/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 8 Find the inverse of $$ \left... [FREE SOLUTION] | 91Ó°ÊÓ

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Find the inverse of $$ \left[\begin{array}{rr} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right] $$

Short Answer

Expert verified
The inverse matrix is \(\begin{bmatrix}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta\end{bmatrix}\).

Step by step solution

01

Understand the Matrix Form

The given matrix \(\begin{bmatrix}\cos \theta & \sin \theta \-\sin \theta & \cos \theta\end{bmatrix}\) is a 2x2 rotation matrix. Rotation matrices are commonly used to perform rotations in Euclidean space. They have a specific property that their inverse is also a rotation matrix.
02

Recall Property of Rotation Matrices

For any rotation matrix \(A\), its inverse \(A^{-1}\) is the same as its transpose, i.e., \(A^{-1} = A^T\). This property is due to orthogonality, where a matrix times its transpose gives the identity matrix.
03

Transpose the Matrix

The transpose of the given matrix is achieved by switching its rows with columns:\[A^T = \begin{bmatrix}\cos \theta & -\sin \theta \\sin \theta & \cos \theta\end{bmatrix}\]
04

Verify the Inverse Property

To ensure the computed inverse is correct, check that the product of the matrix and its transpose equals the identity matrix:\[\begin{bmatrix}\cos \theta & \sin \theta \-\sin \theta & \cos \theta\end{bmatrix}\begin{bmatrix}\cos \theta & -\sin \theta \\sin \theta & \cos \theta\end{bmatrix}= \begin{bmatrix}1 & 0 \0 & 1\end{bmatrix}\]
05

Conclude the Inverse Matrix

The inverse of the given rotation matrix is:\[\begin{bmatrix}\cos \theta & -\sin \theta \\sin \theta & \cos \theta\end{bmatrix}\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Rotation Matrix
A rotation matrix is a specific kind of transformation matrix used to perform rotations in Euclidean space. These matrices are incredibly useful in areas such as computer graphics and robotics.
The general form for a 2D rotation matrix is given by:
  • \[\begin{bmatrix} \cos \theta & -\sin \theta \ \sin \theta & \cos \theta \end{bmatrix}\]
Here, \(\theta\) represents the angle of rotation. The key property of rotation matrices is that they preserve distances and angles, meaning they don't distort objects. They simply rotate them around the origin in the plane.
Another important feature of rotation matrices is that they are orthogonal. This means their inverse is equal to their transpose, which is a significant simplification when solving matrix equations.
Orthogonality
Orthogonality is a concept indicating that two vectors are perpendicular to each other, and the dot product between them is zero. In terms of matrices, orthogonality implies certain properties that are incredibly useful:
An orthogonal matrix is a square matrix whose rows and columns are orthonormal vectors. This means that each vector has a length of 1 and is perpendicular to every other vector.
The relationship between orthogonality and rotation matrices is particularly special. A rotation matrix is orthogonal, so its inverse is equal to its transpose.
  • For an orthogonal matrix \(A\), we have \(A^T = A^{-1}\).
  • This simplifies many calculations, especially in transforming coordinates from one system to another.
Due to these properties, orthogonal matrices, such as rotation matrices, are very efficient computationally, as their norm remains constant under multiplication by their transpose.
Transpose of a Matrix
The transpose of a matrix is a new matrix obtained by flipping its rows and columns. It is symbolized by \(A^T\) when referring to matrix \(A\).
For the given matrix:
  • \[\begin{bmatrix} \cos \theta & \sin \theta \ -\sin \theta & \cos \theta \end{bmatrix},\]
its transpose is calculated by switching the first row with the first column and the second row with the second column, resulting in:
  • \[\begin{bmatrix} \cos \theta & -\sin \theta \ \sin \theta & \cos \theta \end{bmatrix}\]
Once transposed, the matrix differs only in the sign of the off-diagonal elements, which is crucial in maintaining orthogonality.
The transpose operation is especially valuable because when a matrix is symmetric and orthogonal, like the rotation matrices, it simplifies finding the inverse.
Identity Matrix
An identity matrix, denoted as \(I\), is a special kind of matrix that acts as the multiplicative identity in matrix algebra. In 2D, the identity matrix is:
  • \[\begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}\]
When any matrix is multiplied by the identity matrix, it remains unchanged:
  • \(AI = IA = A\) for any matrix \(A\).
Identity matrices are crucial for verifying matrix operations like finding inverses. For instance, when the product of a matrix and its purported inverse gives the identity matrix, it confirms that the inverse is correct.
In the context of the original exercise, multiplying the rotation matrix by its transpose resulted in the identity matrix, thus verifying the inverse. The identity matrix provides a backbone for matrix operations, ensuring consistency and correctness in transformations.

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Most popular questions from this chapter

Let $$ A=\left[\begin{array}{rrr} 3 & -2 & 7 \\ 6 & 5 & 4 \\ 0 & 4 & 9 \end{array}\right] \text { and } B=\left[\begin{array}{rrr} 6 & -2 & 4 \\ 0 & 1 & 3 \\ 7 & 7 & 5 \end{array}\right] $$ Use the row method or column method (as appropriate (a) the first row of \(A B\) (b) the third row of \(A B\) (c) the second column of \(A B\) (d) the first column of \(B A\). (e) the third row of \(A A\) (f) the third column of \(A A\) Answer: (a) [674141] (b) [6367.57] (c) \(\left[\begin{array}{l}41 \\ 21 \\ 67\end{array}\right]\) \((d)\left[\begin{array}{r}6 \\ 6 \\ 63\end{array}\right]\) (c) [245697] (f) \(\left[\begin{array}{l}76 \\ 98 \\ 97\end{array}\right]\)

(a) If \(A\) is a \(3 \times 5\) matrix, then what is the maximum possible number of leading I's in its reduced row echelon form? (b) If \(B\) is a \(3 \times 6\) matrix whose last column has all zeros, then what is the maximum possible number of parameters in the general solution of the linear system with augmented matrix B? (c) If \(C\) is a \(5 \times 3\) matrix, then what is the minimum possible number of rows of zeros in any row echelon form of

Let 0 denote a \(2 \times 2\) matrix, each of whose entries is zero. (a) Is there a \(2 \times 2\) matrix \(A\) such that \(A \neq 0\) and \(A A=0 ?\) Justify your answer. (b) Is there a \(2 \times 2\) matrix \(A\) such that \(A \neq 0\) and \(A A=A^{2}\) Justify your answer.

Use the inversion algorithm to find the inverse of the given matrix, if the inverse exists. $$\left[\begin{array}{lll} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 1 & 1 & 0 \end{array}\right]$$

If the \(n \times n\) matrix \(A\) can be expressed as \(A=L U\), where \(L\) is a lower triangular matrix and \(U\) is an upper triangular matrix, then the linear system \(A \mathbf{x}=\mathbf{b}\) can be expressed as \(L U \mathbf{x}=\mathbf{b}\) and can be solved in two steps: Step 1. Let \(U_{\mathbf{x}=\mathbf{y}, \text { so that } L U_{\mathbf{x}}=\mathbf{b} \text { can be expressed as } L \mathbf{y}=\mathbf{b . ~ S o l v e ~ t h i s ~ s y s t e m .}}\) Step \(2 .\) Solve the system \(U \mathbf{x}=\mathbf{y}\) for \(\mathbf{x}\) In each part, use this two-step method to solve the given system. (a) \(\left[\begin{array}{rrr}1 & 0 & 0 \\ -2 & 3 & 0 \\ 2 & 4 & 1\end{array}\right]\left[\begin{array}{rrr}2 & -1 & 3 \\ 0 & 1 & 2 \\ 0 & 0 & 4\end{array}\right]\left[\begin{array}{l}x_{1} \\ x_{2} \\\ x_{3}\end{array}\right]=\left[\begin{array}{r}1 \\ -2 \\\ 0\end{array}\right]\) (b) \(\left[\begin{array}{rrr}2 & 0 & 0 \\ 4 & 1 & 0 \\ -3 & -2 & 3\end{array}\right]\left[\begin{array}{rrr}3 & -5 & 2 \\ 0 & 4 & 1 \\ 0 & 0 & 2\end{array}\right]\left[\begin{array}{l}x_{1} \\ x_{2} \\\ x_{3}\end{array}\right]=\left[\begin{array}{r}4 \\ -5 \\\ 2\end{array}\right]\)

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