If
\[
f(x)=\frac{1+x}{1-x}
\]
and \(A\) is a matrix such that \(A^{3}=0\), then
\[
f(A)=
\]
Show Hint
An alternative algebraic trick to avoid infinite geometric expansions is to set up a direct matrix equation: let $f(A) = X \implies (I - A)X = I + A$. Since $A^3 = 0$, multiply both sides by $(I + A + A^2)$ to use the factorization rule $(I - A)(I + A + A^2) = I - A^3 = I$. This instantly yields $X = (I + A + A^2)(I + A) = I + 2A + 2A^2$.
Step 1: Understanding the Concept:
This problem involves matrix functions. If a function \( f(x) \) can be expanded as a power series, then \( f(A) \) can be computed by substituting matrix \( A \) into that series. The condition \( A^3 = 0 \) tells us that \( A \) is a nilpotent matrix of index 3, which means all terms in the series involving \( A^3 \) and higher powers will be the zero matrix. Step 2: Key Formula or Approach:
1. \( f(A) = (I + A)(I - A)^{-1} \).
2. Use the geometric series expansion for matrices: \( (I - A)^{-1} = I + A + A^2 + A^3 + \dots \). Step 3: Detailed Explanation:
The given matrix function is \( f(A) = (I + A)(I - A)^{-1} \).
We know the identity \( (I - A)(I + A + A^2) = I + A + A^2 - A - A^2 - A^3 = I - A^3 \).
Since it is given that \( A^3 = 0 \), we have:
\[ (I - A)(I + A + A^2) = I \]
This implies that the inverse of \( (I - A) \) is precisely \( I + A + A^2 \).
Now, substitute this into the expression for \( f(A) \):
\[ f(A) = (I + A)(I + A + A^2) \]
Perform the matrix multiplication:
\[ f(A) = I(I + A + A^2) + A(I + A + A^2) \]
\[ f(A) = I + A + A^2 + A + A^2 + A^3 \]
Collect like terms:
\[ f(A) = I + 2A + 2A^2 + A^3 \]
Applying the condition \( A^3 = 0 \):
\[ f(A) = I + 2A + 2A^2 \] Step 4: Final Answer:
The result of the matrix function evaluation is \( I + 2A + 2A^2 \), which is option (A).