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 value of the expression \( I + A + A^2 + \ldots + A^{k-1} \) given that \( A^k = 0 \) (the null matrix).

### Step-by-Step Solution:

1. **Recognize the Series**:
We have the series \( S = I + A + A^2 + \ldots + A^{k-1} \). This is a finite geometric series where the first term is \( I \) (the identity matrix) and the common ratio is \( A \).

2. **Use the Formula for the Sum of a Geometric Series**:
The sum of a geometric series can be expressed as:
\[ S = \frac{a(1 - r^n)}{1 - r} \]
where \( a \) is the first term, \( r \) is the common ratio, and \( n \) is the number of terms. In our case:
- \( a = I \)
- \( r = A \)
- \( n = k \)

Thus, we can write:
\[ S = I \frac{I - A^k}{I - A} \]

3. **Substituting \( A^k = 0 \)**:
Since \( A^k = 0 \), we substitute this into the equation:
\[ S = I \frac{I - 0}{I - A} = \frac{I}{I - A} \]

4. **Final Result**:
Therefore, we have:
\[ S = (I - A)^{-1} \]
This means that the expression \( I + A + A^2 + \ldots + A^{k-1} \) simplifies to:
\[ S = (I - A)^{-1} \]

### Conclusion:
The final result is:
\[ I + A + A^2 + \ldots + A^{k-1} = (I - A)^{-1} \]