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Q43E

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Found in: Page 74

### Linear Algebra With Applications

Book edition 5th
Author(s) Otto Bretscher
Pages 442 pages
ISBN 9780321796974

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# Use the formula derived in Exercise ${\mathbf{2}}{\mathbf{.}}{\mathbf{1}}{\mathbf{.}}{\mathbf{13}}$ to find the inverse of the rotation matrix localid="1659346816315" ${\mathbf{A}}{\mathbf{=}}\mathbf{\left[}\begin{array}{cc}\mathbf{cos\theta }& \mathbf{-}\mathbf{sin\theta }\\ \mathbf{sin\theta }& \mathbf{cos\theta }\end{array}\mathbf{\right]}$. Interpret the linear transformation defined by ${{\mathbf{A}}}^{\mathbf{-}\mathbf{1}}$geometrically. Explain.

The inverse of the matrix is, ${\mathrm{A}}^{-1}=\left(\begin{array}{cc}\mathrm{cos\theta }& \mathrm{sin\theta }\\ -\mathrm{sin\theta }& \mathrm{cos\theta }\end{array}\right)$ , the geometrical interpretation is,

.

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## Step 1: Compute the inverse of the matrix.

Consider the matrix.

$\mathrm{A}=\left(\begin{array}{cc}\mathrm{cos\theta }& -\mathrm{sin\theta }\\ \mathrm{sin\theta }& \mathrm{cos\theta }\end{array}\right)$

The inverse of the matrix is,

${\mathrm{A}}^{-1}=\frac{1}{\mathrm{cos\theta }×\mathrm{cos\theta }-\mathrm{sin\theta }×-\mathrm{sin}\theta }\left(\begin{array}{cc}\mathrm{cos\theta }& \mathrm{sin\theta }\\ -\mathrm{sin\theta }& \mathrm{cos\theta }\end{array}\right)\phantom{\rule{0ex}{0ex}}⇒{\mathrm{A}}^{-1}=\frac{1}{{\mathrm{cos}}^{2}\theta +{\mathrm{sin}}^{2}\theta }\left(\begin{array}{cc}\mathrm{cos\theta }& \mathrm{sin\theta }\\ -\mathrm{sin\theta }& \mathrm{cos\theta }\end{array}\right)\phantom{\rule{0ex}{0ex}}\therefore {\mathrm{A}}^{-1}=\left(\begin{array}{cc}\mathrm{cos\theta }& \mathrm{sin\theta }\\ -\mathrm{sin\theta }& \mathrm{cos\theta }\end{array}\right)$

## Step 2: Interpret the linear transformation.

The matrix $A$ represents the rotation by an angle $\theta$ and scaling by a factor $r$.

The inverse matrix ${A}^{-1}$ represents the rotation by an angle $-\theta$ and scaling by a factor $\frac{1}{r}$.The geometrical representation is,

## Step 3: Final answer.

The inverse matrix is, ${\mathrm{A}}^{-1}=\left(\begin{array}{cc}\mathrm{cos\theta }& \mathrm{sin\theta }\\ -\mathrm{sin\theta }& \mathrm{cos\theta }\end{array}\right)$ with geometrical interpretation of linear

transformation is,

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