# Linear combination

1. Jun 14, 2010

### Telemachus

1. The problem statement, all variables and given/known data
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?

2. Jun 14, 2010

### vela

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

3. Jun 14, 2010

### 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.

4. Jun 14, 2010

### Staff: Mentor

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.

5. Jun 14, 2010

### vela

Staff Emeritus
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."

6. Jun 15, 2010

### Telemachus

Thanks Mark and vela.