Let $$A = \begin{bmatrix}0 & 2 & -3 \\-2 & 0 & 1 \\ 3 & -1 & 0 \end{bmatrix}$$ and B be a matrix such that $$B(I- A)=I+A.$$ Then the sumof the diagonal elements of $$B^{T}B$$ is equal to _________

We need to find the sum of diagonal elements of $$B^TB$$. Given $$A = \begin{bmatrix}0&2&-3\\-2&0&1\\3&-1&0\end{bmatrix}$$ and $$B(I-A) = I+A$$.

Since $$A^T = \begin{bmatrix}0&-2&3\\2&0&-1\\-3&1&0\end{bmatrix} = -A$$, it follows that A is skew-symmetric ($$A^T = -A$$). Substituting this result into $$B(I-A) = I+A$$ gives $$B = (I+A)(I-A)^{-1}$$.

For a skew-symmetric matrix A, one computes $$B^T = ((I-A)^{-1})^T(I+A)^T = ((I-A)^T)^{-1}(I+A^T) = (I-A^T)^{-1}(I-A) = (I+A)^{-1}(I-A)$$ and therefore $$B^TB = (I+A)^{-1}(I-A)(I+A)(I-A)^{-1}$$.

Since $$(I-A)(I+A) = I - A^2 + A - A = I - A^2$$ and $$(I+A)(I-A) = I - A^2 - A + A = I - A^2$$, it follows that $$(I-A)$$ and $$(I+A)$$ commute. Therefore, $$B^TB = (I+A)^{-1}(I+A)(I-A)(I-A)^{-1} = I \cdot I = I$$, showing that B is an orthogonal matrix ($$B^TB = I$$).

Finally, the trace of $$B^TB = I$$ is $$\text{tr}(B^TB) = \text{tr}(I_3) = 1 + 1 + 1 = 3$$.

The answer is 3.