1. Understand the problem:
Given a square matrix A satisfying \( A^2 = A \), we need to find \( (I + A)^3 \).
2. Expand \( (I + A)^3 \):
Using the binomial expansion:
\[ (I + A)^3 = I^3 + 3I^2A + 3IA^2 + A^3 \]
3. Simplify using \( A^2 = A \):
Since \( A^2 = A \), then \( A^3 = A^2 = A \). Thus:
\[ (I + A)^3 = I + 3A + 3A + A = I + 7A \]
Correct Answer: (B) 7A + I
Given that \( A \) is a square matrix such that \( A^2 = A \). This means \( A \) is an idempotent matrix. We want to find \( (I + A)^3 \).
Let's expand \( (I + A)^3 \):
\[ (I + A)^3 = (I + A)(I + A)(I + A) \] \[ = (I + 2A + A^2)(I + A) \quad \text{(using } (I + A)^2 = I + 2A + A^2\text{)} \] \[ = (I + 2A + A)(I + A) \quad \text{(since } A^2 = A\text{)} \] \[ = (I + 3A)(I + A) \] \[ = I + 3A + A + 3A^2 \] \[ = I + 4A + 3A \quad \text{(since } A^2 = A\text{)} \] \[ = I + 7A \]Therefore, \( (I + A)^3 = I + 7A \).
In an experiment to determine the figure of merit of a galvanometer by half deflection method, a student constructed the following circuit. He applied a resistance of \( 520 \, \Omega \) in \( R \). When \( K_1 \) is closed and \( K_2 \) is open, the deflection observed in the galvanometer is 20 div. When \( K_1 \) is also closed and a resistance of \( 90 \, \Omega \) is removed in \( S \), the deflection becomes 13 div. The resistance of galvanometer is nearly:
A wooden block of mass M lies on a rough floor. Another wooden block of the same mass is hanging from the point O through strings as shown in the figure. To achieve equilibrium, the coefficient of static friction between the block on the floor and the floor itself is