JAMIA 2023 — Mathematics PYQ

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Question

JAMIA 2023 — Mathematics PYQ

JAMIA | Mathematics | 2023

If Z is an idempotent matrix, then (I+Z)”

Choose the correct answer:

Explanation

1. Definition of Idempotent Matrix

An $n \times n$ matrix $Z$ is called idempotent if:

Z^2 = Z

From this property, it follows that for any integer $k \geq 1$ :

Z^k = Z

2. Binomial Expansion

Using the Binomial Theorem for matrices (which is applicable here because the Identity matrix $I$ and $Z$ always commute, i.e., $IZ = ZI = Z$ ), we can expand $(I + Z)^n$ :

(I + Z)^n = \binom{n}{0} I^n + \binom{n}{1} I^{n-1} Z + \binom{n}{2} I^{n-2} Z^2 + \dots + \binom{n}{n} Z^n

Since $I^k = I$ and $Z^k = Z$ (for $k \geq 1$ ), we can simplify the expression:

(I + Z)^n = I + \binom{n}{1} Z + \binom{n}{2} Z + \dots + \binom{n}{n} Z

3. Factoring the Matrix $Z$

Now, factor out $Z$ from all terms except the Identity matrix:

(I + Z)^n = I + Z \left[ \binom{n}{1} + \binom{n}{2} + \dots + \binom{n}{n} \right]

We know from the properties of binomial coefficients that the sum of the $n$ -th row is:

\sum_{k=0}^{n} \binom{n}{k} = 2^n

Therefore:

\binom{n}{1} + \binom{n}{2} + \dots + \binom{n}{n} = 2^n - \binom{n}{0}

\binom{n}{1} + \binom{n}{2} + \dots + \binom{n}{n} = 2^n - 1

4. Final Result

Substituting this sum back into our equation:

(I + Z)^n = I + (2^n - 1)Z

Summary:

If $Z^2 = Z$ , then for any positive integer $n$ :

(I + Z)^n = I + (2^n - 1)Z