Question
If \[ A = \begin{bmatrix} 0 & -x \\ x & 0 \end{bmatrix}, \quad B = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} \] and \(x^2 = -1\), show that \[ (A + B)^2 = A^2 + B^2. \]
Solution
Step 1: Compute \(A^2\)
\[ A^2 = \begin{bmatrix} 0 & -x \\ x & 0 \end{bmatrix} \begin{bmatrix} 0 & -x \\ x & 0 \end{bmatrix} = \begin{bmatrix} -x^2 & 0 \\ 0 & -x^2 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \] (using \(x^2=-1\))Step 2: Compute \(B^2\)
\[ B^2 = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \]Step 3: Compute \(AB\) and \(BA\)
\[ AB = \begin{bmatrix} -x & 0 \\ 0 & x \end{bmatrix}, \quad BA = \begin{bmatrix} x & 0 \\ 0 & -x \end{bmatrix} \] \[ AB + BA = O \]Step 4: Expand \((A+B)^2\)
\[ (A+B)^2 = A^2 + AB + BA + B^2 \] \[ = A^2 + B^2 \]Final Result
\[
(A + B)^2 = A^2 + B^2
\]
Hence proved.