Chapter 5: Problem 20
Determine whether the given matrix is orthogonal. If it is, find its inverse. $$\left[\begin{array}{rrrr}\frac{1}{2} & -\frac{1}{2} & \frac{1}{2} & \frac{1}{2} \\\\\frac{1}{2} & \frac{1}{2} & \frac{1}{2} & -\frac{1}{2} \\ -\frac{1}{2} & \frac{1}{2} & \frac{1}{2} & \frac{1}{2} \\\\\frac{1}{2} & \frac{1}{2} & -\frac{1}{2} & \frac{1}{2}\end{array}\right]$$
Short Answer
Step by step solution
Definition of Orthogonal Matrix
Transpose the Matrix
Calculate the Transpose
Compute \( A^T A \)
Result of \( A^T A \)
Inverse of Orthogonal Matrix
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Transpose of a Matrix
Let's take a look at how this applies to a matrix. If you have a matrix \( A \) with elements arranged in a 4x4 grid, the element \( a_{ij} \) from the original matrix becomes the element \( a_{ji} \) in the transposed matrix. This means the element in the first row and second column of \( A \) becomes the element in the second row and first column of \( A^T \).
For example, take this matrix:
- \( A = \begin{bmatrix} 1 & 2 \ 3 & 4 \end{bmatrix} \)
- \( A^T = \begin{bmatrix} 1 & 3 \ 2 & 4 \end{bmatrix} \)
Identity Matrix
The identity matrix is a special type of square matrix where the diagonal elements are all 1's, and all other elements are 0's. For a 3x3 identity matrix, it looks like this:
- \( I_3 = \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix} \)
- \( A \cdot I_4 = A \)
Matrix Multiplication
For two matrices \( A \) and \( B \), you multiply them to form a product matrix \( C \). The element \( c_{ij} \) in matrix \( C \) is calculated by taking the sum of the products of elements from the \( i \)-th row of \( A \) and the \( j \)-th column of \( B \).
Let's consider two matrices:
- \( A = \begin{bmatrix} 1 & 2 \ 3 & 4 \end{bmatrix} \)
- \( B = \begin{bmatrix} 5 & 6 \ 7 & 8 \end{bmatrix} \)
- \( C_{11} = 1 \cdot 5 + 2 \cdot 7 \)
- \( C_{12} = 1 \cdot 6 + 2 \cdot 8 \)
- \( C_{21} = 3 \cdot 5 + 4 \cdot 7 \)
- \( C_{22} = 3 \cdot 6 + 4 \cdot 8 \)
- \( \begin{bmatrix} 19 & 22 \ 43 & 50 \end{bmatrix} \)