If $$A = \begin{bmatrix} \cos\theta & i\sin\theta \\ i\sin\theta & \cos\theta \end{bmatrix}$$, $$(\theta = \frac{\pi}{24})$$ and $$A^5 = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$$, where $$i = \sqrt{-1}$$, then which one of the following is not true?

We have the matrix

$$A=\begin{bmatrix}\cos\theta & i\sin\theta \\ i\sin\theta & \cos\theta\end{bmatrix},\qquad\theta=\frac{\pi}{24},\qquad i=\sqrt{-1}.$$

To make the powers of $$A$$ easy to handle, write the $$2\times2$$ identity matrix as $$I$$ and introduce the matrix

$$J=\begin{bmatrix}0&1\\1&0\end{bmatrix}.$$

Observe that $$J^2=I$$ because

$$\begin{bmatrix}0&1\\1&0\end{bmatrix} \begin{bmatrix}0&1\\1&0\end{bmatrix} =\begin{bmatrix}1&0\\0&1\end{bmatrix}=I.$$

Now rewrite $$A$$ in terms of $$I$$ and $$J$$:

$$A=\cos\theta\,I+i\sin\theta\,J.$$

Since $$I$$ and $$J$$ commute ($$IJ=JI$$), we may use the standard trigonometric-exponential identity

$$\bigl(\cos\theta\,I+i\sin\theta\,J\bigr)^n=\cos(n\theta)\,I+i\sin(n\theta)\,J,$$

which is the matrix analogue of $$\bigl(\cos\theta+i\sin\theta\bigr)^n=\cos(n\theta)+i\sin(n\theta)$$ (De Moivre’s formula).

Putting $$n=5$$ we obtain

$$A^5=\cos(5\theta)\,I+i\sin(5\theta)\,J =\begin{bmatrix} \cos(5\theta) & i\sin(5\theta)\\ i\sin(5\theta) & \cos(5\theta) \end{bmatrix}.$$

Comparing with $$A^5=\begin{bmatrix}a&b\\c&d\end{bmatrix}$$ gives

$$a=d=\cos(5\theta),\qquad b=c=i\sin(5\theta).$$

Square each entry:

$$a^2=d^2=\cos^2(5\theta),$$

$$b^2=c^2=(i\sin(5\theta))^2=i^2\sin^2(5\theta)=-\sin^2(5\theta).$$

Because $$\theta=\dfrac{\pi}{24}$$,

$$5\theta=\frac{5\pi}{24},\qquad 10\theta=\frac{10\pi}{24}=\frac{5\pi}{12}=75^{\circ}.$$

With these preparations we can check each option.

Option A:

$$a^2+b^2=\cos^2(5\theta)+\bigl(-\sin^2(5\theta)\bigr) =\cos^2(5\theta)-\sin^2(5\theta) =\cos\bigl(2\cdot5\theta\bigr) =\cos(10\theta) =\cos75^{\circ}\approx0.2588.$$

This value satisfies $$0\le a^2+b^2\le1$$, so Option A is true.

Option B:

$$a^2-d^2=\cos^2(5\theta)-\cos^2(5\theta)=0,$$

so Option B is true.

Option C:

$$a^2-c^2=\cos^2(5\theta)-\bigl(-\sin^2(5\theta)\bigr) =\cos^2(5\theta)+\sin^2(5\theta)=1,$$

so Option C is true.

Option D:

$$a^2-b^2=\cos^2(5\theta)-\bigl(-\sin^2(5\theta)\bigr) =\cos^2(5\theta)+\sin^2(5\theta)=1,$$

whereas Option D claims this difference equals $$\dfrac12$$. The claim is therefore false.

Exactly one statement must be “not true,” and we have just seen that Option D is the only false one.

Hence, the correct answer is Option D.