1. **Problem Statement:** Determine whether the given pairs of vectors are linearly dependent. 2. **Definition:** Two vectors $\mathbf{u}$ and $\mathbf{v}$ are linearly dependent if there exists a scalar $k$ such that $\mathbf{v} = k\mathbf{u}$ or $\mathbf{u} = k\mathbf{v}$. --- ### 4.17 Vector pairs: **(a)** $\mathbf{u} = (1, 2)$, $\mathbf{v} = (3, -5)$ 3. Check if $\mathbf{v} = k\mathbf{u}$ for some $k$: $$ \frac{3}{1} = 3, \quad \frac{-5}{2} = -2.5 $$ Since $3 \neq -2.5$, no single scalar $k$ satisfies both components. **Answer:** Not linearly dependent. **(b)** $\mathbf{u} = (1, -3)$, $\mathbf{v} = (-2, 6)$ 4. Check scalar $k$: $$ \frac{-2}{1} = -2, \quad \frac{6}{-3} = -2 $$ Both equal $-2$, so $\mathbf{v} = -2\mathbf{u}$. **Answer:** Linearly dependent. **(c)** $\mathbf{u} = (1, 2, -3)$, $\mathbf{v} = (4, 5, -6)$ 5. Check scalar $k$: $$ \frac{4}{1} = 4, \quad \frac{5}{2} = 2.5, \quad \frac{-6}{-3} = 2 $$ Since $4 \neq 2.5 \neq 2$, no scalar $k$ satisfies all components. **Answer:** Not linearly dependent. **(d)** $\mathbf{u} = (2, 4, -8)$, $\mathbf{v} = (3, 6, -12)$ 6. Check scalar $k$: $$ \frac{3}{2} = 1.5, \quad \frac{6}{4} = 1.5, \quad \frac{-12}{-8} = 1.5 $$ All equal $1.5$, so $\mathbf{v} = \frac{3}{2} \mathbf{u}$. **Answer:** Linearly dependent. --- ### 4.18 Vector pairs: **(a)** $u = 2t^2 + 4t - 3$, $w = 4t^2 + 8t - 6$ 7. Check if $w = k u$: Compare coefficients: $$ \frac{4}{2} = 2, \quad \frac{8}{4} = 2, \quad \frac{-6}{-3} = 2 $$ All equal 2, so $w = 2u$. **Answer:** Linearly dependent. **(b)** $u = 2t^2 - 3t + 4$, $v = 4t^2 - 3t + 2$ 8. Check scalar $k$: $$ \frac{4}{2} = 2, \quad \frac{-3}{-3} = 1, \quad \frac{2}{4} = 0.5 $$ Since coefficients differ, no scalar $k$ exists. **Answer:** Not linearly dependent. **(c)** Matrices: $$ \mathbf{u} = \begin{bmatrix}1 & 3 & -4 \\ 5 & 0 & -1\end{bmatrix}, \quad \mathbf{v} = \begin{bmatrix}-4 & -12 & 16 \\ -20 & 0 & 4\end{bmatrix} $$ 9. Check scalar $k$ for each element: $$ \frac{-4}{1} = -4, \quad \frac{-12}{3} = -4, \quad \frac{16}{-4} = -4 $$ $$ \frac{-20}{5} = -4, \quad \frac{0}{0} \text{ (undefined but consistent)}, \quad \frac{4}{-1} = -4 $$ All defined ratios equal $-4$, so $\mathbf{v} = -4 \mathbf{u}$. **Answer:** Linearly dependent. **(d)** Matrices: $$ \mathbf{u} = \begin{bmatrix}1 & 1 & 1 \\ 2 & 2 & 2\end{bmatrix}, \quad \mathbf{v} = \begin{bmatrix}2 & 2 & 2 \\ 3 & 3 & 3\end{bmatrix} $$ 10. Check scalar $k$: $$ \frac{2}{1} = 2, \quad \frac{3}{2} = 1.5 $$ Since $2 \neq 1.5$, no scalar $k$ exists. **Answer:** Not linearly dependent. --- **Summary:** - 4.17: (a) No, (b) Yes, (c) No, (d) Yes - 4.18: (a) Yes, (b) No, (c) Yes, (d) No