$$Let A = \begin{pmatrix} \cos\theta & 0 & -\sin\theta \\ 0 & 1 & 0 \\ \sin\theta & 0 & \cos\theta \end{pmatrix}.$$ If for some $$\theta \in (0,\pi), A^2 = A$$ T, then the sum of the diagonal elements of the matrix $$(A + I)^3 + (A - I)^3 - 6A is \text{equal to}:$$

Step 1: Calculate $$A^2$$ and $$A^T$$

First, we need to calculate the square of matrix $$A$$ , denoted as $$A^2$$ , by multiplying $$A$$ by itself. We also need to find the transpose of matrix $$A$$ , denoted as $$A^T$$ , by swapping its rows and columns.

Step 2: Equate $$A^2$$ and $$A^T$$ to find $$\theta$$

Given that $$A^2 = A^T$$ , we equate the corresponding elements of the two matrices. This gives us a system of trigonometric equations. Solving these equations for $$\theta$$ within the given interval $$(0, \pi)$$ yields $$\theta = \frac{2\pi}{3}$$ .

Step 3: Simplify the expression $$(A + I)^3 + (A - I)^3 - 6A$$

We expand the cubic terms $$(A+I)^3$$ and $$(A-I)^3$$ using the binomial expansion formula $$(x+y)^3 = x^3 + 3x^2y + 3xy^2 + y^3$$ and $$(x-y)^3 = x^3 - 3x^2y + 3xy^2 - y^3$$ . Since $$I$$ is the identity matrix, $$AI = A$$ and $$I^n = I$$ . After expansion, many terms cancel out, simplifying the expression to $$2A^3$$ .

Step 4: Calculate $$A^3$$

Since we found that $$A^2 = A^T$$ , we can substitute $$A^T$$ for $$A^2$$ in the expression for $$A^3$$ . Thus, $$A^3 = A^2 \cdot A = A^T \cdot A$$ . Performing this matrix multiplication, we find that $$A^T A$$ results in the identity matrix $$I$$ .

Step 5: Calculate the final expression and sum of diagonal elements

Substituting $$A^3 = I$$ into the simplified expression $$2A^3$$ , we get $$2I$$ . The identity matrix $$I$$ has ones on its diagonal, so $$2I$$ has twos on its diagonal. The sum of the diagonal elements of $$2I$$ is $$2+2+2=6$$ .

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