Question

If $A=\left[\begin{array}{cc}cos\theta & sin\theta \\ -sin\theta & cos\theta \end{array}\right]$
Then prove that $A^m=\left[\begin{array}{cc}cos\theta & sin\theta \\ -sin\theta & cos\theta \end{array}\right]$ where $nϵN$

Answer 1

We shall prove the result by using principle of mathematical induction.
We have P(n): If $A=\left[\begin{array}{cc}cos\theta & sin\theta \\ -sin\theta & cos\theta \end{array}\right],$ then $A^n=\left[\begin{array}{cc}cos\theta & sin\theta \\ -sin\theta & cos\theta \end{array}\right],nϵNP\left(1\right):A=\left[\begin{array}{cc}cos\theta & sin\theta \\ -sin\theta & cos\theta \end{array}\right],$ $So{A}^1=\left[\begin{array}{cc}cos\theta & sin\theta \\ -sin\theta & cos\theta \end{array}\right]$
Therefore, the result is true for n = 1
Let the result be true for n = k. So
$P\left(k\right):A=\left[\begin{array}{cc}cos\theta & sin\theta \\ -sin\theta & cos\theta \end{array}\right]$, $Then{A}^k=\left[\begin{array}{cc}cosk\theta & sink\theta \\ -sink\theta & cosk\theta \end{array}\right]-------\left(1\right)$
Now, we prove that the result holds for n = k + 1
$Now{A}^{k+1}=A.A^k\left[\begin{array}{cc}cos\theta & sin\theta \\ -sin\theta & cos\theta \end{array}\right]\left[\begin{array}{cc}cosk\theta & sink\theta \\ -sink\theta & cosk\theta \end{array}\right]Using\left(\right(1\left)\right)$
$=\left[\begin{array}{cc}cos\theta cosk\theta -sin\theta sin\theta & cos\theta sink+sin\theta cosk\theta \\ -sin\theta cosk\theta +cos\theta sink\theta & -sin\theta sink\theta +cos\theta cosk\theta \end{array}\right]$
$=\left[\begin{array}{cc}cos(\theta +k\theta )& sin(\theta +k\theta )\\ -sin(\theta +k\theta )& cos(\theta +k\theta )\end{array}\right]=\left[\begin{array}{cc}cos(k+1)\theta & sin(k+1)\theta \\ -sin(k+1)\theta & cos(k+1)\theta \end{array}\right]$
Therefore, the result is true n = k + 1
Thus by pricniple of mathematical induction, we have $A^n=\left[\begin{array}{cc}cosn\theta & sin\theta \\ -sinn\theta & cosn\theta \end{array}\right],$ holds for all natural numbers.