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A is a square matrix and I is an identit...

A is a square matrix and I is an identity matrix of the same order. If `A^(3)=O`, then inverse of matrix `(I-A)` is

A

`I+A`

B

`I-A+A^(2)`

C

`A+A^(2)`

D

`I+A+A^(2)`

Text Solution

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The correct Answer is:
To find the inverse of the matrix \( I - A \) given that \( A^3 = O \) (where \( O \) is the null matrix), we can follow these steps: ### Step-by-Step Solution: 1. **Start with the given condition**: We know that \( A^3 = O \). This means that \( A \) is a nilpotent matrix of index 3. 2. **Rewrite the equation**: We can express \( A^3 = O \) as \( -A^3 = -O \). This is simply multiplying both sides by -1, which doesn't change the equality. 3. **Add the identity matrix \( I \)**: We add the identity matrix \( I \) to both sides: \[ I - A^3 = I \] 4. **Recognize the identity**: We can use the identity \( a^3 - b^3 = (a - b)(a^2 + ab + b^2) \). In our case, let \( a = I \) and \( b = A \): \[ I^3 - A^3 = (I - A)(I^2 + IA + A^2) \] Since \( I^3 = I \) and \( A^3 = O \), we have: \[ I - O = (I - A)(I^2 + A + A^2) \] Thus, we can write: \[ I = (I - A)(I^2 + A + A^2) \] 5. **Finding the inverse**: To find the inverse of \( I - A \), we can rearrange the equation: \[ I - A = I (I^2 + A + A^2)^{-1} \] Therefore, the inverse of \( I - A \) is: \[ (I - A)^{-1} = I^2 + A + A^2 \] 6. **Final result**: Since \( I^2 = I \), we can simplify this to: \[ (I - A)^{-1} = I + A + A^2 \] ### Conclusion: Thus, the inverse of the matrix \( I - A \) is: \[ (I - A)^{-1} = I + A + A^2 \]
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