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GO Classes DPP | GATE CS, DA | Linear Algebra | Matrix Multiplication

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7 7 votes

Let $\theta=\dfrac{2\pi}{1013}$. Now consider the matrix $A=
\begin{bmatrix}
\cos\theta & \sin\theta\\
-\sin\theta & \cos\theta
\end{bmatrix}$.

Then the matrix $A^{2026}$ is

  1. $\begin{bmatrix}
    \cos\theta & \sin\theta\\
    -\sin\theta & \cos\theta
    \end{bmatrix}$
     
  2. $\begin{bmatrix}
    1 & 0\\
    0 & 1
    \end{bmatrix}$
     
  3. $\begin{bmatrix}
    \cos^{2}\theta & \sin^{2}\theta\\
    -\sin^{2}\theta & \cos^{2}\theta
    \end{bmatrix}$
     
  4. $\begin{bmatrix}
    0 & 1\\
    -1 & 0
    \end{bmatrix}$

2 Answers

4 4 votes

We have $A=
\begin{bmatrix}
\cos\theta & \sin\theta\\
-\sin\theta & \cos\theta
\end{bmatrix}$.


Now,
$A^2=
\begin{bmatrix}
\cos\theta & \sin\theta\\
-\sin\theta & \cos\theta
\end{bmatrix}
\begin{bmatrix}
\cos\theta & \sin\theta\\
-\sin\theta & \cos\theta
\end{bmatrix}
=
\begin{bmatrix}
\cos2\theta & \sin2\theta\\
-\sin2\theta & \cos2\theta
\end{bmatrix}$.


Similarly,

$A^3=
\begin{bmatrix}
\cos3\theta & \sin3\theta\\
-\sin3\theta & \cos3\theta
\end{bmatrix}$,

and in general,

$A^n=
\begin{bmatrix}
\cos n\theta & \sin n\theta\\
-\sin n\theta & \cos n\theta
\end{bmatrix}$.

Therefore,

$A^{2026}=
\begin{bmatrix}
\cos(2026\theta) & \sin(2026\theta)\\
-\sin(2026\theta) & \cos(2026\theta)
\end{bmatrix}$.

Now $\theta=\dfrac{2\pi}{1013}$, so

$2026\theta=2026\cdot\dfrac{2\pi}{1013}=4\pi$.

Hence,
$A^{2026}=
\begin{bmatrix}
\cos4\pi & \sin4\pi\\
-\sin4\pi & \cos4\pi
\end{bmatrix}$.

Since $\cos4\pi=1$ and $\sin4\pi=0$, we get

$A^{2026}=
\begin{bmatrix}
1 & 0\\
0 & 1
\end{bmatrix}$.

So, the correct option is B.

 

$\underline{\textbf{Alternate Solution : }}$

We use the fact that

$A=
\begin{bmatrix}
\cos\theta & \sin\theta\\
-\sin\theta & \cos\theta
\end{bmatrix}$

represents a rotation matrix. Hence, for any positive integer $n$,

$A^n=
\begin{bmatrix}
\cos(n\theta) & \sin(n\theta)\\
-\sin(n\theta) & \cos(n\theta)
\end{bmatrix}$.

Therefore,

$A^{2026}=
\begin{bmatrix}
\cos(2026\theta) & \sin(2026\theta)\\
-\sin(2026\theta) & \cos(2026\theta)
\end{bmatrix}$.

Now $\theta=\dfrac{2\pi}{1013}$, so

$2026\theta=2026\cdot\dfrac{2\pi}{1013}=4\pi$.

Hence,
$A^{2026}=
\begin{bmatrix}
\cos4\pi & \sin4\pi\\
-\sin4\pi & \cos4\pi
\end{bmatrix}$.

Since $\cos4\pi=1$ and $\sin4\pi=0$, we get

$A^{2026}=
\begin{bmatrix}
1 & 0\\
0 & 1
\end{bmatrix}$.

So, the correct option is B.

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