If `A` is non-singular and `(A-2I)(A-4I)=O ,t h e n1/6A+4/3A^(-1)` is equal to `O I` b. `2I` c. `6I` d. `I`

Options

A. `O`
B. `I`
C. `2I`
D. `6I`

Explanation

To solve the problem, we need to find the value of \( \frac{1}{6}A + \frac{4}{3}A^{-1} \) given that \( (A - 2I)(A - 4I) = O \) where \( A \) is a non-singular matrix.

### Step-by-step Solution:

1. **Start with the given equation**:
\[ (A - 2I)(A - 4I) = O \]

2. **Expand the left-hand side**:
\[ A^2 - 4A - 2A + 8I = O \]
This simplifies to:
\[ A^2 - 6A + 8I = O \]

3. **Rearranging the equation**:
\[ A^2 - 6A + 8I = 0 \implies A^2 - 6A = -8I \]

4. **Multiply both sides by \( A^{-1} \)** (since \( A \) is non-singular):
\[ A^{-1}(A^2 - 6A) = A^{-1}(-8I) \]
This gives us:
\[ A + 8A^{-1} = -8A^{-1} \]

5. **Rearranging the equation**:
\[ A + 8A^{-1} = 6I \]

6. **Now, isolate \( A \) and \( A^{-1} \)**:
\[ A + 8A^{-1} = 6I \implies A = 6I - 8A^{-1} \]

7. **Divide the entire equation by 6**:
\[ \frac{1}{6}A + \frac{8}{6}A^{-1} = I \]

8. **Simplify the coefficients**:
\[ \frac{1}{6}A + \frac{4}{3}A^{-1} = I \]

9. **Thus, we conclude**:
\[ \frac{1}{6}A + \frac{4}{3}A^{-1} = I \]

### Final Answer:
The value of \( \frac{1}{6}A + \frac{4}{3}A^{-1} \) is \( I \).