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Thread: matrices problem

  1. #1
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    matrices problem

    Question:

    The transpose of a matrix $\displaystyle M=\begin{pmatrix} a & b \\ c & d \end{pmatrix}$ is the matrix $\displaystyle M^{T}=\begin{pmatrix} a & c \\ b & d \end{pmatrix} $ and M is said to be orthogonal when $\displaystyle M^{T}M=I$, where I is the unit matrix. Given that the matrix

    $\displaystyle N=\begin{pmatrix} \frac{2}{\sqrt{5}} &\frac{1}{\sqrt{5}} \\ -\frac{1}{\sqrt{5}} & k \end{pmatrix} $

    is orthogonal, find the value of k. Describe geometrically the transformation of x-y plane which is represented by N. Under a transformation S of the real plane into a itself, a point $\displaystyle P = (x,y) $ is mapped onto the point $\displaystyle S(P) = (ax+by, cx+dy) $. Show that when M is orthogonal, the distance between any two points P and Q is the same as the distance between their images S(P) and S(Q).

    My attempt:



    $\displaystyle N=\begin{pmatrix} \frac{2}{\sqrt{5}} &\frac{1}{\sqrt{5}} \\ -\frac{1}{\sqrt{5}} & k \end{pmatrix} $

    $\displaystyle N^{T}=\begin{pmatrix} \frac{2}{\sqrt{5}} &-\frac{1}{\sqrt{5}} \\ \frac{1}{\sqrt{5}} & k \end{pmatrix} $



    $\displaystyle N^{T}N=I$

    $\displaystyle \begin{pmatrix} \frac{2}{\sqrt{5}} &\frac{1}{\sqrt{5}} \\ -\frac{1}{\sqrt{5}} & k \end{pmatrix}\begin{pmatrix} \frac{2}{\sqrt{5}} &-\frac{1}{\sqrt{5}} \\ \frac{1}{\sqrt{5}} & k \end{pmatrix} = \begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix} $

    $\displaystyle \begin{pmatrix} 5 &\frac{2}{5}- \frac{k}{\sqrt{5}} \\ \frac{2}{5}- \frac{k}{\sqrt{5}} & k^{2} - \frac{1}{\sqrt{5}} \end{pmatrix}= \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} $

    $\displaystyle \frac{2}{5}- \frac{k}{\sqrt{5}} =0 $

    $\displaystyle k = \frac{2\sqrt{5}}{5} $



    The transformation represented by N is a rotation by angle 26.6 (degrees) clockwise



    I am having difficulty answering the last part of the question

    Please I would like some help with the last part of the question

    $\begin{pmatrix}a &b \\ c & d\end{pmatrix}\begin{pmatrix}x\\ y\end{pmatrix} =\begin{pmatrix}ax+by\\ cx+dy \end{pmatrix}$
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  2. #2
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    Re: matrices problem

    If the absolute value of the determinant of the transformation is 1 then distances are preserved.

    $M^T M = I$

    $|M^T M|= |M^T||M| = |M|^2 = |I| = 1$

    $|M|=\pm 1$

    $|~|M|~|=1$
    Thanks from bigmansouf
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