Given: $M$ is a $3 \times 3$ non-zero, skew-symmetric real matrix, meaning $M^T = -M$.
Key properties of skew-symmetric matrices:
Option (A): $M$ is invertible
Since $M$ is $3 \times 3$ (odd dimension) and skew-symmetric: $$\det(M) = \det(M^T) = \det(-M) = (-1)^3\det(M) = -\det(M)$$
This gives $\det(M) = -\det(M)$, so $\det(M) = 0$.
Therefore, $M$ is not invertible.
Option (A) is FALSE
Option (B): The matrix $I + M$ is invertible
Eigenvalues of $M$ are purely imaginary or zero. If $\lambda$ is an eigenvalue of $M$, then $\lambda = i\alpha$ for some real $\alpha$ (or $\lambda = 0$).
Eigenvalues of $I + M$ are $1 + \lambda = 1 + i\alpha$.
For $I + M$ to be singular, we need $1 + \lambda = 0$, i.e., $\lambda = -1$.
But $-1$ is real, which cannot be an eigenvalue of a skew-symmetric matrix (eigenvalues must be purely imaginary or zero).
Therefore, $I + M$ has no zero eigenvalues and is invertible.
Option (B) is TRUE
Option (C): There exists non-zero real $\alpha$ such that $\alpha I + M$ is not invertible
For $\alpha I + M$ to be singular, we need an eigenvalue of $M$ equal to $-\alpha$.
Since eigenvalues of $M$ are purely imaginary or zero, we need $-\alpha$ to be purely imaginary or zero.
For real non-zero $\alpha$, $-\alpha$ is real and non-zero, so it cannot be an eigenvalue of $M$.
Option (C) is FALSE
Option (D): All eigenvalues of $M$ are real
Eigenvalues of skew-symmetric real matrices are purely imaginary or zero, not real (except 0).
Option (D) is FALSE
Answer: (B) the matrix $I + M$ is invertible