Let A be a non - singular symmetric matrix of order 3. If `A^(T)=A^(2)-I`, then `(A-I)^(-1)` is equal to

Options

A. A
B. 2A
C. `A-I`
D. `2A-I`

Answer

To solve the problem, we start with the given equation involving the non-singular symmetric matrix \( A \):

\[ A^T = A^2 - I \]

Since \( A \) is symmetric, we have \( A^T = A \). Therefore, we can rewrite the equation as:

\[ A = A^2 - I \]

Now, we can rearrange this equation:

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

This is a quadratic equation in terms of \( A \). We can factor this equation or use the quadratic formula to find \( A \). However, we are interested in finding \( (A - I)^{-1} \).

### Step 1: Rearranging the equation

We can rearrange the equation \( A^2 - A - I = 0 \) to express \( A^2 \):

\[ A^2 = A + I \]

### Step 2: Finding \( (A - I)^{-1} \)

To find \( (A - I)^{-1} \), we can manipulate the equation we have. We know:

\[ A^2 = A + I \]

Subtract \( I \) from both sides:

\[ A^2 - I = A \]

Now, we can express \( A \) in terms of \( A - I \):

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

### Step 3: Inverting \( A - I \)

We can now express \( A \) in terms of \( (A - I) \):

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

To find \( (A - I)^{-1} \), we can use the property of inverses. Since we have \( A = (A - I) + I \), we can express this in terms of inverses:

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

From the equation \( A = (A - I) + I \), we can see that if we multiply both sides by \( (A - I)^{-1} \):

\[ (A - I)^{-1}A = (A - I)^{-1}((A - I) + I) \]

This simplifies to:

\[ (A - I)^{-1}A = I + (A - I)^{-1}I \]

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

\[ (A - I)^{-1}A = I + (A - I)^{-1} \]

### Final Step: Conclusion

From the above manipulations, we can conclude that:

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

Thus, the answer is:

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