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Algebra of Matrices — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsAlgebra of Matrices5 Marks
Question
If $\text{A}=\begin{bmatrix}\cos2\theta&\sin2\theta\\-\sin2\theta&\cos2\theta\end{bmatrix},$ find $A^2.$
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Answer

Given: $\text{A}=\begin{bmatrix}\cos2\theta&\sin2\theta\\-\sin2\theta&\cos2\theta\end{bmatrix}$
Now,
$\text{A}^2=\text{A.A}$
$=\begin{bmatrix}\cos2\theta&\sin2\theta\\-\sin2\theta&\cos2\theta\end{bmatrix}\begin{bmatrix}\cos2\theta&\sin2\theta\\-\sin2\theta&\cos2\theta\end{bmatrix}$
$=\begin{bmatrix}\cos^22\theta-\sin^22\theta&\cos2\theta\sin^2+\cos2\theta\sin^2\theta\\-\cos2\theta\sin^2\theta-\sin^2\theta\cos^2\theta&-\sin^22\theta+\cos^22\theta\end{bmatrix}$
$=\begin{bmatrix}\cos4\theta&2\sin^2\theta\cos^2\theta\\-2\sin^2\cos2\theta&\cos4\theta\end{bmatrix}$
$\begin{Bmatrix}\text{ since }\cos^2\theta-\sin^2\theta=\cos2\theta\end{Bmatrix}$
$=\begin{bmatrix}\cos4\theta&\sin4\theta\\-\sin4\theta&\cos4\theta\end{bmatrix}$
$\begin{Bmatrix}\text{ since }\sin^2\theta=2\sin\theta\cos\theta\end{Bmatrix}$
Hence,
$\text{A}^2=\begin{bmatrix}\cos4\theta&\sin4\theta\\-\sin4\theta&\cos4\theta\end{bmatrix}$

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