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

Options

A. `I+A`
B. `I-A+A^(2)`
C. `A+A^(2)`
D. `I+A+A^(2)`

Answer

To find the inverse of the matrix \( I - A \) given that \( A^3 = O \) (where \( O \) is the zero matrix), we can use the properties of matrices and some algebraic identities. Here’s a step-by-step solution:

### Step 1: Understand the given condition
We know that \( A^3 = O \). This means that \( A \) is a nilpotent matrix of index 3.

### Step 2: Use the identity for the difference of cubes
Recall the algebraic identity for the difference of cubes:
\[ x^3 - y^3 = (x - y)(x^2 + xy + y^2) \]
In our case, we can let \( x = I \) (the identity matrix) and \( y = A \). Thus, we have:
\[ I^3 - A^3 = (I - A)(I^2 + IA + A^2) \]
Since \( I^3 = I \) and \( A^3 = O \), we can rewrite this as:
\[ I - O = (I - A)(I^2 + A + A^2) \]
This simplifies to:
\[ I = (I - A)(I^2 + A + A^2) \]

### Step 3: Rearranging to find the inverse
From the equation \( I = (I - A)(I^2 + A + A^2) \), we can see that \( I^2 + A + A^2 \) acts as the inverse of \( I - A \). Therefore, we can conclude:
\[ (I - A)^{-1} = I^2 + A + A^2 \]

### Step 4: Final expression
Since \( I^2 = I \), we can simplify our final expression:
\[ (I - A)^{-1} = I + A + A^2 \]

Thus, the inverse of the matrix \( I - A \) is:
\[ (I - A)^{-1} = I + A + A^2 \]