gamma \end{pmatrix} \]

\[ B^{-3} = (B^{-1})^3 = \begin{pmatrix} \cos^6 \alpha & 0 & 0 \\ 0 & \cos^6 \beta & 0 \\ 0 & 0 & \cos^6 \gamma \end{pmatrix} \]

### Step 3: Calculate \( A^{-1} B^{-1} \)

\[ A^{-1} B^{-1} = \begin{pmatrix} \sin^2 \alpha \cos^2 \alpha & 0 & 0 \\ 0 & \sin^2 \beta \cos^2 \beta & 0 \\ 0 & 0 & \sin^2 \gamma \cos^2 \gamma \end{pmatrix} \]

### Step 4: Calculate \( A^{-2} B^{-2} \)

\[ A^{-2} = (A^{-1})^2 = \begin{pmatrix} \sin^4 \alpha & 0 & 0 \\ 0 & \sin^4 \beta & 0 \\ 0 & 0 & \sin^4 \gamma \end{pmatrix} \]

\[ B^{-2} = (B^{-1})^2 = \begin{pmatrix} \cos^4 \alpha & 0 & 0 \\ 0 & \cos^4 \beta & 0 \\ 0 & 0 & \cos^4 \gamma \end{pmatrix} \]

### Step 5: Substitute into \( C \)

Now substituting these matrices into the expression for \( C \):

\[ C = \begin{pmatrix} \sin^{10} \alpha + \cos^{10} \alpha & 0 & 0 \\ 0 & \sin^{10} \beta + \cos^{10} \beta & 0 \\ 0 & 0 & \sin^{10} \gamma + \cos^{10} \gamma \end{pmatrix} \]

Now, we need to calculate \( 5 A^{-1} B^{-1} (A^{-3} + B^{-3}) \) and \( 10 A^{-2} B^{-2} (A^{-1} + B^{-1}) \).

### Step 6: Simplifying \( C \)

Using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \):

\[ \sin^{10} \theta + \cos^{10} \theta = (\sin^2 \theta + \cos^2 \theta)(\sin^8 \theta + \cos^8 \theta) = 1 \cdot (\sin^8 \theta + \cos^8 \theta) \]

Thus, we can conclude that:

\[ C = I \]

Where \( I \) is the identity matrix.

### Step 7: Find the Determinant of \( C \)

The determinant of the identity matrix \( I \) is:

\[ |C| = 1 \]

### Conclusion

Thus, the final answer is:

\[ \boxed{1} \]