1. Compute \( A^2 \):
\[ A^2 = A \cdot A = \begin{bmatrix} 0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0 \end{bmatrix} \cdot \begin{bmatrix} 0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0 \end{bmatrix}. \]
2. Perform matrix multiplication:
\[ A^2 = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = I, \]
where \( I \) is the identity matrix.
3. Since \( A^2 = I \), this confirms the property \( A^2 = I \). The other options are incorrect:
Match List-I with List-II
List-I (Matrix) | List-II (Inverse of the Matrix) |
---|---|
(A) \(\begin{bmatrix} 1 & 7 \\ 4 & -2 \end{bmatrix}\) | (I) \(\begin{bmatrix} \tfrac{2}{15} & \tfrac{1}{10} \\[6pt] -\tfrac{1}{15} & \tfrac{1}{5} \end{bmatrix}\) |
(B) \(\begin{bmatrix} 6 & -3 \\ 2 & 4 \end{bmatrix}\) | (II) \(\begin{bmatrix} \tfrac{1}{5} & -\tfrac{2}{15} \\[6pt] -\tfrac{1}{10} & \tfrac{7}{30} \end{bmatrix}\) |
(C) \(\begin{bmatrix} 5 & 2 \\ -5 & 4 \end{bmatrix}\) | (III) \(\begin{bmatrix} \tfrac{1}{15} & \tfrac{7}{30} \\[6pt] \tfrac{2}{15} & -\tfrac{1}{30} \end{bmatrix}\) |
(D) \(\begin{bmatrix} 7 & 4 \\ 3 & 6 \end{bmatrix}\) | (IV) \(\begin{bmatrix} \tfrac{2}{15} & -\tfrac{1}{15} \\[6pt] \tfrac{1}{6} & \tfrac{1}{6} \end{bmatrix}\) |
A quantity \( X \) is given by: \[ X = \frac{\epsilon_0 L \Delta V}{\Delta t} \] where:
- \( \epsilon_0 \) is the permittivity of free space,
- \( L \) is the length,
- \( \Delta V \) is the potential difference,
- \( \Delta t \) is the time interval.
The dimension of \( X \) is the same as that of: