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What is the inverse of \[ \begin{pmatrix} 1&0&0\\0&\cos x &\sin x\\ 0 &\sin x &-\cos x \end{pmatrix} \] Please help me to solve the above problem.

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2 Answers 2

Let $A_x :=\left[ \begin{array}{cc} \cos x & \sin x \\ \sin x & -\cos x \end{array} \right]$, and let $R_x :=\left[ \begin{array}{cc} \cos x & -\sin x \\ \sin x & \cos x \end{array} \right]$ be the matrix of the rotation by angle $x$ in the plane (that is, for all ${\bf v}$ in $\mathbb R^2$, $\ R_x\cdot {\bf v}$ is the rotated version of $\bf v$), we have that $$A_x = R_x\cdot \left[ \begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array} \right]\ \text{ and }\ (R_x)^{-1} = R_{-x} = \left[ \begin{array}{cc} \cos x & \sin x \\ -\sin x & \cos x \end{array} \right]\text{, so} $$ $$(A_x)^{-1} = \left[ \begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array} \right]\cdot R_{-x}$$ So, by easy matrix multiplication, one can verify that the additional $1$ in the additional dimension is not hurting much, ie. the requested inverse is: $$ \left[ \begin{array}{ccc} 1&0&0\\ 0 &\cos x & \sin x \\ 0 & \sin x & -\cos x \end{array} \right] $$

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Maybe it is too early in the morning, but should the signs on sin x both be positive (maybe since this is a rotation matrix, it does not matter)? –  Amzoti Sep 19 '12 at 14:32
    
Oh, sorry, it does matter. As the next answer calculates, it is not the same.. –  Berci Sep 19 '12 at 15:34
    
...and we thus see that the given matrix is in fact orthogonal. –  J. M. Sep 20 '12 at 18:25

Implement the formula $\def\adj{\operatorname{adj}}A^{-1}=\frac{1}{\det A}\cdot \adj A$

Find $\det A$

$\det A=\begin{vmatrix} 1&0&0\\0&\cos x &\sin x\\ 0 &\sin x &-\cos x \end{vmatrix}=-1$

Find $\adj A$

$A_{11}=(-1)^{1+1}\left\lvert \begin{array}{cc} \cos x & \sin x \\ \sin x & -\cos x \end{array} \right\rvert=-1$

$A_{12}=(-1)^{1+2}\left\lvert \begin{array}{cc} 0 & \sin x \\ 0 & -\cos x \end{array} \right\rvert=0$

$A_{13}=(-1)^{1+3}\left\lvert \begin{array}{cc} 0 & \cos x \\ 0 & \sin x \end{array} \right\rvert=0$

$A_{21}=(-1)^{2+1}\left\lvert \begin{array}{cc} 0 & 0 \\ \sin x & -\cos x \end{array} \right\rvert=0$

$A_{22}=(-1)^{2+2}\left\lvert \begin{array}{cc} 1 & 0 \\ 0 & -\cos x \end{array} \right\rvert=-\cos x$

$A_{23}=(-1)^{2+3}\left\lvert \begin{array}{cc} 1 & 0 \\ 0 & \sin x \end{array} \right\rvert=-\sin x$

$A_{31}=(-1)^{3+1}\left\lvert \begin{array}{cc} 0 & 0 \\ \cos x & \sin x \end{array} \right\rvert=0$

$A_{32}=(-1)^{3+2}\left\lvert \begin{array}{cc} 1 & 0 \\ 0 & \sin x \end{array} \right\rvert=-\sin x$

$A_{33}=(-1)^{3+3}\left\lvert \begin{array}{cc} 1 & 0 \\ 0 & \cos x \end{array} \right\rvert=\cos x$

$A^{-1}=\left\lvert \begin{array}{ccc} 1&0&0\\ 0 &\cos x & \sin x \\ 0 & \sin x & -\cos x \end{array} \right\rvert$

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