1. **Problem Statement:** Find the value(s) of $h$ for which the vectors in each set are linearly dependent. Vectors are linearly dependent if the determinant of the matrix formed by placing them as columns is zero. --- **11. Vectors:** $$\mathbf{v_1} = \begin{bmatrix}1 \\ -1 \\ 4\end{bmatrix}, \mathbf{v_2} = \begin{bmatrix}3 \\ -5 \\ 7\end{bmatrix}, \mathbf{v_3} = \begin{bmatrix}-1 \\ 5 \\ h\end{bmatrix}$$ Form matrix $A = [\mathbf{v_1} \ \mathbf{v_2} \ \mathbf{v_3}]$: $$A = \begin{bmatrix}1 & 3 & -1 \\ -1 & -5 & 5 \\ 4 & 7 & h \end{bmatrix}$$ Calculate $\det(A)$: $$\det(A) = 1 \cdot \begin{vmatrix} -5 & 5 \\ 7 & h \end{vmatrix} - 3 \cdot \begin{vmatrix} -1 & 5 \\ 4 & h \end{vmatrix} + (-1) \cdot \begin{vmatrix} -1 & -5 \\ 4 & 7 \end{vmatrix}$$ Calculate minors: $$= 1((-5)(h) - 5(7)) - 3((-1)(h) - 5(4)) - 1((-1)(7) - (-5)(4))$$ $$= 1(-5h - 35) - 3(-h - 20) - 1(-7 + 20)$$ $$= -5h - 35 + 3h + 60 - 13$$ $$= (-5h + 3h) + (-35 + 60 - 13) = -2h + 12$$ Set determinant to zero for linear dependence: $$-2h + 12 = 0 \implies h = 6$$ --- **12. Vectors:** $$\mathbf{v_1} = \begin{bmatrix}2 \\ -4 \\ 1\end{bmatrix}, \mathbf{v_2} = \begin{bmatrix}-6 \\ 7 \\ -3\end{bmatrix}, \mathbf{v_3} = \begin{bmatrix}8 \\ h \\ 4\end{bmatrix}$$ Matrix: $$A = \begin{bmatrix}2 & -6 & 8 \\ -4 & 7 & h \\ 1 & -3 & 4 \end{bmatrix}$$ Calculate $\det(A)$: $$= 2 \cdot \begin{vmatrix}7 & h \\ -3 & 4 \end{vmatrix} - (-6) \cdot \begin{vmatrix} -4 & h \\ 1 & 4 \end{vmatrix} + 8 \cdot \begin{vmatrix} -4 & 7 \\ 1 & -3 \end{vmatrix}$$ Calculate minors: $$= 2(7 \cdot 4 - h \cdot (-3)) + 6(-4 \cdot 4 - h \cdot 1) + 8(-4 \cdot (-3) - 7 \cdot 1)$$ $$= 2(28 + 3h) + 6(-16 - h) + 8(12 - 7)$$ $$= 2(28 + 3h) + 6(-16 - h) + 8(5)$$ $$= 56 + 6h - 96 - 6h + 40$$ $$= (56 - 96 + 40) + (6h - 6h) = 0$$ Since determinant is zero for all $h$, vectors are always linearly dependent. --- **13. Vectors:** $$\mathbf{v_1} = \begin{bmatrix}1 \\ 5 \\ -3\end{bmatrix}, \mathbf{v_2} = \begin{bmatrix}-2 \\ -9 \\ 6\end{bmatrix}, \mathbf{v_3} = \begin{bmatrix}3 \\ h \\ -9\end{bmatrix}$$ Matrix: $$A = \begin{bmatrix}1 & -2 & 3 \\ 5 & -9 & h \\ -3 & 6 & -9 \end{bmatrix}$$ Calculate $\det(A)$: $$= 1 \cdot \begin{vmatrix} -9 & h \\ 6 & -9 \end{vmatrix} - (-2) \cdot \begin{vmatrix} 5 & h \\ -3 & -9 \end{vmatrix} + 3 \cdot \begin{vmatrix} 5 & -9 \\ -3 & 6 \end{vmatrix}$$ Calculate minors: $$= 1((-9)(-9) - h \cdot 6) + 2(5 \cdot (-9) - h \cdot (-3)) + 3(5 \cdot 6 - (-9)(-3))$$ $$= 1(81 - 6h) + 2(-45 + 3h) + 3(30 - 27)$$ $$= 81 - 6h - 90 + 6h + 9$$ $$= (81 - 90 + 9) + (-6h + 6h) = 0$$ Determinant is zero for all $h$, so vectors are always linearly dependent. --- **14. Vectors:** $$\mathbf{v_1} = \begin{bmatrix}1 \\ -3 \\ 4\end{bmatrix}, \mathbf{v_2} = \begin{bmatrix}-6 \\ 8 \\ 7\end{bmatrix}, \mathbf{v_3} = \begin{bmatrix}4 \\ -2 \\ h\end{bmatrix}$$ Matrix: $$A = \begin{bmatrix}1 & -6 & 4 \\ -3 & 8 & -2 \\ 4 & 7 & h \end{bmatrix}$$ Calculate $\det(A)$: $$= 1 \cdot \begin{vmatrix} 8 & -2 \\ 7 & h \end{vmatrix} - (-6) \cdot \begin{vmatrix} -3 & -2 \\ 4 & h \end{vmatrix} + 4 \cdot \begin{vmatrix} -3 & 8 \\ 4 & 7 \end{vmatrix}$$ Calculate minors: $$= 1(8h - (-2)(7)) + 6((-3)h - (-2)(4)) + 4((-3)(7) - 8(4))$$ $$= 1(8h + 14) + 6(-3h + 8) + 4(-21 - 32)$$ $$= 8h + 14 + 6(-3h + 8) + 4(-53)$$ $$= 8h + 14 - 18h + 48 - 212$$ $$= (8h - 18h) + (14 + 48 - 212) = -10h - 150$$ Set determinant to zero: $$-10h - 150 = 0 \implies -10h = 150 \implies h = -15$$ --- **Final answers:** - 11: $h = 6$ - 12: All $h$ (always dependent) - 13: All $h$ (always dependent) - 14: $h = -15$