2. Let \( A=\left(\begin{array}{lll}2 & 4 & 6 \\ 1 & 3 & 5 \\ 0 & 1 & 2\end{array}\right) \) (a) Calculate the transpose of matrix \( A \), denoted as \( A^{T} \). (b) Multiply matrix \( A \) by the scalar 3 to find \( 3 A \). (c) Verify if \( A \) is a symmetric matrix.

Let's go through each part of the problem step by step. ### (a) Calculate the transpose of matrix \( A \), denoted as \( A^{T} \). The transpose of a matrix is obtained by swapping its rows and columns. For the given matrix \( A \): \[ A = \begin{pmatrix} 2 & 4 & 6 \\ 1 & 3 & 5 \\ 0 & 1 & 2 \end{pmatrix} \] The transpose \( A^{T} \) is: \[ A^{T} = \begin{pmatrix} 2 & 1 & 0 \\ 4 & 3 & 1 \\ 6 & 5 & 2 \end{pmatrix} \] ### (b) Multiply matrix \( A \) by the scalar 3 to find \( 3 A \). To multiply a matrix by a scalar, we multiply each element of the matrix by that scalar. Thus, for \( 3A \): \[ 3A = 3 \cdot \begin{pmatrix} 2 & 4 & 6 \\ 1 & 3 & 5 \\ 0 & 1 & 2 \end{pmatrix} = \begin{pmatrix} 3 \cdot 2 & 3 \cdot 4 & 3 \cdot 6 \\ 3 \cdot 1 & 3 \cdot 3 & 3 \cdot 5 \\ 3 \cdot 0 & 3 \cdot 1 & 3 \cdot 2 \end{pmatrix} \] Calculating each element: \[ 3A = \begin{pmatrix} 6 & 12 & 18 \\ 3 & 9 & 15 \\ 0 & 3 & 6 \end{pmatrix} \] ### (c) Verify if \( A \) is a symmetric matrix. A matrix is symmetric if it is equal to its transpose, i.e., \( A = A^{T} \). From our calculations: \[ A = \begin{pmatrix} 2 & 4 & 6 \\ 1 & 3 & 5 \\ 0 & 1 & 2 \end{pmatrix} \] \[ A^{T} = \begin{pmatrix} 2 & 1 & 0 \\ 4 & 3 & 1 \\ 6 & 5 & 2 \end{pmatrix} \] Since \( A \neq A^{T} \), we conclude that \( A \) is **not** a symmetric matrix. ### Summary of Results: - \( A^{T} = \begin{pmatrix} 2 & 1 & 0 \\ 4 & 3 & 1 \\ 6 & 5 & 2 \end{pmatrix} \) - \( 3A = \begin{pmatrix} 6 & 12 & 18 \\ 3 & 9 & 15 \\ 0 & 3 & 6 \end{pmatrix} \) - \( A \) is not a symmetric matrix.