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Determinants — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDeterminants3 Marks
Question
If $\text{A}=\begin{bmatrix}\cos\theta&\sin\theta\\-\sin\theta&\cos\theta \end{bmatrix},$ then for any natural number, find the value of Det $(A^n)$.
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Answer

Let $\text{A}=\begin{bmatrix}\cos\theta&\sin\theta\\-\sin\theta&\cos\theta \end{bmatrix}$
Then, $\text{A}^2=\begin{bmatrix}\cos\theta&\sin\theta\\-\sin\theta&\cos\theta \end{bmatrix}\begin{bmatrix}\cos\theta&\sin\theta\\-\sin\theta&\cos\theta \end{bmatrix}$
$=\begin{bmatrix}\cos^2\theta-\sin^2\theta&\cos\theta\sin\theta+\sin\theta\cos\theta\\-\sin\theta\cos\theta-\cos\theta\sin\theta&-\sin^2\theta+\cos^2\theta \end{bmatrix}$
$=\begin{bmatrix}\cos2\theta&\sin2\theta\\-\sin2\theta&\cos2\theta \end{bmatrix}$
Similarly, $\text{A}^{\text{n}}=\begin{bmatrix}\cos(\text{n}\theta)&\sin(\text{n}\theta)\\-\sin(\text{n}\theta)&\cos(\text{n}\theta) \end{bmatrix}$
Therefore,
$|\text{A}^{\text{n}}|=\begin{vmatrix}\cos(\text{n}\theta)&\sin(\text{n}\theta)\\-\sin(\text{n}\theta)&\cos(\text{n}\theta) \end{vmatrix}$
$=\cos^2(\text{n}\theta)+\sin^2(\text{n}\theta)$
$=1$
Hence, det $(A^n) = 1$

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