https://file.aiquickdraw.com/mathbot/file/1732403902286tgdswluj.jpg

Let's solve this step by step:

The problem states that:

1. \( A = \begin{bmatrix} 1 & -1 \\ 2 & -1 \end{bmatrix} \)  
2. \( B = \begin{bmatrix} a & 1 \\ b & -1 \end{bmatrix} \)  
3. The condition is \( (A + B)^2 = A^2 + B^2 \).

---

### Step 1: Expand the matrix equation \( (A + B)^2 = A^2 + B^2 \)

The left-hand side:
\[
(A + B)^2 = (A + B)(A + B) = A^2 + AB + BA + B^2.
\]

Thus, the equation becomes:
\[
A^2 + AB + BA + B^2 = A^2 + B^2.
\]

Canceling \( A^2 \) and \( B^2 \) (since they appear on both sides):
\[
AB + BA = 0.
\]

---

### Step 2: Calculate \( AB + BA \)

#### Compute \( AB \):
\[
A = \begin{bmatrix} 1 & -1 \\ 2 & -1 \end{bmatrix}, \quad B = \begin{bmatrix} a & 1 \\ b & -1 \end{bmatrix}.
\]

\[
AB = \begin{bmatrix} 1 & -1 \\ 2 & -1 \end{bmatrix} \cdot \begin{bmatrix} a & 1 \\ b & -1 \end{bmatrix}.
\]

Performing the multiplication:
\[
AB = \begin{bmatrix} (1)(a) + (-1)(b) & (1)(1) + (-1)(-1) \\ (2)(a) + (-1)(b) & (2)(1) + (-1)(-1) \end{bmatrix}
= \begin{bmatrix} a - b & 2 \\ 2a - b & 3 \end{bmatrix}.
\]

#### Compute \( BA \):
\[
BA = \begin{bmatrix} a & 1 \\ b & -1 \end{bmatrix} \cdot \begin{bmatrix} 1 & -1 \\ 2 & -1 \end{bmatrix}.
\]

Performing the multiplication:
\[
BA = \begin{bmatrix} (a)(1) + (1)(2) & (a)(-1) + (1)(-1) \\ (b)(1) + (-1)(2) & (b)(-1) + (-1)(-1) \end{bmatrix}
= \begin{bmatrix} a + 2 & -a - 1 \\ b - 2 & -b + 1 \end{bmatrix}.
\]

#### Add \( AB \) and \( BA \):
\[
AB + BA = \begin{bmatrix} a - b & 2 \\ 2a - b & 3 \end{bmatrix} + \begin{bmatrix} a + 2 & -a - 1 \\ b - 2 & -b + 1 \end{bmatrix}.
\]

Adding element-wise:
\[
AB + BA = \begin{bmatrix} (a - b) + (a + 2) & 2 + (-a - 1) \\ (2a - b) + (b - 2) & 3 + (-b + 1) \end{bmatrix}.
\]

Simplify:
\[
AB + BA = \begin{bmatrix} 2a - b + 2 & 1 - a \\ 2a - 2 & 4 - b \end{bmatrix}.
\]

---

### Step 3: Set \( AB + BA = 0 \)

Equating \( AB + BA \) to the zero matrix:
\[
\begin{bmatrix} 2a - b + 2 & 1 - a \\ 2a - 2 & 4 - b \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}.
\]

This gives four equations:
1. \( 2a - b + 2 = 0 \),
2. \( 1 - a = 0 \),
3. \( 2a - 2 = 0 \),
4. \( 4 - b = 0 \).

---

### Step 4: Solve the system of equations

From equation (2):
\[
1 - a = 0 \implies a = 1.
\]

From equation (3):
\[
2a - 2 = 0 \implies 2(1) - 2 = 0 \implies a = 1 \; \text{(consistent)}.
\]

From equation (4):
\[
4 - b = 0 \implies b = 4.
\]

From equation (1):
\[
2a - b + 2 = 0 \implies 2(1) - 4 + 2 = 0 \implies 0 = 0 \; \text{(consistent)}.
\]

---

### Final Answer:
\[
\boxed{C. \; a = 1, b = 4.}
\]

Would you like further details or clarifications? Here are some related questions:

1. Can you explain how matrix multiplication works in this example?  
2. Why does \( AB + BA = 0 \) simplify the problem?  
3. How can we interpret the condition \( (A + B)^2 = A^2 + B^2 \) geometrically?  
4. What are the properties of matrices that satisfy \( AB + BA = 0 \)?  
5. How would the solution change if \( A \) or \( B \) were non-square?

**Tip**: Always double-check matrix dimensions and consistency when solving equations involving matrices.

If A = [[1, -1], [2, -1]] and B = [[a, 1], [b, -1]], and (A + B)^2 = A^2 + B^2, find the values of a and b.

Learn how to solve the matrix equation (A+B)^2 = A^2 + B^2 where A and B are 2x2 matrices. This problem involves concepts of matrix addition, multiplication, and properties. Find the values of a and b step by step.

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