The determinant of a $3 \times 3$ skew-symmetric matrix is always zero. A matrix has full rank if and only if its determinant is non-zero. Therefore, if the determinant is zero, the rank is less than the dimension of the matrix. For the given matrix A, $| A|=0$, so its rank is at most 2.
If $A = \begin{bmatrix} 0 & -h & -g \\ h & 0 & -f \\ g & f & 0 \end{bmatrix}$, then correct statement is - (1) A is an invertible matrix (2) A is an orthogonal matrix (3) $|A|=0$ (4) Rank (A) = 3
Asked in Asst. Statistical Officer (Agri. Deptt.) 2024 · Topic: Mechanics (Newton's Laws of Motion and their applications)
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