- #1
Telemachus
- 820
- 30
Homework Statement
Let V=\mathbb(R)^3. Find all values of k for which the vector u is a linear combination of the vectors given below:
v_1=(2,3,5); v_2=(3,7,8); v_3=(1,-6,1) y u=(7,-2,k)
\begin{Bmatrix}{ 2\lambda_1+3\lambda_2+\lambda_3=7} \\3\lambda_1+7\lambda_2-6\lambda_3=-2 \\5\lambda_1+8\lambda_2+\lambda_3=k \end{matrix}
\begin{bmatrix}{2}&{3}&{1}&{7}\\{3}&{7}&{-6}&{-2}\\{5}&{8}&{3}&{k}\end{bmatrix}\rightarrow{<br /> \begin{bmatrix}{2}&{3}&{1}&{7}\\{0}&{5/2}&{-15/2}&{-25/2}\\{0}&{1/2}&{1/2}&{k-35/2}\end{bmatrix}}\rightarrow{<br /> \begin{bmatrix}{2}&{3}&{1}&{7}\\{0}&{5}&{-15}&{-25}\\{0}&{0}&{32}&{k-35}\end{bmatrix}}\begin{bmatrix}{2}&{3}&{1}&{7}\\{0}&{1}&{-3}&{-5}\\{0}&{0}&{3}&{k-30}\end{bmatrix}\rightarrow{\begin{bmatrix}{2}&{3}&{0}&{\displaystyle\frac{-k}{3}+10}\\{0}&{1}&{0}&{k-35}\\{0}&{0}&{3}&{k-30}\end{bmatrix}}\rightarrow{\begin{bmatrix}{2}&{0}&{0}&{\displaystyle\frac{-10}{3}k+115}\\{0}&{1}&{0}&{k-35}\\{0}&{0}&{1}&{k/3-10}\end{bmatrix}}\rightarrow{\begin{bmatrix}{1}&{0}&{0}&{\displaystyle\frac{-5}{3}k+115/2}\\{0}&{1}&{0}&{k-35}\\{0}&{0}&{1}&{k/3-10}\end{bmatrix}}
\begin{Bmatrix}{ \lambda_1=\displaystyle\frac{-5}{3}k+\displaystyle\frac{115}{2}} \\ \lambda_2=k-35 \\ \lambda_3=\displaystyle\frac{k}{3}-10 \end{matrix}
Is that right?
