We are asked to find the expression for $f(A)$ where $f(x) = \frac{1+x}{1-x}$ and $A$ is a matrix satisfying $A^3 = 0$.

$f(A)$ Expression Derivation

We can express $f(x)$ using its Taylor series expansion around $x=0$. The geometric series is $\frac{1}{1-x} = 1 + x + x^2 + x^3 + \dots$ for $|x|<1$. Using this, we can write $f(x)$ as:

$f(x) = (1+x) \times \frac{1}{1-x}$

Substituting the geometric series:

$f(x) = (1+x) (1 + x + x^2 + x^3 + \dots)$

Expanding this product:

$f(x) = (1 + x + x^2 + x^3 + \dots) + (x + x^2 + x^3 + x^4 + \dots)$

Combine like terms:

$f(x) = 1 + 2x + 2x^2 + 2x^3 + \dots$

Alternatively, we can write $f(x) = \frac{-(1-x) + 2}{1-x} = -1 + \frac{2}{1-x}$. $f(x) = -1 + 2(1 + x + x^2 + x^3 + \dots)$ $f(x) = -1 + 2 + 2x + 2x^2 + 2x^3 + \dots$ $f(x) = 1 + 2x + 2x^2 + 2x^3 + \dots$

Applying Nilpotent Property $A^3=0$

Now, we apply this series expansion to the matrix $A$. We replace $x$ with $A$ and $1$ with the identity matrix $I$:

$f(A) = I + 2A + 2A^2 + 2A^3 + \dots$

We are given that $A^3 = 0$. This property implies that all higher powers of $A$ are also the zero matrix:

• $A^4 = A \cdot A^3 = A \cdot 0 = 0$
• $A^5 = A \cdot A^4 = A \cdot 0 = 0$
• And so on for all $A^n$ where $n \ge 3$.

Substituting $A^3 = 0$ and higher powers into the series for $f(A)$:

$f(A) = I + 2A + 2A^2 + 2(0) + 2(0) + \dots$

This simplifies the expression to:

$f(A) = I + 2A + 2A^2$

Final Result for $f(A)$

Comparing this result with the given options, we find that $I + 2A + 2A^2$ corresponds to option 1 ($1 + 2A + 2A^2$, where '1' represents the identity matrix $I$).