Let A be a `nxxn` matrix such that`A ^(n) = alpha A,` where `alpha ` is a
real number different from 1 and - 1. The matrix `A + I_(n)` is

Options

A. singular
B. invertible
C. scalar matrix
D. None of these

Explanation

To solve the problem, we need to analyze the given condition for the matrix \( A \) and determine the nature of the matrix \( A + I_n \).

### Step-by-Step Solution:

1. **Given Condition**: We have \( A^n = \alpha I_n \), where \( \alpha \) is a real number different from 1 and -1.

2. **Assuming Order of Matrix**: Let's assume \( A \) is a \( 2 \times 2 \) matrix for simplicity. We can write:
\[ A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \]

3. **Calculating \( A^2 \)**: We need to find \( A^2 \) to understand the behavior of \( A \):
\[ A^2 = A \cdot A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} a^2 + bc & ab + bd \\ ac + dc & bc + d^2 \end{pmatrix} \]

4. **Using the Given Condition**: Since \( A^n = \alpha I_n \), we can deduce that:
\[ A^2 = \alpha^{2/n} I_n \]
This means that the eigenvalues of \( A \) must satisfy the equation \( \lambda^n = \alpha \).

5. **Eigenvalues of \( A \)**: The eigenvalues \( \lambda \) of \( A \) can be expressed as:
\[ \lambda = \alpha^{1/n} \text{ or } \lambda = \alpha^{1/n} \text{ (with multiplicity)} \]

6. **Characteristic Polynomial**: The characteristic polynomial of \( A \) can be expressed as:
\[ p(x) = x^2 - (a + d)x + (ad - bc) \]
The roots of this polynomial are the eigenvalues.

7. **Condition for Invertibility**: A matrix is invertible if and only if its determinant is non-zero. The determinant of \( A \) is given by:
\[ \text{det}(A) = ad - bc \]
Since \( \alpha \) is not equal to 1 or -1, the eigenvalues are non-zero, which implies that \( \text{det}(A) \neq 0 \).

8. **Analyzing \( A + I_n \)**: Now we consider the matrix \( A + I_n \):
\[ A + I_n = \begin{pmatrix} a + 1 & b \\ c & d + 1 \end{pmatrix} \]
The eigenvalues of \( A + I_n \) will be \( \lambda + 1 \) where \( \lambda \) are the eigenvalues of \( A \).

9. **Conclusion**: Since the eigenvalues of \( A \) are non-zero, \( \lambda + 1 \) will also be non-zero. Therefore, \( A + I_n \) is invertible.

### Final Answer:
The matrix \( A + I_n \) is **invertible**.