If A = \left[\begin{array}{ccc}cosx&-sinx\\sinx&cosx\end{array}\right], then show that (A^{-1} )^{-1}

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If A = \left[\begin{array}{ccc}cosx&-sinx\\sinx&cosx\end{array}\right], then show that (A^{-1} )^{-1}

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Phúc Điền 4 years 2021-08-09T06:20:12+00:00 1 Answers 13 views 0

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  1. thanhcong
    0
    2021-08-09T06:21:57+00:00
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    Given:

    The matrix is:

    A=\begin{bmatrix}\cos x&-\sin x\\\sin x&\cos x\end{bmatrix}

    To show:

    (A^{-1})^{-1}=A

    Solution:

    If a matrix is:

    M=\begin{bmatrix}a&b\\c&d\end{bmatrix}

    Then,

    M^{-1}=\dfrac{1}{ad-bc}\begin{bmatrix}d&-b\\-c&a\end{bmatrix}

    We have,

    A=\begin{bmatrix}\cos x&-\sin x\\\sin x&\cos x\end{bmatrix}

    Using the above formula, we get

    A^{-1}=\dfrac{1}{(\cos x)(\cos x)-(-\sin x)(\sin x)}\begin{bmatrix}\cos x&\sin x\\-\sin x&\cos x\end{bmatrix}

    A^{-1}=\dfrac{1}{\cos^2x+\sin^2x}\begin{bmatrix}\cos x&\sin x\\-\sin x&\cos x\end{bmatrix}

    A^{-1}=\dfrac{1}{1}\begin{bmatrix}\cos x&\sin x\\-\sin x&\cos x\end{bmatrix}

    A^{-1}=\begin{bmatrix}\cos x&\sin x\\-\sin x&\cos x\end{bmatrix}

    Now, the inverse of A^{-1} is:

    (A^{-1})^{-1}=\dfrac{1}{(\cos x)(\cos x)-(\sin x)(-\sin x)}\begin{bmatrix}\cos x&-\sin x\\\sin x&\cos x\end{bmatrix}

    (A^{-1})^{-1}=\dfrac{1}{\cos^2x+\sin^2x}\begin{bmatrix}\cos x&-\sin x\\\sin x&\cos x\end{bmatrix}

    (A^{-1})^{-1}=\dfrac{1}{1}A

    (A^{-1})^{-1}=A

    Hence proved.

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