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  • If A = [cos alpha sin x] and A-1 = A’, find value of α.

If A= \begin{bmatrix} \cos \alpha &\sin \alpha \\-\sin \alpha &\cos \alpha \end{bmatrix} and A^{-1} = A', find value of \alpha

Answers (1)

Given,A= \begin{bmatrix} \cos \alpha &\sin \alpha \\-\sin \alpha &\cos \alpha \end{bmatrix}

We know that in transpose of a matrix, the rows of the matrix become the columns.

\begin{aligned} &\therefore \mathrm{A}^{\prime}=\left[\begin{array}{cc} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{array}\right]^{\mathrm{T}}=\left[\begin{array}{cc} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{array}\right]\\ &\text { Inverse of a matrix }\\ &A=A^{-1}=\frac{\operatorname{adj}(A)}{|A|}\\ &\text { Clearly }|\mathrm{A}|=\left|\begin{array}{cc} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{array}\right|\\ &\therefore|\mathrm{A}|=\cos ^{2} \alpha+\sin ^{2} \alpha=1_{\{\mathrm{using} \text { trigonometric identity }} \end{aligned}

Adj(A) is given by the transpose of the cofactor matrix.

\\\therefore \operatorname{adj}(\mathrm{A})=\left[\begin{array}{cc}\cos \alpha & -(-\sin \alpha) \\ -\sin \alpha & \cos \alpha\end{array}\right]^{\mathrm{T}}=\left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha\end{array}\right]$ \\$\therefore \mathrm{A}^{-1}=\frac{\operatorname{adj}(\mathrm{A})}{|\mathrm{A}|}=1\left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha\end{array}\right]=\left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha\end{array}\right]$
According to question:
\\A^{\prime}=A^{-1}$ \\$\therefore\left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha\end{array}\right]=\left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha\end{array}\right]$
since both the matrices are equal irrespective of the value of $\alpha$.
$\therefore \alpha$  can be any real number

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