Find k for Linear Combination of (2,3,5), (3,7,8), (1,-6,1)

[Telemachus]

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 $$u=(7,-2,k)$$

$$\begin{Bmatrix}{ 2\lambda_1+3\lambda_2+\lambda_3=7} \\3\lambda_1+7\lambda_2-6\lambda_3=-2 \\5\lambda_1+8\lambda_2+\lambda_3=k \end{matrix}$$

$$\begin{bmatrix}{2}&{3}&{1}&{7}\\{3}&{7}&{-6}&{-2}\\{5}&{8}&{3}&{k}\end{bmatrix}\rightarrow{ \begin{bmatrix}{2}&{3}&{1}&{7}\\{0}&{5/2}&{-15/2}&{-25/2}\\{0}&{1/2}&{1/2}&{k-35/2}\end{bmatrix}}\rightarrow{ \begin{bmatrix}{2}&{3}&{1}&{7}\\{0}&{5}&{-15}&{-25}\\{0}&{0}&{32}&{k-35}\end{bmatrix}}$$$$\begin{bmatrix}{2}&{3}&{1}&{7}\\{0}&{1}&{-3}&{-5}\\{0}&{0}&{3}&{k-30}\end{bmatrix}\rightarrow{\begin{bmatrix}{2}&{3}&{0}&{\displaystyle\frac{-k}{3}+10}\\{0}&{1}&{0}&{k-35}\\{0}&{0}&{3}&{k-30}\end{bmatrix}}\rightarrow{\begin{bmatrix}{2}&{0}&{0}&{\displaystyle\frac{-10}{3}k+115}\\{0}&{1}&{0}&{k-35}\\{0}&{0}&{1}&{k/3-10}\end{bmatrix}}\rightarrow{\begin{bmatrix}{1}&{0}&{0}&{\displaystyle\frac{-5}{3}k+115/2}\\{0}&{1}&{0}&{k-35}\\{0}&{0}&{1}&{k/3-10}\end{bmatrix}}$$

$$\begin{Bmatrix}{ \lambda_1=\displaystyle\frac{-5}{3}k+\displaystyle\frac{115}{2}} \\ \lambda_2=k-35 \\ \lambda_3=\displaystyle\frac{k}{3}-10 \end{matrix}$$

Is that right?

 

[vela]

You seem to have made a mistake in the second step.

 

[Telemachus]

Oh shhhhhhh... ok. Anyway. What does it mean when I got this kind of solution in terms of vectorial spaces? I mean, when its undetermined. And if I get an incompatible solution, what would it mean?

Bye and thanks.

 

[Mark44]

Your matrix is really an augmented matrix, with the first three columns being the three given vectors, and the fourth column being the vector u. If you end up with a row whose first three entries are 0 and the fourth is an some expression involving k, the system is inconsistent if the expression involving k is nonzero. The system is consistent if and only if the expression involving k is equal to zero.

 

[vela]

And if your initial work had been correct, that would mean for any value of k you could find a linear combination that would work, so your answer should have been "all real numbers."

 

[Telemachus]

Thanks Mark and vela.