Question

If \(A\) is a square matrix such that \(A^2 = A\) and \((I + A)^n = I + \lambda A\), then \(\lambda\) is equal to

• A. \(2n - 1\)
• B. \(2^n - 1\)
• C. \(2n + 1\)
• D. None of these

Hint:

For an idempotent matrix \(A\), \(A^k = A\) for all positive integers \(k\).

Answer 1

The Correct Option is B

To solve the problem, we start by analyzing the given conditions about the square matrix \(A\). 

1. The matrix \(A\) satisfies the condition \(A^2 = A\). This means \(A\) is an idempotent matrix. Idempotent matrices have the property that their eigenvalues can only be 0 or 1.
2. We are also given that \((I + A)^n = I + \lambda A\). This equation holds for some scalar \(\lambda\).
3. To find \(\lambda\), let's expand \((I + A)^n\) using the binomial theorem: \((I + A)^n = \sum_{k=0}^{n} \binom{n}{k} I^{n-k} A^k\)Given the property of the identity matrix \(I^{n-k} = I\) and the idempotent property \(A^k = A\) for \(k \geq 1\), the expression simplifies to: \(I^n + \binom{n}{1} IA + \binom{n}{2} A^2 + \cdots + \binom{n}{n} A^n = I + nA\)but more generally considering \(A^2 = A\), \((I + A)^n = I + \sum_{k=1}^{n} \binom{n}{k} A\)
4. Recognizing that: \(\sum_{k=1}^{n} \binom{n}{k} = 2^n - 1\)(because \(\sum_{k=0}^{n} \binom{n}{k} = 2^n\) and \(\binom{n}{0} = 1\)), we equate this to the form in the problem \((I + \lambda A)\): \(I + \lambda A = I + (2^n - 1)A\)Therefore, \(\lambda = 2^n - 1\).

Thus, the value of \(\lambda\) is \(2^n - 1\).