We know that $A = I_2 - MM^T$ and $M^TM = I_1$, which implies that $M$ is a unit vector. The matrix $A$ is a projection matrix, and for a projection matrix, the eigenvalues are either 0 or 1.
In this case, the eigenvalue $\lambda$ of $A$ can be either 0 or 1. Therefore, the sum of squares of all possible values of $\lambda$ is:
$0^2 + 1^2 = 1 + 1 = 2.$
Let $A = \begin{bmatrix} \cos \theta & 0 & -\sin \theta \\ 0 & 1 & 0 \\ \sin \theta & 0 & \cos \theta \end{bmatrix}$. 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 equal to
Let $ A $ be a $ 3 \times 3 $ matrix such that $ | \text{adj} (\text{adj} A) | = 81.
$ If $ S = \left\{ n \in \mathbb{Z}: \left| \text{adj} (\text{adj} A) \right|^{\frac{(n - 1)^2}{2}} = |A|^{(3n^2 - 5n - 4)} \right\}, $ then the value of $ \sum_{n \in S} |A| (n^2 + n) $ is:
Match List-I with List-II: List-I