Animated Solution for Mathematics - Matrices and Determinants: Let $\theta =\frac{\pi }{5}$ and $A=\left(\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right)$. If $B=A+A^4$ then $det(B)$ :

The matrix $A$ is defined as a standard rotation matrix: This matrix represents a pure rotation by an angle $\theta$ in a 2D plane.

The structure of $A$ follows the form $\left(\begin{array}{cc}X& Y\\ -Y& X\end{array}\right)$, which is characteristic of rotation-scaling transformations.

In linear algebra, raising a rotation matrix to a power $n$ is equivalent to rotating by $n\theta$. Therefore, $A^4$ is a rotation matrix corresponding to the angle $4\theta$:

When we define $B=A+A^4$, the resulting matrix $B$ inherits the same structural form. Let $X=\cos \theta +\cos 4\theta$ and $Y=\sin \theta +\sin 4\theta$. Then, the matrix $B$ is expressed as:

The determinant of a matrix with the structure $\left(\begin{array}{cc}X& Y\\ -Y& X\end{array}\right)$ is given by $X^2+Y^2$. Substituting our values for $X$ and $Y$, we calculate:

Expanding these terms, we obtain:

Using the trigonometric identity ${\cos }^2\alpha +{\sin }^2\alpha =1$ and the compound angle formula $\cos (A-B)=\cos A\cos B+\sin A\sin B$, the expression simplifies to:

We are given $\theta =\frac{\pi }{5}$, which corresponds to $36^{\circ }$. Consequently, $3\theta =108^{\circ }$. We must evaluate the expression $2+2\cos (108^{\circ })$.

Recalling that $\cos (108^{\circ })=-\sin (18^{\circ })=\frac{1-\sqrt{5}}{4}$, we substitute this into our determinant formula:

Simplifying the arithmetic:

The final value is $\frac{5-\sqrt{5}}{2}$, which is approximately $1.382$. This value lies within the interval $(1,2)$.