If A is non-singular matrix and `(A+I)(A-I)``= 0` then `A+A^(-1)=` . . .

Options

A. 2A
B. 0
C. `I`
D. 3`I`

Explanation

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

\[ (A + I)(A - I) = 0 \]

### Step 1: Expand the equation

Expanding the left-hand side, we have:

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

This simplifies to:

\[ A^2 - I = 0 \]

### Step 2: Rearranging the equation

From the equation \(A^2 - I = 0\), we can rearrange it to:

\[ A^2 = I \]

### Step 3: Express \(A\) in terms of \(A^{-1}\)

Since \(A\) is a non-singular matrix, we can take the inverse of both sides. We know that if \(A^2 = I\), then multiplying both sides by \(A^{-1}\) gives:

\[ A \cdot A = I \implies A^{-1} \cdot A^2 = A^{-1} \cdot I \]

This simplifies to:

\[ A = A^{-1} \]

### Step 4: Calculate \(A + A^{-1}\)

Now, we need to find \(A + A^{-1}\):

\[ A + A^{-1} = A + A \]

Since \(A = A^{-1}\), we can write:

\[ A + A^{-1} = 2A \]

### Final Answer

Thus, we conclude that:

\[ A + A^{-1} = 2A \]