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MATRICES — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSMATRICES1 Mark
Question
If $\begin{bmatrix}\cos\frac{2\pi}{7}&-\sin\frac{2\pi}{7}\\\sin\frac{2\pi}{7}&\cos\frac{2\pi}{7}\end{bmatrix}^\text{k}=\begin{bmatrix}1&0\\0&1\end{bmatrix},$ then the least positive integral value of k is:
  1. 3
  2. 4
  3. 6
  4. 7
✓

Answer

  1. 7
Solution:
$\text{A}=\begin{bmatrix}\cos\frac{2\pi}{7}&-\sin\frac{2\pi}{7}\\\sin\frac{2\pi}{7}&\cos\frac{2\pi}{7}\end{bmatrix}$
$\Rightarrow\text{A}^2=\text{A}\times\text{A}$
$\Rightarrow\text{A}^2=\begin{bmatrix}\cos\frac{2\pi}{7}&-\sin\frac{2\pi}{7}\\\sin\frac{2\pi}{7}&\cos\frac{2\pi}{7}\end{bmatrix}\begin{bmatrix}\cos\frac{2\pi}{7}&-\sin\frac{2\pi}{7}\\\sin\frac{2\pi}{7}&\cos\frac{2\pi}{7}\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}\cos^2\frac{2\pi}{7}-\sin^2\frac{2\pi}{7}&\Big(-2\cos\frac{2\pi}{7}-\sin\frac{2\pi}{7}\Big)\\2\cos\frac{2\pi}{7}\sin\frac{2\pi}{7}&\cos\frac{2\pi}{7}-\sin^2\frac{2\pi}{7}\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}\cos\frac{4\pi}{7}&-\sin\frac{4\pi}{7}\\\sin\frac{4\pi}{7}&\cos\frac{4\pi}{7}\end{bmatrix}$$\begin{bmatrix}\because\ \cos^2\theta-\sin^2\theta=\cos2\theta\\2\sin\theta\cos\theta=\sin2\theta\end{bmatrix}$
$\Rightarrow\text{A}^3=\text{A}^2\times\text{A}$
$\Rightarrow\text{A}^3=\begin{bmatrix}\cos\frac{4\pi}{7}&-\sin\frac{4\pi}{7}\\\sin\frac{4\pi}{7}&\cos\frac{4\pi}{7}\end{bmatrix}\begin{bmatrix}\cos\frac{2\pi}{7}&-\sin\frac{2\pi}{7}\\\sin\frac{2\pi}{7}&\cos\frac{2\pi}{7}\end{bmatrix}$
$\Rightarrow\text{A}^3=\begin{bmatrix}\Big(\cos\frac{4\pi}{7}\cos\frac{2\pi}{7}-\sin\frac{4\pi}{7}\sin\frac{2\pi}{7}\Big)&\Big(-\cos\frac{4\pi}{7}\sin\frac{2\pi}{7}-\sin\frac{4\pi}{7}\cos\frac{2\pi}{7}\Big)\\\Big(\sin\frac{4\pi}{7}\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}\sin\frac{2\pi}{7}\Big)&\Big(-\sin\frac{2\pi}{7}\sin\frac{4\pi}{7}+\cos\frac{4\pi}{7}\cos\frac{2\pi}{7}\Big)\end{bmatrix}$
$\Rightarrow\text{A}^3=\begin{bmatrix}\cos\frac{6\pi}{7}&-\sin\frac{6\pi}{7}\\\sin\frac{6\pi}{7}&\cos\frac{6\pi}{7}\end{bmatrix}$$\begin{bmatrix}\because\ \cos\text{(A+B)}=\cos\text{A}\cos\text{B}-\sin\text{A}\sin\text{B}\\\sin\text{(A+B)}=\sin\text{A}\cos\text{B}+\cos\text{A}\sin\text{B}\end{bmatrix}$
Now we check if the pattern is same for k = 6.
Here,
$\text{A}^6=\text{A}^3.\text{A}^3$
$\Rightarrow\text{A}^6=\begin{bmatrix}\cos\frac{6\pi}{7}&-\sin\frac{6\pi}{7}\\\sin\frac{6\pi}{7}&\cos\frac{6\pi}{7}\end{bmatrix}\begin{bmatrix}\cos\frac{6\pi}{7}&-\sin\frac{6\pi}{7}\\\sin\frac{6\pi}{7}&\cos\frac{6\pi}{7}\end{bmatrix}$
$\Rightarrow\text{A}^6=\begin{bmatrix}\cos\frac{12\pi}{7}&-\sin\frac{12\pi}{7}\\\sin\frac{12\pi}{7}&\cos\frac{12\pi}{7}\end{bmatrix}$
Now, we check if the pattern is same for k = 7.
Here,
$\text{A}^7=\text{A}^6\times\text{A}$
$\Rightarrow\text{A}^7=\begin{bmatrix}\cos\frac{6\pi}{7}&-\sin\frac{6\pi}{7}\\\sin\frac{6\pi}{7}&\cos\frac{6\pi}{7}\end{bmatrix}\begin{bmatrix}\cos\frac{2\pi}{7}&-\sin\frac{2\pi}{7}\\\sin\frac{2\pi}{7}&\cos\frac{2\pi}{7}\end{bmatrix}$
$\Rightarrow\text{A}^7=\begin{bmatrix}\cos\frac{14\pi}{7}&-\sin\frac{14\pi}{7}\\\sin\frac{14\pi}{7}&\cos\frac{14\pi}{7}\end{bmatrix}$
$\Rightarrow\text{A}^7=\begin{bmatrix}\cos2\pi&-\sin2\pi\\\sin2\pi&\cos2\pi\end{bmatrix}$$\begin{bmatrix}\because\ \frac{14\pi}{7}=2\pi\end{bmatrix}$
$=\begin{bmatrix}1&0\\0&1\end{bmatrix}$
So, the least positive integral value of k is 7.

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