\[ \vec{a}_{1}=\left[\begin{array}{l}1 \\ 3\end{array}\right], \vec{a}_{2}=\left[\begin{array}{c}4 \\ 12\end{array}\right], \vec{a}_{3}=\left[\begin{array}{c}7 \\ 21\end{array}\right], \vec{a}_{4}=\left[\begin{array}{c}8 \\ 24\end{array}\right] \] Describe all solutions of \( \vec{a}_{1} x_{1}+\vec{a}_{2} x_{2}+\vec{a}_{3} x_{3}+\vec{a}_{4} x_{4}=\overrightarrow{0} \)

To solve the equation \[ \vec{a}_{1} x_{1} + \vec{a}_{2} x_{2} + \vec{a}_{3} x_{3} + \vec{a}_{4} x_{4} = \overrightarrow{0}, \] we first express the vectors \(\vec{a}_{1}, \vec{a}_{2}, \vec{a}_{3}, \vec{a}_{4}\) in terms of their components: \[ \vec{a}_{1} = \begin{bmatrix} 1 \\ 3 \end{bmatrix}, \quad \vec{a}_{2} = \begin{bmatrix} 4 \\ 12 \end{bmatrix}, \quad \vec{a}_{3} = \begin{bmatrix} 7 \\ 21 \end{bmatrix}, \quad \vec{a}_{4} = \begin{bmatrix} 8 \\ 24 \end{bmatrix}. \] Next, we can observe that each vector is a scalar multiple of \(\vec{a}_{1}\): - \(\vec{a}_{2} = 4 \vec{a}_{1}\) - \(\vec{a}_{3} = 7 \vec{a}_{1}\) - \(\vec{a}_{4} = 8 \vec{a}_{1}\) This means that all vectors \(\vec{a}_{1}, \vec{a}_{2}, \vec{a}_{3}, \vec{a}_{4}\) are linearly dependent, and they span a one-dimensional space (the line through \(\vec{a}_{1}\)). We can rewrite the equation as: \[ x_{1} \begin{bmatrix} 1 \\ 3 \end{bmatrix} + x_{2} \begin{bmatrix} 4 \\ 12 \end{bmatrix} + x_{3} \begin{bmatrix} 7 \\ 21 \end{bmatrix} + x_{4} \begin{bmatrix} 8 \\ 24 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}. \] This can be simplified to: \[ \begin{bmatrix} x_{1} + 4x_{2} + 7x_{3} + 8x_{4} \\ 3x_{1} + 12x_{2} + 21x_{3} + 24x_{4} \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}. \] From the first component, we have: \[ x_{1} + 4x_{2} + 7x_{3} + 8x_{4} = 0. \] From the second component, we have: \[ 3x_{1} + 12x_{2} + 21x_{3} + 24x_{4} = 0. \] Notice that the second equation is just 3 times the first equation, which means it does not provide any new information. Therefore, we only need to solve the first equation. We can express \(x_{1}\) in terms of the other variables: \[ x_{1} = -4x_{2} - 7x_{3} - 8x_{4}. \] Letting \(x_{2}, x_{3}, x_{4}\) be free variables, we can denote them as: \[ x_{2} = t_1, \quad x_{3} = t_2, \quad x_{4} = t_3, \] where \(t_1, t_2, t_3\) are any real numbers. Then we can express \(x_{1}\) as: \[ x_{1} = -4t_1 - 7t_2 - 8t_3. \] Thus, the general solution can be written as: \[ \begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \end{bmatrix} = \begin{bmatrix} -4t_1 - 7t_2 - 8t_3 \\ t_1 \\ t_2 \\ t_3 \end{bmatrix}, \] where \(t_1, t_2, t_3 \in \mathbb{R}\). In conclusion, the solution set is a three-dimensional subspace of \(\mathbb{R}^4\), parameterized by the free variables \(t_1, t_2, t_3\).