If `A=[a b0a]` is nth root of `I_2,` then choose the correct statements: If `n` is odd, `a=1,b=0` If `n` is odd, `a=-1,b=0` If `n` is even, `a=1,b=0` If `n` is even, `a=-1,b=0` a. i, ii, iii, iv b. ii, iii, iv c. i, ii, iii, iv d. i, iii, iv

Options

A. i, ii, iii
B. ii, iii, iv
C. i, ii, iii, iv
D. i, iii, iv

Explanation

To solve the problem, we need to determine the conditions under which the matrix \( A = \begin{pmatrix} a & b \\ 0 & a \end{pmatrix} \) is an \( n \)-th root of the identity matrix \( I_2 \). This means that \( A^n = I_2 \).

### Step-by-Step Solution:

1. **Define the Matrix A**:
\[ A = \begin{pmatrix} a & b \\ 0 & a \end{pmatrix} \]

2. **Calculate \( A^2 \)**:
\[ A^2 = A \cdot A = \begin{pmatrix} a & b \\ 0 & a \end{pmatrix} \cdot \begin{pmatrix} a & b \\ 0 & a \end{pmatrix} = \begin{pmatrix} a^2 & ab + ba \\ 0 & a^2 \end{pmatrix} = \begin{pmatrix} a^2 & 2ab \\ 0 & a^2 \end{pmatrix} \]

3. **Calculate \( A^3 \)**:
\[ A^3 = A^2 \cdot A = \begin{pmatrix} a^2 & 2ab \\ 0 & a^2 \end{pmatrix} \cdot \begin{pmatrix} a & b \\ 0 & a \end{pmatrix} = \begin{pmatrix} a^3 & a^2b + 2ab \\ 0 & a^3 \end{pmatrix} = \begin{pmatrix} a^3 & 3a^2b \\ 0 & a^3 \end{pmatrix} \]

4. **General Form of \( A^n \)**:
By observing the pattern, we can generalize:
\[ A^n = \begin{pmatrix} a^n & na^{n-1}b \\ 0 & a^n \end{pmatrix} \]

5. **Set \( A^n = I_2 \)**:
We need \( A^n = I_2 = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \). This gives us two equations:
\[ a^n = 1 \quad \text{and} \quad na^{n-1}b = 0 \]

6. **Analyze the equations**:
- From \( a^n = 1 \), we have:
- \( a = 1 \) or \( a = -1 \) (since \( a^n = 1 \) for any integer \( n \)).
- From \( na^{n-1}b = 0 \):
- If \( n \neq 0 \), then \( b = 0 \) (since \( a^{n-1} \) cannot be zero if \( a \) is either 1 or -1).

7. **Determine conditions based on n**:
- **If \( n \) is odd**:
- \( a = 1 \) and \( b = 0 \) is valid.
- \( a = -1 \) and \( b = 0 \) is also valid.
- **If \( n \) is even**:
- \( a = 1 \) and \( b = 0 \) is valid.
- \( a = -1 \) and \( b = 0 \) is also valid.

### Conclusion:
Thus, we can conclude:
- If \( n \) is odd, both \( a = 1, b = 0 \) and \( a = -1, b = 0 \) are valid.
- If \( n \) is even, both \( a = 1, b = 0 \) and \( a = -1, b = 0 \) are valid.

### Correct Statements:
- If \( n \) is odd, \( a = 1, b = 0 \) (True).
- If \( n \) is odd, \( a = -1, b = 0 \) (True).
- If \( n \) is even, \( a = 1, b = 0 \) (True).
- If \( n \) is even, \( a = -1, b = 0 \) (True).

Thus, the correct answer is **option d: i, iii, iv**.