Chapter 3 - Determinants - 3.2 Properties of Determinants - Problems - Page 220: 48

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Suppose $A=\begin{bmatrix} \alpha x-\beta y &\beta x-\alpha y\\ \beta x+\alpha y & \alpha x+\beta y \end{bmatrix}\\ B=\begin{bmatrix} \alpha x &\beta x-\alpha y\\ \beta x & \alpha x+\beta y \end{bmatrix}\\ C=\begin{bmatrix} -\beta y &\beta x-\alpha y\\ \alpha y & \alpha x+\beta y \end{bmatrix}$ According to property P6, we have $\det (A)=\det(B)+\det (C)\\ \begin{bmatrix} \alpha x-\beta y &\beta x-\alpha y\\ \beta x+\alpha y & \alpha x+\beta y \end{bmatrix}=\begin{bmatrix} \alpha x &\beta x-\alpha y\\ \beta x & \alpha x+\beta y \end{bmatrix}+\begin{bmatrix} -\beta y &\beta x-\alpha y\\ \alpha y & \alpha x+\beta y \end{bmatrix}$ Using property P2: $=\begin{bmatrix} \alpha x &\beta x\\ \alpha x & \beta x \end{bmatrix}+\begin{bmatrix} \alpha x& -\alpha y\\ \beta x & \beta y \end{bmatrix}+\begin{bmatrix} -\beta y &\beta x\\ \alpha y & \alpha x \end{bmatrix}+\begin{bmatrix} -\beta y &-\alpha y\\ \alpha y & \beta y \end{bmatrix}\\ =x\begin{bmatrix} \alpha &\beta \\ \alpha & \beta \end{bmatrix}+x\begin{bmatrix} \alpha & -\alpha y\\ \beta & \beta y \end{bmatrix}+y\begin{bmatrix} -\beta &\beta x\\ \alpha & \alpha x \end{bmatrix}+y\begin{bmatrix} -\beta &-\alpha \\ \alpha & \beta \end{bmatrix}\\ =x^2\begin{bmatrix} \alpha &\beta \\ \beta & \alpha \end{bmatrix}+xy\begin{bmatrix} \alpha & -\alpha \\ \beta & \beta \end{bmatrix}+yx\begin{bmatrix} -\beta &\beta \\ \alpha & \alpha \end{bmatrix}+y^2\begin{bmatrix} -\beta &-\alpha \\ \alpha & \beta \end{bmatrix}\\ =x^2\begin{bmatrix} \alpha &\beta \\ \beta & \alpha \end{bmatrix}+xy\begin{bmatrix} \alpha & -\alpha \\ \beta & \beta \end{bmatrix}+yx\begin{bmatrix} -\beta &\beta \\ \alpha & \alpha \end{bmatrix}y^2\begin{bmatrix} \beta &\alpha \\ \alpha & \beta \end{bmatrix}\\ =x^2\begin{bmatrix} \alpha &\beta \\ \beta & \alpha \end{bmatrix}+xy\begin{bmatrix} \alpha & -\alpha \\ \beta & \beta \end{bmatrix}-yx\begin{bmatrix} \beta &-\beta \\ \alpha & \alpha \end{bmatrix}-y^2\begin{bmatrix} \beta &\alpha \\ \alpha & \beta \end{bmatrix}$ Using property P1: $=x^2\begin{bmatrix} \alpha &\beta \\ \beta & \alpha \end{bmatrix}+xy\begin{bmatrix} \alpha & -\alpha \\ \beta & \beta \end{bmatrix}+xy\begin{bmatrix} \alpha & \alpha \\\beta &-\beta \\ \end{bmatrix}+y^2\begin{bmatrix} \alpha & \beta \\\beta &\alpha \\ \end{bmatrix}\\ =x^2\begin{bmatrix} \alpha &\beta \\ \beta & \alpha \end{bmatrix}+xy\begin{bmatrix} \alpha & -\alpha \\ \beta & \beta \end{bmatrix}-xy\begin{bmatrix} \alpha & -\alpha \\\beta &-\beta \\ \end{bmatrix}+y^2\begin{bmatrix} \alpha & \beta \\\beta &\alpha \\ \end{bmatrix}\\ =(x^2+y^2)\begin{bmatrix} \alpha & \beta \\\beta &\alpha \\ \end{bmatrix}$