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Mathematics · Ch 1 — Applications of Matrices and Determinants

Application of Matrices to Geometry

1.2.4

Application of Matrices to Geometry

Matrices give a compact way to describe geometric transformations of the plane, and one of the most important is the rotation of the axes. (In the Tamil Nadu Samacheer Kalvi Class 12 Maths syllabus this topic follows the same NCERT/CBSE curriculum treatment of matrices and orthogonal transformations.)

Let OO be the origin and let x′Oxx'Ox and y′Oyy'Oy be the original xx- and yy-axes. A point PP has coordinates (x,y)(x,y) in this system. Now rotate both axes about OO through an angle θ\theta to obtain the new axes X′OXX'OX and Y′OYY'OY; with respect to these new axes the same point PP has coordinates (X,Y)(X,Y).

Figure 1.1Rotation of the coordinate axes through an angle $\theta$ (old axes $x'Ox,y'Oy$ to new axes $X'OX,Y'OY$)
Fig. 1.1 — Rotation of the coordinate axes through an angle $\theta$ (old axes $x'Ox,y'Oy$ to new axes $X'OX,Y'OY$)

Drawn by us to help you understand the concept clearly, and verified to make sure it's accurate. For exams, practice from your textbook's own diagram.

What this figure shows. The original axes x′Oxx'Ox and y′Oyy'Oy and the new axes X′OXX'OX, Y′OYY'OY obtained by rotating about the origin OO through an angle θ\theta. A point PP is dropped perpendicularly onto both systems; the construction points L,M,N,T,QL,M,N,T,Q give the projections used to derive x=Xcos⁡θ−Ysin⁡θx=X\cos\theta-Y\sin\theta and $y …

Reading the perpendicular projections in the figure gives

x=Xcos⁡θ−Ysin⁡θ,y=Xsin⁡θ+Ycos⁡θ.x = X\cos\theta - Y\sin\theta,\qquad y = X\sin\theta + Y\cos\theta.

In matrix form,

[xy]=[cos⁡θ−sin⁡θsin⁡θcos⁡θ][XY].\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix} \begin{bmatrix} X \\ Y \end{bmatrix}.

The matrix W=[cos⁡θ−sin⁡θsin⁡θcos⁡θ]W=\begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix} is the rotation matrix. Its determinant is

∣W∣=cos⁡2θ+sin⁡2θ=1,|W| = \cos^2\theta + \sin^2\theta = 1,

so WW is non-singular and invertible, with

W−1=[cos⁡θsin⁡θ−sin⁡θcos⁡θ]=WT.W^{-1} = \begin{bmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{bmatrix} = W^{T}.

Hence the reverse transformation is

X=xcos⁡θ+ysin⁡θ,Y=−xsin⁡θ+ycos⁡θ.X = x\cos\theta + y\sin\theta,\qquad Y = -x\sin\theta + y\cos\theta. …

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