
Step 1: Find the collector current (\( I_C \)). The collector current (\( I_C \)) is determined by: \[ I_C = \frac{V_{CC}}{R_C}, \] where: - \( V_{CC} = 1 \, \text{V} \), - \( R_C = 1 \, \text{k}\Omega = 1000 \, \Omega \). Substitute the values: \[ I_C = \frac{1}{1000} = 1 \, \text{mA}. \]
Step 2: Relate base current (\( I_B \)) and collector current (\( I_C \)). The current gain (\( \beta \)) of the transistor is given by: \[ \beta = \frac{I_C}{I_B}. \] Rearrange to solve for \( I_B \): \[ I_B = \frac{I_C}{\beta}. \] Substitute \( I_C = 1 \, \text{mA} = 10^{-3} \, \text{A} \) and \( \beta = 100 \): \[ I_B = \frac{10^{-3}}{100} = 10 \, \mu \text{A}. \]
Final Answer: The base current is: \[ \boxed{I_B = 10 \, \mu \text{A}}. \]

Let \( a \in \mathbb{R} \) and \( A \) be a matrix of order \( 3 \times 3 \) such that \( \det(A) = -4 \) and \[ A + I = \begin{bmatrix} 1 & a & 1 \\ 2 & 1 & 0 \\ a & 1 & 2 \end{bmatrix} \] where \( I \) is the identity matrix of order \( 3 \times 3 \).
If \( \det\left( (a + 1) \cdot \text{adj}\left( (a - 1) A \right) \right) \) is \( 2^m 3^n \), \( m, n \in \{ 0, 1, 2, \dots, 20 \} \), then \( m + n \) is equal to: