If \(A = \begin{bmatrix} 1 & 2 \\ -1 & -2 \end{bmatrix}\), \(B = \begin{bmatrix} 2 & a \\ -1 & b \end{bmatrix}\) and if \((A + B)^2 = A^2 + B^2\), find value of \(a\) and \(b\).

Answer: \(a = 2\), \(b = -1\)

\((A + B)^2 = (A + B)(A + B) = A^2 + AB + BA + B^2\). Since \((A + B)^2 = A^2 + B^2\), we need \(AB + BA = O\).

\(AB = \begin{bmatrix} 1 & 2 \\ -1 & -2 \end{bmatrix}\begin{bmatrix} 2 & a \\ -1 & b \end{bmatrix} = \begin{bmatrix} 2 - 2 & a + 2b \\ -2 + 2 & -a - 2b \end{bmatrix} = \begin{bmatrix} 0 & a + 2b \\ 0 & -a - 2b \end{bmatrix}\)

\(BA = \begin{bmatrix} 2 & a \\ -1 & b \end{bmatrix}\begin{bmatrix} 1 & 2 \\ -1 & -2 \end{bmatrix} = \begin{bmatrix} 2 - a & 4 - 2a \\ -1 - b & -2 - 2b \end{bmatrix}\)

\(AB + BA = \begin{bmatrix} 2 - a & 4 - a + 2b \\ -1 - b & -2 - a - 4b \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}\)

Equating corresponding elements: \(2 - a = 0 \Rightarrow a = 2\) and \(-1 - b = 0 \Rightarrow b = -1\).

Check: \(4 - 2 + 2(-1) = 0\) and \(-2 - 2 - 4(-1) = 0\). Both hold, so \(a = 2\), \(b = -1\).