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Question: If \(A = \begin{bmatrix} a & b \\ b & a \end{bmatrix}\) and \(A^{2} = \begin{bmatrix} \alpha & \beta...

If A=[abba]A = \begin{bmatrix} a & b \\ b & a \end{bmatrix} and A2=[αββα]A^{2} = \begin{bmatrix} \alpha & \beta \\ \beta & \alpha \end{bmatrix}then

A

α=a2+b2,β=ab\alpha = a^{2} + b^{2},\beta = ab

B

α=a2+b2,β=2ab\alpha = a^{2} + b^{2},\beta = 2ab

C

α=a2+b2,β=a2−b2\alpha = a^{2} + b^{2},\beta = a^{2} - b^{2}

D

α=2ab,β=a2+b2\alpha = 2ab,\beta = a^{2} + b^{2}

Answer

α=a2+b2,β=2ab\alpha = a^{2} + b^{2},\beta = 2ab

Explanation

Solution

A2=[αββα]=[abba][abba]A^{2} = \begin{bmatrix} \alpha & \beta \\ \beta & \alpha \end{bmatrix} = \begin{bmatrix} a & b \\ b & a \end{bmatrix}\begin{bmatrix} a & b \\ b & a \end{bmatrix} =[a2+b22ab2aba2+b2]= \begin{bmatrix} a^{2} + b^{2} & 2ab \\ 2ab & a^{2} + b^{2} \end{bmatrix}.

On comparing, we get, α=a2+b2,β=2ab\alpha = a^{2} + b^{2},\beta = 2ab