If A is non-singular and `(A-2I)(A-4I)=0`, then `(1)/(6)(A)+(4)/(3)(A^(-1))=`

Options

A. `I`
B. `0`
C. `2I`
D. `6I`

Answer

To solve the problem, we start with the equation given:

\[ (A - 2I)(A - 4I) = 0 \]

### Step 1: Expand the equation
We can expand the left-hand side:

\[ A^2 - 4AI - 2AI + 8I = 0 \]

This simplifies to:

\[ A^2 - 6AI + 8I = 0 \]

### Step 2: Rearrange the equation
Rearranging gives us:

\[ A^2 - 6A + 8I = 0 \]

### Step 3: Isolate \(A^2\)
We can isolate \(A^2\):

\[ A^2 = 6A - 8I \]

### Step 4: Multiply both sides by \(A^{-1}\)
Since \(A\) is non-singular, we can multiply both sides by \(A^{-1}\):

\[ A^{-1}A^2 = A^{-1}(6A - 8I) \]

This simplifies to:

\[ A = 6I - 8A^{-1} \]

### Step 5: Rearrange to find \(A^{-1}\)
Rearranging gives us:

\[ 8A^{-1} = 6I - A \]

Dividing by 8:

\[ A^{-1} = \frac{6}{8}I - \frac{1}{8}A = \frac{3}{4}I - \frac{1}{8}A \]

### Step 6: Substitute \(A^{-1}\) into the expression
Now we substitute \(A^{-1}\) into the expression we need to evaluate:

\[ \frac{1}{6}A + \frac{4}{3}A^{-1} \]

Substituting \(A^{-1}\):

\[ \frac{1}{6}A + \frac{4}{3}\left(\frac{3}{4}I - \frac{1}{8}A\right) \]

### Step 7: Simplify the expression
This becomes:

\[ \frac{1}{6}A + \left(I - \frac{1}{6}A\right) \]

Combining the terms:

\[ \frac{1}{6}A - \frac{1}{6}A + I = I \]

### Final Result
Thus, we conclude that:

\[ \frac{1}{6}A + \frac{4}{3}A^{-1} = I \]