Problem 25
Question
If \(A\) is an orthogonal matrix, then \(A^{-1}\) equals a. \(A^{T}\) b. \(A\) c. \(A^{2}\) d. none of these
Step-by-Step Solution
Verified Answer
The answer is (a) \( A^{T} \).
1Step 1: Understanding Orthogonal Matrices
A matrix is orthogonal if its columns are orthonormal vectors. A fundamental property of orthogonal matrices is that the transpose of the matrix is equal to its inverse. We need to establish this property mathematically.
2Step 2: Defining Orthogonal Matrix Property
If matrix \( A \) is orthogonal, then \( A^T A = I \), where \( I \) is the identity matrix. This relationship implies that the inverse of \( A \) must satisfy \( A A^{-1} = I \).
3Step 3: Applying Transpose Property
Given the property \( A^T A = I \), we can see that multiplying \( A \) from the right gives \( A^T A = A^T \), which implies \( A^{-1} = A^T \).
4Step 4: Choosing the Correct Option
Since we have established that \( A^{-1} = A^T \), option (a) \( A^{T} \) is the correct answer. It aligns with the property that the inverse of an orthogonal matrix is its transpose.
Key Concepts
Matrix TransposeInverse MatrixOrthonormal VectorsIdentity Matrix
Matrix Transpose
The transpose of a matrix is an important concept in linear algebra. When you transpose a matrix, you essentially "flip" it over its diagonal. This means the rows of the original matrix become the columns of the transposed matrix, and vice versa. If a matrix is denoted by \( A \), its transpose is written as \( A^T \).
Here’s a simple example:
Here’s a simple example:
- For a matrix \( A = \begin{bmatrix} 1 & 2 \ 3 & 4 \end{bmatrix} \), the transpose would be \( A^T = \begin{bmatrix} 1 & 3 \ 2 & 4 \end{bmatrix} \).
Inverse Matrix
A matrix's inverse can be thought of as the "undoing" of a matrix operation. If you have a matrix \( A \), its inverse is denoted as \( A^{-1} \). The relationship between a matrix and its inverse is such that when you multiply them together, you get the identity matrix \( I \). Mathematically, this is expressed as \( A A^{-1} = A^{-1} A = I \).
Not every matrix has an inverse. A matrix must be square (same number of rows and columns) and have a non-zero determinant for its inverse to exist. Orthogonal matrices are a special case where the inverse has a unique property. For orthogonal matrices, the inverse is the same as the transpose. This simplifies many calculations because you can substitute the transpose straight away.
Ultimately, understanding inverses is key to solving many matrix equations. They allow you to reverse or manipulate matrix transformations in mathematical and physical applications.
Not every matrix has an inverse. A matrix must be square (same number of rows and columns) and have a non-zero determinant for its inverse to exist. Orthogonal matrices are a special case where the inverse has a unique property. For orthogonal matrices, the inverse is the same as the transpose. This simplifies many calculations because you can substitute the transpose straight away.
Ultimately, understanding inverses is key to solving many matrix equations. They allow you to reverse or manipulate matrix transformations in mathematical and physical applications.
Orthonormal Vectors
Orthonormal vectors are the building blocks of orthogonal matrices. When we say vectors are orthonormal, we mean they have two specific properties: they are orthogonal to each other, and each vector has a magnitude of 1.
Orthogonal means that the vectors are at right angles (90 degrees) to each other. If \( \mathbf{u} \) and \( \mathbf{v} \) are two vectors, they are orthogonal if their dot product is zero: \( \mathbf{u} \cdot \mathbf{v} = 0 \). Normalization to a magnitude of 1 is achieved by dividing each vector by its magnitude.
Orthogonal means that the vectors are at right angles (90 degrees) to each other. If \( \mathbf{u} \) and \( \mathbf{v} \) are two vectors, they are orthogonal if their dot product is zero: \( \mathbf{u} \cdot \mathbf{v} = 0 \). Normalization to a magnitude of 1 is achieved by dividing each vector by its magnitude.
- Example: If \( \mathbf{u} = \begin{bmatrix} 3 \ 4 \end{bmatrix} \), its magnitude is 5. The normalized vector is \( \mathbf{u} = \begin{bmatrix} \frac{3}{5} \ \frac{4}{5} \end{bmatrix} \).
Identity Matrix
The identity matrix is like the number 1 in matrix operations. It is a square matrix with ones on the main diagonal and zeros elsewhere. For any matrix \( A \), when you multiply it by an identity matrix \( I \), the result is the original matrix. This is expressed as \( AI = IA = A \).
In the 2x2 case, the identity matrix looks like this:
Overall, the identity matrix is a key concept in understanding matrix algebra and how transformations interact with each other.
In the 2x2 case, the identity matrix looks like this:
- \( I = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} \)
Overall, the identity matrix is a key concept in understanding matrix algebra and how transformations interact with each other.
Other exercises in this chapter
Problem 23
If \(D_{1}\) and \(D_{2}\) are two \(3 \times 3\) diagonal matrices, then which of the following is/are true? a. \(D_{1} D_{2}\) is diagonal matrix b. \(D_{1} D
View solution Problem 24
If \(A\) and \(B\) are symmetric and commute, then which of the following is/are symmetric? a. \(A^{-1} B\) b. \(A B^{-1}\) c. \(A^{-1} B^{-1}\) d. None of thes
View solution Problem 25
A skew-symmetric matrix \(A\) satisfies the relation \(A^{2}+I=O\), where \(I\) is a unit matrix then \(A\) is a. idempotent b. orthogonal c. of even order d. o
View solution Problem 26
If \(Z\) is an idempotent matrix, th?n \((I+Z)^{n}\) a. \(I+2^{n} Z\) b. \(I+\left(2^{n}-1\right) Z\) c. \(I-\left(2^{\prime \prime}-1\right) Z\) d. none of the
View solution