1. **State the problem:** Determine if the vectors $\vec{v}_1 = \begin{bmatrix} -16 \\ 8 \\ -4 \end{bmatrix}$, $\vec{v}_2 = \begin{bmatrix} 16 \\ 20 \\ -4 \end{bmatrix}$, and $\vec{v}_3 = \begin{bmatrix} -4 \\ -12 \\ 3 \end{bmatrix}$ are linearly independent or dependent. 2. **Recall the definition:** Vectors are linearly dependent if there exist scalars $c_1, c_2, c_3$, not all zero, such that $$c_1 \vec{v}_1 + c_2 \vec{v}_2 + c_3 \vec{v}_3 = \vec{0}.$$ If the only solution is $c_1 = c_2 = c_3 = 0$, they are linearly independent. 3. **Given equation:** $$\vec{0} = 1 \cdot \vec{v}_1 + 1 \cdot \vec{v}_2 + 0 \cdot \vec{v}_3 = \begin{bmatrix} -16 \\ 8 \\ -4 \end{bmatrix} + \begin{bmatrix} 16 \\ 20 \\ -4 \end{bmatrix} + \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} = \begin{bmatrix} -16 + 16 + 0 \\ 8 + 20 + 0 \\ -4 - 4 + 0 \end{bmatrix} = \begin{bmatrix} 0 \\ 28 \\ -8 \end{bmatrix}.$$ 4. **Check the sum:** The sum is $\begin{bmatrix} 0 \\ 28 \\ -8 \end{bmatrix} \neq \vec{0}$, so the given linear combination does not equal the zero vector. 5. **Interpretation:** Since the provided combination does not produce the zero vector, the example does not show linear dependence. 6. **Test linear dependence by solving:** Set up the equation: $$c_1 \vec{v}_1 + c_2 \vec{v}_2 + c_3 \vec{v}_3 = \vec{0}$$ which is $$c_1 \begin{bmatrix} -16 \\ 8 \\ -4 \end{bmatrix} + c_2 \begin{bmatrix} 16 \\ 20 \\ -4 \end{bmatrix} + c_3 \begin{bmatrix} -4 \\ -12 \\ 3 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}.$$ This gives the system: 1. $-16 c_1 + 16 c_2 - 4 c_3 = 0$ 2. $8 c_1 + 20 c_2 - 12 c_3 = 0$ 3. $-4 c_1 - 4 c_2 + 3 c_3 = 0$ 7. **Solve the system:** From equation 1: $$-16 c_1 + 16 c_2 - 4 c_3 = 0 \implies -4 c_3 = 16 c_1 - 16 c_2 \implies c_3 = \frac{16 c_2 - 16 c_1}{4} = 4 c_2 - 4 c_1.$$ 8. Substitute $c_3 = 4 c_2 - 4 c_1$ into equation 2: $$8 c_1 + 20 c_2 - 12 (4 c_2 - 4 c_1) = 0$$ $$8 c_1 + 20 c_2 - 48 c_2 + 48 c_1 = 0$$ $$ (8 + 48) c_1 + (20 - 48) c_2 = 0$$ $$56 c_1 - 28 c_2 = 0$$ Divide both sides by 28: $$\cancel{28} 2 c_1 - \cancel{28} c_2 = 0 \implies 2 c_1 - c_2 = 0 \implies c_2 = 2 c_1.$$ 9. Substitute $c_2 = 2 c_1$ into expression for $c_3$: $$c_3 = 4 (2 c_1) - 4 c_1 = 8 c_1 - 4 c_1 = 4 c_1.$$ 10. Substitute $c_2 = 2 c_1$ and $c_3 = 4 c_1$ into equation 3: $$-4 c_1 - 4 (2 c_1) + 3 (4 c_1) = 0$$ $$-4 c_1 - 8 c_1 + 12 c_1 = 0$$ $$(-4 - 8 + 12) c_1 = 0$$ $$0 \cdot c_1 = 0$$ This is true for all $c_1$. 11. **Conclusion:** Since $c_1$ can be any scalar, there exist nonzero scalars $c_1, c_2 = 2 c_1, c_3 = 4 c_1$ such that the linear combination equals zero. Therefore, the vectors $\vec{v}_1, \vec{v}_2, \vec{v}_3$ are linearly dependent. **Final answer:** The vectors are linearly dependent because there exist nontrivial scalars satisfying the zero vector equation.