Find all 2 by 2 matrices ([a, b], [c, d]) that are commute with the 2 by 2 matrix ([1, 2], [-1, -1]) 

Let the matrix \( A = \begin{pmatrix} 1 & 2 \\ -1 & -1 \end{pmatrix} \). We are tasked with finding all \( 2 \times 2 \) matrices \( B = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \) that commute with \( A \), meaning that:

\[
AB = BA
\]

### Step 1: Compute \( AB \)
The product of \( A \) and \( B \) is:

\[
AB = \begin{pmatrix} 1 & 2 \\ -1 & -1 \end{pmatrix} \begin{pmatrix} a & b \\ c & d \end{pmatrix}
= \begin{pmatrix} 1 \cdot a + 2 \cdot c & 1 \cdot b + 2 \cdot d \\ -1 \cdot a - 1 \cdot c & -1 \cdot b - 1 \cdot d \end{pmatrix}
= \begin{pmatrix} a + 2c & b + 2d \\ -a - c & -b - d \end{pmatrix}
\]

### Step 2: Compute \( BA \)
The product of \( B \) and \( A \) is:

\[
BA = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} 1 & 2 \\ -1 & -1 \end{pmatrix}
= \begin{pmatrix} a \cdot 1 + b \cdot (-1) & a \cdot 2 + b \cdot (-1) \\ c \cdot 1 + d \cdot (-1) & c \cdot 2 + d \cdot (-1) \end{pmatrix}
= \begin{pmatrix} a - b & 2a - b \\ c - d & 2c - d \end{pmatrix}
\]

### Step 3: Set \( AB = BA \)
Now, set the matrices equal to each other:

\[
\begin{pmatrix} a + 2c & b + 2d \\ -a - c & -b - d \end{pmatrix} = \begin{pmatrix} a - b & 2a - b \\ c - d & 2c - d \end{pmatrix}
\]

This gives us the following system of equations by equating corresponding elements:

1. \( a + 2c = a - b \)  (top left entry)
2. \( b + 2d = 2a - b \)  (top right entry)
3. \( -a - c = c - d \)  (bottom left entry)
4. \( -b - d = 2c - d \)  (bottom right entry)

### Step 4: Solve the system of equations
#### From equation (1):
\[
a + 2c = a - b \quad \Rightarrow \quad 2c = -b \quad \Rightarrow \quad b = -2c
\]

#### From equation (2):
\[
b + 2d = 2a - b \quad \Rightarrow \quad (-2c) + 2d = 2a + 2c \quad \Rightarrow \quad 2d = 2a + 4c \quad \Rightarrow \quad d = a + 2c
\]

#### From equation (3):
\[
-a - c = c - d \quad \Rightarrow \quad -a - c = c - (a + 2c) \quad \Rightarrow \quad -a - c = c - a - 2c \quad \Rightarrow \quad -a - c = -a - c
\]
This equation is always true, so it gives no new information.

#### From equation (4):
\[
-b - d = 2c - d \quad \Rightarrow \quad (-2c) - (a + 2c) = 2c - (a + 2c) \quad \Rightarrow \quad -3c - a = 0 \quad \Rightarrow \quad a = -3c
\]

### Step 5: Conclusion
Now, substitute \( a = -3c \) into the expressions for \( b \) and \( d \):
\[
b = -2c, \quad d = a + 2c = -3c + 2c = -c
\]

Thus, the matrix \( B \) is given by:
\[
B = \begin{pmatrix} -3c & -2c \\ c & -c \end{pmatrix}
= c \begin{pmatrix} -3 & -2 \\ 1 & -1 \end{pmatrix}
\]
where \( c \) is any real number.

Therefore, all matrices that commute with \( A = \begin{pmatrix} 1 & 2 \\ -1 & -1 \end{pmatrix} \) are scalar multiples of \( \begin{pmatrix} -3 & -2 \\ 1 & -1 \end{pmatrix} \).

---

Would you like more details on any part of the solution? Here are 5 related questions you could explore:

1. What are the properties of matrices that commute with a given matrix?
2. How can the concept of commutative matrices be applied in quantum mechanics?
3. Can two different matrices commute with the same matrix?
4. What are the eigenvalues and eigenvectors of the matrix \( A \)?
5. How does the solution change if we consider \( 3 \times 3 \) matrices?

**Tip:** The condition \( AB = BA \) is often used to identify diagonalizable or symmetric matrices in linear algebra.

Find all 2 by 2 matrices ([a, b], [c, d]) that commute with the 2 by 2 matrix ([1, 2], [-1, -1]).

Explore how to find all 2x2 matrices that commute with a given matrix using matrix multiplication and solving systems of linear equations. This detailed solution covers key concepts in linear algebra, including the commutative property of matrices and solving for matrix entries.

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