⟶ [ 1 ⟶ [ -1 ⟶ [ 5
u = -3 u2 = 9 u3 = -7 <--- u1, u2, and u3 are vectors
-2] -6] h]
are linearly independent?
x1⃗=(1,−1,5),x2⃗=(−3,9,−7),x3⃗=(−2,−6,h)k1x1⃗+k2x2⃗+k3x3⃗=0⃗k1=k2=k3=0k1(1,−1,5)+k2(−3,9,−7)+k3(−2,−6,h)==(0,0,0)k1−3k2−2k3=0−k1+9k2−6k3=05k1−7k2+hk3=0Δ=∣1−3−2−19−65−7h∣=9h+90−14++90−3h−42=6h+124≠0h≠−1246h≠−623\vec{x_1}=(1,-1,5), \vec{x_2}=(-3,9,-7), \vec{x_3}=(-2,-6,h)\\ k_1\vec{x_1}+k_2\vec{x_2}+k_3\vec{x_3}=\vec{0}\\ k_1=k_2=k_3=0\\ k_1(1,-1,5)+k_2(-3,9,-7)+k_3(-2,-6,h)=\\=(0,0,0)\\ k_1-3k_2-2k_3=0\\ -k_1+9k_2-6k_3=0\\ 5k_1-7k_2+hk_3=0\\ \Delta= \begin{vmatrix} 1& -3&-2 \\ -1 & 9&-6\\ 5&-7&h \end{vmatrix}=9h+90-14+\\+90-3h-42=6h+124\neq0\\ h\neq-\frac{124}{6}\\ h\neq-\frac{62}{3}\\x1=(1,−1,5),x2=(−3,9,−7),x3=(−2,−6,h)k1x1+k2x2+k3x3=0k1=k2=k3=0k1(1,−1,5)+k2(−3,9,−7)+k3(−2,−6,h)==(0,0,0)k1−3k2−2k3=0−k1+9k2−6k3=05k1−7k2+hk3=0Δ=∣∣1−15−39−7−2−6h∣∣=9h+90−14++90−3h−42=6h+124=0h=−6124h=−362
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