Let A =`[{:(1,0),(0,-1):}]and B =[{:(1,x),(0,1):}]` If AB = BA, then what is the value of x ?

Options

A. -1
B. 0
C. 1
D. Any real number

Explanation

To solve the problem, we need to find the value of \( x \) such that the matrices \( A \) and \( B \) commute, meaning \( AB = BA \).

Given:
\[ A = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}, \quad B = \begin{pmatrix} 1 & x \\ 0 & 1 \end{pmatrix} \]

### Step 1: Calculate \( AB \)

To find \( AB \), we multiply matrix \( A \) by matrix \( B \):

\[ AB = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \begin{pmatrix} 1 & x \\ 0 & 1 \end{pmatrix} \]

Calculating the elements:
- First row, first column: \( 1 \cdot 1 + 0 \cdot 0 = 1 \)
- First row, second column: \( 1 \cdot x + 0 \cdot 1 = x \)
- Second row, first column: \( 0 \cdot 1 + (-1) \cdot 0 = 0 \)
- Second row, second column: \( 0 \cdot x + (-1) \cdot 1 = -1 \)

Thus, we have:
\[ AB = \begin{pmatrix} 1 & x \\ 0 & -1 \end{pmatrix} \]

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

Now, we calculate \( BA \):

\[ BA = \begin{pmatrix} 1 & x \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \]

Calculating the elements:
- First row, first column: \( 1 \cdot 1 + x \cdot 0 = 1 \)
- First row, second column: \( 1 \cdot 0 + x \cdot (-1) = -x \)
- Second row, first column: \( 0 \cdot 1 + 1 \cdot 0 = 0 \)
- Second row, second column: \( 0 \cdot 0 + 1 \cdot (-1) = -1 \)

Thus, we have:
\[ BA = \begin{pmatrix} 1 & -x \\ 0 & -1 \end{pmatrix} \]

### Step 3: Set \( AB \) equal to \( BA \)

Since we want \( AB = BA \), we set the two matrices equal to each other:

\[ \begin{pmatrix} 1 & x \\ 0 & -1 \end{pmatrix} = \begin{pmatrix} 1 & -x \\ 0 & -1 \end{pmatrix} \]

### Step 4: Compare the elements

From the equality of the matrices, we can compare the corresponding elements:
1. From the first row, second column: \( x = -x \)

### Step 5: Solve for \( x \)

Solving the equation \( x = -x \):
\[ 2x = 0 \implies x = 0 \]

Thus, the value of \( x \) is:
\[ \boxed{0} \]