91Ó°ÊÓ

A symmetric matrix is a square matrix that is equal to its transpose. This means that the elements are mirrored across the main diagonal. To check if a matrix is symmetric, verify that each element at position \(i,j\) is the same as the element at position \(j,i\). For instance, if we have matrix \(A\), check that \(A_{ij} = A_{ji}\) for all \(i, j\).

Here's an example:
\[ \begin{array}{ccc} 1 & 2 & 3 \ 2 & 4 & 5 \ 3 & 5 & 6 \ \end{array} \]
This matrix is symmetric because it is identical to its transpose:
\[ \begin{array}{ccc} 1 & 2 & 3 \ 2 & 4 & 5 \ 3 & 5 & 6 \ \end{array} \]
One of the matrices provided in the problem that turned out to be symmetric was matrix (b):
\[ \begin{array}{rrr} 1 & 2 & -3 \ 2 & 1 & 4 \ -3 & 4 & 7 \ \end{array} \].

A skew-symmetric matrix is a square matrix that is equal to the negative of its transpose. This means for a matrix \(A\) to be skew-symmetric, \(A = -A^T\). Therefore, for all elements, the relationship \(A_{ij} = -A_{ji}\) must hold true. Additionally, all diagonal elements of a skew-symmetric matrix are zero because \(A_{ii} = -A_{ii}\) implies \(A_{ii} = 0\).

Consider the following example:
\[ \begin{array}{ccc} 0 & -2 & -3 \ 2 & 0 & -4 \ 3 & 4 & 0 \ \end{array} \]
To determine if it's skew-symmetric, we transpose and negate the matrix:
\[ \begin{array}{ccc} 0 & 2 & 3 \ -2 & 0 & 4 \ -3 & -4 & 0 \ \end{array} \]
Now, by negating:
\[ \begin{array}{ccc} 0 & -2 & -3 \ 2 & 0 & -4 \ 3 & 4 & 0 \ \end{array} \]
Since we get the original matrix, it's verified that this matrix is skew-symmetric. From the problem, matrix (c) was confirmed to be skew-symmetric.

An orthogonal matrix is a square matrix whose transpose is equal to its inverse. Mathematically, a matrix \(A\) is orthogonal if and only if \(A^T \times A = I\), where \(I\) is the identity matrix. This property ensures that the matrix preserves the dot product, meaning the lengths and angles in the space remain constant under the transformation represented by the matrix.

Let's verify if a given matrix is orthogonal. Consider:
\[ \begin{array}{ccc} 1 & 0 & 0 \ 0 & \frac{1}{\text{√2}} & -\frac{1}{\text{√2}} \ 0 & \frac{1}{\text{√2}} & \frac{1}{\text{√2}} \ \end{array} \]
First, find its transpose:
\[ \begin{array}{ccc} 1 & 0 & 0 \ 0 & \frac{1}{\text{√2}} & \frac{1}{\text{√2}} \ 0 & -\frac{1}{\text{√2}} & \frac{1}{\text{√2}} \ \end{array} \]
Then, multiply the original matrix by its transpose:
\[ \begin{array}{ccc} 1 & 0 & 0 \ 0 & \frac{1}{\text{√2}} & -\frac{1}{\text{√2}} \ 0 & \frac{1}{\text{√2}} & \frac{1}{\text{√2}} \ \end{array} \times \begin{array}{ccc} 1 & 0 & 0 \ 0 & \frac{1}{\text{√2}} & \frac{1}{\text{√2}} \ 0 & -\frac{1}{\text{√2}} & \frac{1}{\text{√2}} \ \end{array} = \begin{array}{ccc} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \ \end{array} \]
Since the result is the identity matrix, this proves the matrix is orthogonal. In the exercise, matrix (a) is such a case.