If `A^(k)=0` (null matrix) for some value of k, then `l+A+A^(2)+…A^(k-1)` is equal to

Options

A. 0
B. `(1-A)^(-1)`
C. `(1+A)^(k)`
D. none of these

Answer

To solve the problem, we need to find the sum of the series \( I + A + A^2 + \ldots + A^{k-1} \) given that \( A^k = 0 \) (the null matrix).

### Step-by-Step Solution:

1. **Understanding the Series**:
The series we need to evaluate is:
\[ S = I + A + A^2 + \ldots + A^{k-1} \]
where \( I \) is the identity matrix.

2. **Recognizing the Geometric Series**:
This series can be recognized as a finite geometric series with the first term \( I \) and the common ratio \( A \). The number of terms in this series is \( k \) (from \( 0 \) to \( k-1 \)).

3. **Using the Formula for the Sum of a Geometric Series**:
The formula for the sum of a finite geometric series is:
\[ S_n = a \frac{1 - r^n}{1 - r} \]
where \( a \) is the first term, \( r \) is the common ratio, and \( n \) is the number of terms.

In our case, \( a = I \), \( r = A \), and \( n = k \).

4. **Applying the Formula**:
Plugging in the values, we get:
\[ S = I \frac{1 - A^k}{I - A} \]

5. **Substituting \( A^k = 0 \)**:
Since it is given that \( A^k = 0 \) (the null matrix), we can substitute this into our equation:
\[ S = I \frac{1 - 0}{I - A} = \frac{I}{I - A} \]

6. **Final Result**:
Thus, the sum \( I + A + A^2 + \ldots + A^{k-1} \) is equal to:
\[ S = \frac{I}{I - A} \]