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

Options

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

Answer

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

### Step-by-Step Solution:

1. **Identify the Sum**:
We start with the sum \( S = I + A + A^2 + \ldots + A^{k-1} \).

2. **Multiply by A**:
Next, we multiply the entire sum \( S \) by \( A \):
\[ AS = A + A^2 + A^3 + \ldots + A^k \]

3. **Substitute \( A^k = 0 \)**:
Since \( A^k = 0 \), we can substitute this into our equation:
\[ AS = A + A^2 + A^3 + \ldots + 0 = A + A^2 + A^3 + \ldots + A^{k-1} \]

4. **Rearranging the Equation**:
Now we can rearrange the equation:
\[ AS = S - I \]
This is because \( S = I + A + A^2 + \ldots + A^{k-1} \).

5. **Final Equation**:
We now have:
\[ AS + I = S \]
Rearranging gives:
\[ S - AS = I \]

6. **Factor out S**:
We can factor \( S \) out:
\[ S(I - A) = I \]

7. **Solve for S**:
If \( I - A \) is invertible, we can multiply both sides by \( (I - A)^{-1} \):
\[ S = I(I - A)^{-1} = (I - A)^{-1} \]

### Conclusion:
Thus, the result is:
\[ S = (I - A)^{-1} \]