If A = [cos θ, i sin θ; i sin θ, cos θ], θ = π/24 and A⁵ = [a b; c d], Where i = √−1, Then Which One of the Following Is Not True

JEE Main 2019. Previous Year Question.

Problem

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

Key insight. Multiplying $A$ by itself follows exactly the same pattern as the cosine and sine double-angle formulas — because $i^2=-1$ makes the $(i\sin\theta)^2$ term behave like $-\sin^2\theta$ in the expansion. That means $A^n$ has the same shape as $A$ itself, just with $\theta$ replaced by $n\theta$, so there’s no need to multiply matrices five times by hand.

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Approach

Compute $A^2$ directly to spot the pattern: it comes out in exactly the same form as $A$, but with $\theta$ replaced by $2\theta$. By induction, $A^n=\begin{pmatrix}\cos(n\theta)&i\sin(n\theta)\\i\sin(n\theta)&\cos(n\theta)\end{pmatrix}$, so $A^5$’s entries are immediate. Check each of the four statements directly using $a=d=\cos5\theta$ and $b=c=i\sin5\theta$.

Solution

Step 1 — Compute A² to find the pattern

$$A^2_{(1,1)} = \cos\theta\cos\theta+i\sin\theta\cdot i\sin\theta = \cos^2\theta+i^2\sin^2\theta = \cos^2\theta-\sin^2\theta = \cos2\theta$$

$$A^2_{(1,2)} = \cos\theta\cdot i\sin\theta+i\sin\theta\cdot\cos\theta = 2i\sin\theta\cos\theta = i\sin2\theta$$

By symmetry, $A^2_{(2,1)}=A^2_{(1,2)}$ and $A^2_{(2,2)}=A^2_{(1,1)}$, so:

$$A^2 = \begin{pmatrix}\cos2\theta & i\sin2\theta\\ i\sin2\theta & \cos2\theta\end{pmatrix}$$

Step 2 — Generalise: Aⁿ has the same shape with nθ

By the same computation repeated (or by induction), for any positive integer $n$:

$$A^n = \begin{pmatrix}\cos(n\theta) & i\sin(n\theta)\\ i\sin(n\theta) & \cos(n\theta)\end{pmatrix}$$

Step 3 — Identify a, b, c, d for A⁵

$$a=\cos5\theta,\quad b=i\sin5\theta,\quad c=i\sin5\theta,\quad d=\cos5\theta$$

In particular, $a=d$ and $b=c$.

Step 4 — Check option (A): a² − d² = 0

Since $a=d$, $a^2-d^2=0$ always. True.

Step 5 — Check option (B): a² − c² = 1

$$a^2-c^2 = \cos^25\theta-(i\sin5\theta)^2 = \cos^25\theta-i^2\sin^25\theta = \cos^25\theta+\sin^25\theta = 1$$

True (this holds for any $\theta$, not just $\theta=\pi/24$).

Step 6 — Check option (C): 0 < a² + b² < 1

$$a^2+b^2 = \cos^25\theta+(i\sin5\theta)^2 = \cos^25\theta-\sin^25\theta = \cos10\theta$$

With $\theta=\dfrac{\pi}{24}$: $10\theta=\dfrac{10\pi}{24}=\dfrac{5\pi}{12}=75°$, and $\cos75°\approx0.259$, which indeed satisfies $0<0.259<1$. True.

Step 7 — Check option (D): a² − b² = 1/2

Since $b=c$, this is the same expression as Step 5:

$$a^2-b^2 = a^2-c^2 = 1$$

But option (D) claims this equals $\dfrac12$ — it doesn’t; it’s always exactly $1$, regardless of $\theta$. This is the false statement.

Answer

Common mistakes

• Not noticing that $b=c$, and re-deriving option (D) from scratch instead of reusing the work from option (B). Since $b$ and $c$ are literally equal here, $a^2-b^2$ and $a^2-c^2$ are the same expression — spotting this symmetry immediately flags that (B) and (D) can’t both be correct as stated, since they claim different values for the same quantity.
• Forgetting the sign flip from $i^2=-1$ when squaring $b=i\sin(5\theta)$. $b^2=(i\sin5\theta)^2=-\sin^25\theta$, so $a^2-b^2=\cos^25\theta-(-\sin^25\theta)=\cos^25\theta+\sin^25\theta=1$, not $\cos^25\theta-\sin^25\theta$ — mixing this up would incorrectly suggest (D) depends on $\theta$.

Practise next

• If $B=\begin{pmatrix}\cos\phi & i\sin\phi\\ i\sin\phi & \cos\phi\end{pmatrix}$ and $B^3=\begin{pmatrix}p&q\\r&s\end{pmatrix}$, find $p^2-r^2$ and $p^2-q^2$, using the same $A^n$ pattern to confirm which quantities are always $1$ regardless of $\phi$.