\#1.) Is \( \left[\begin{array}{c}4 \\ -3 \\ 1\end{array}\right] \) an eigenvector of \( \left[\begin{array}{ccc}3 & 7 & 9 \\ -4 & -5 & 1 \\ 2 & 4 & 4\end{array}\right] \) ? If so, find the eigenvalue.

To determine if \( \left[\begin{array}{c}4 \\ -3 \\ 1\end{array}\right] \) is an eigenvector of the matrix \( \left[\begin{array}{ccc}3 & 7 & 9 \\ -4 & -5 & 1 \\ 2 & 4 & 4\end{array}\right] \), we need to check if there exists a scalar \( \lambda \) such that: \[ A \mathbf{v} = \lambda \mathbf{v} \] where \( A = \left[\begin{array}{ccc}3 & 7 & 9 \\ -4 & -5 & 1 \\ 2 & 4 & 4\end{array}\right] \) and \( \mathbf{v} = \left[\begin{array}{c}4 \\ -3 \\ 1\end{array}\right] \). Let's calculate \( A \mathbf{v} \): \[ A \mathbf{v} = \left[\begin{array}{ccc}3 & 7 & 9 \\ -4 & -5 & 1 \\ 2 & 4 & 4\end{array}\right] \left[\begin{array}{c}4 \\ -3 \\ 1\end{array}\right] \] Calculating each component: 1. First row: \( 3(4) + 7(-3) + 9(1) = 12 - 21 + 9 = 0 \) 2. Second row: \( -4(4) + -5(-3) + 1(1) = -16 + 15 + 1 = 0 \) 3. Third row: \( 2(4) + 4(-3) + 4(1) = 8 - 12 + 4 = 0 \) Thus, we have: \[ A \mathbf{v} = \left[\begin{array}{c}0 \\ 0 \\ 0\end{array}\right] \] This means that \( A \mathbf{v} = 0 \cdot \mathbf{v} \). Therefore, \( \lambda = 0 \) and \( \left[\begin{array}{c}4 \\ -3 \\ 1\end{array}\right] \) is indeed an eigenvector of the matrix with eigenvalue 0.