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(D) Given $A = \begin{bmatrix} 1 & -\alpha \ \alpha & \beta \end{bmatrix}$ and $AA^{T} = I_{2}$.First,find the transpose $A^{T} = \begin{bmatrix} 1 &

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(D) Given $A = \begin{bmatrix} 1 & -\alpha \ \alpha & \beta \end{bmatrix}$ and $AA^{T} = I_{2}$.First,find the transpose $A^{T} = \begin{bmatrix} 1 & \alpha \ -\alpha & \beta \end{bmatrix}$.Now,compute the product $AA^{T}$:$AA^{T} = \begin{bmatrix} 1 & -\alpha \ \alpha & \beta \end{bmatrix} \begin{bmatrix} 1 & \alpha \ -\alpha & \beta \end{bmatrix} = \begin{bmatrix} 1 + \alpha^{2} & \alpha - \alpha\beta \ \alpha - \alpha\beta & \alpha^{2} + \beta^{2} \end{bmatrix}$.Since $AA^{T} = I_{2} = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}$,we equate the corresponding elements:$1 + \alpha^{2} = 1 \Rightarrow \alpha^{2} = 0 \Rightarrow \alpha = 0$.$\alpha - \alpha\beta = 0 \Rightarrow 0 - 0\beta = 0$ (which is always true).$\alpha^{2} + \beta^{2} = 1 \Rightarrow 0 + \beta^{2} = 1 \Rightarrow \beta^{2} = 1$.We need to find the value of $\alpha^{4} + \beta^{4}$:$\alpha^{4} + \beta^{4} = (0)^{2} + (1)^{2} = 0 + 1 = 1$.

If for the matrix $A = \begin{bmatrix} 1 & -\alpha \\ \alpha & \beta \end{bmatrix}$,$AA^{T} = I_{2}$,then the value of $\alpha^{4} + \beta^{4}$ is ....... .

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