To address the problem, let's break it down into steps:
### Understanding the Matrix Equation:
Given matrices [tex]\(A\)[/tex] and [tex]\(B\)[/tex] such that
[tex]\[
(A + B)^2 = A^2 + 2AB + B^2,
\][/tex]
and assuming [tex]\(AB = BA = I_2\)[/tex], we need to deduce that [tex]\(B = A^{-1}\)[/tex].
### Step-by-Step Solution:
1. Matrix Expansion:
The left-hand side of the given equation is the square of the sum of two matrices [tex]\(A\)[/tex] and [tex]\(B\)[/tex]. Using the distributive property of matrix multiplication, we can expand [tex]\((A + B)^2\)[/tex] as follows:
[tex]\[
(A + B)^2 = (A + B)(A + B) = A^2 + AB + BA + B^2.
\][/tex]
2. Simplifying the Expression:
The equation simplifies to:
[tex]\[
A^2 + AB + BA + B^2 = A^2 + 2AB + B^2.
\][/tex]
3. Canceling Out Terms:
Since [tex]\(A^2 + 2AB + B^2\)[/tex] appears on both sides of the equation, we compare each corresponding term:
[tex]\[
A^2 + AB + BA + B^2 = A^2 + 2AB + B^2.
\][/tex]
4. Identifying the Relationship:
By removing the common terms [tex]\(A^2\)[/tex] and [tex]\(B^2\)[/tex] from both sides, we get:
[tex]\[
AB + BA = 2AB.
\][/tex]
5. Simplifying Further:
Rearrange the equation to isolate like terms:
[tex]\[
BA = 2AB - AB.
\][/tex]
[tex]\[
BA = AB.
\][/tex]
Given [tex]\(AB = BA = I_2\)[/tex] and this identifies [tex]\(B\)[/tex] as the inverse of [tex]\(A\)[/tex], therefore:
[tex]\[
B = A^{-1}.
\][/tex]
### Finding [tex]\(B\)[/tex]:
Now, given [tex]\( A = \begin{bmatrix} 1 & 2 \\ 9 & -1 \end{bmatrix} \)[/tex], we need to find the matrix [tex]\( B \)[/tex], which is the inverse of [tex]\( A \)[/tex].
Based on the provided result (which is treated as correct and trustworthy), the inverse of matrix [tex]\( A \)[/tex] is:
[tex]\[
B = A^{-1} = \begin{bmatrix} 0.05263158 & 0.10526316 \\ 0.47368421 & -0.05263158 \end{bmatrix}.
\][/tex]
So, the matrix [tex]\( B \)[/tex] which is the inverse of [tex]\( A \)[/tex], is:
[tex]\[
B = \begin{bmatrix} 0.05263158 & 0.10526316 \\ 0.47368421 & -0.05263158 \end{bmatrix}.
\][/tex]
Thus, we have successfully deduced that [tex]\( B = A^{-1} \)[/tex] and have found the explicit form of [tex]\( B \)[/tex] as requested.
