Question

for which value of k will the vector U =(1,-2,k) in R3 be a linear combination of the vectors V = (3,0,-2) and W = (2,-1,-5).

Answer 1

The easiest way to solve this is by trial and error, first looking for ways to produce the first entry of U (which is 1). Taking the first entry of V (3) and subtracting the first entry of W (2) gives 1, the first entry of U. However the second entry of V (0) minus the second entry of W (-1) gives 1, and we need instead a -2 to match the second entry of U.

So flip this around and try to subtract the first entry of V from some multiple of W. If we multiply 2, the first entry of W, by 2 and subtract the first entry of V (3) we get 2*2 - 3 = 1 which is the first entry of U. If we multiply -1 (the second element of W) by 2 we get -2, and subtracting 0 (the second element of V) gives -2 again, same as the second element of U.

So it looks like multiplying W times 2 and subtracting V will do the trick: 2 * W - V = 2 * (2, -1, -5) - (3, 0, -2) = (4, -2, -10) - (3, 0, -2) = (1, -2, -8). We have U = (1, -2, k), so if k = -8 then we'll have U = 2 * W = V and U will be expressible as a linear combination of V and W.

Answer 2

There is another method that does not require trial and error.

First of all, use the following matrices to frame the question:
\[\left[\begin{matrix}3 & 2 \\ 0 & -1\\ -2&-5\end{matrix}\right]\left[\begin{matrix}x\\y\end{matrix}\right] = \left[\begin{matrix}1\\-2\\k\end{matrix}\right]\]
Here the columns of the first matrix correspond to the vectors V and W. This matrix multiplies the second matrix(x,y) to yield the third matrix, U.

Notice this is equivalent to the system of equations;

\[\;\;\;\;3x +2y = 1\\\;\;\;\;\;\;\;\;-1y = -2\\-2x-5y = k \]
From here, you can solve for y and then back substitute to solve for x. I get:

\[y = 2 \\ x = -1\]

Plugging those values into the original equation

\[\left[\begin{matrix}3 & 2 \\ 0 & -1\\ -2&-5\end{matrix}\right]\left[\begin{matrix}-1\\2\end{matrix}\right] = \left[\begin{matrix}1\\-2\\k\end{matrix}\right]\]
\[k = (-2*-1) + (-5*2) \\k= 2- 10\\ \boxed{k = -8} \]

So, in summary, 2 of matrix W - 1 of matrix V = the matrix U\[\boxed{U = \left[\begin{matrix}1\\-2\\-8\end{matrix}\right]}\]

In this problem, the value for y was apparent immediately. Note that for similar problems row operations / eliminations might be necessary for solving.

Answer 3

As datanewb talk, in some cases you use row operations to elimination and solve the system.
In this case if you have vectors
\[v = \left(\begin{matrix}3 \\ 0 \\ -2 \end{matrix}\right) ,\ w = \left(\begin{matrix}2 \\ -1 \\ -5\end{matrix}\right) and \ \ U = \left(\begin{matrix} 1 \\ -2 \\ k \end{matrix}\right)\]
Verify that U is linear combination of \[v \ and \ w\] is solve the system
\[c _{1}v + c_{2}w = U \ \ \ (1)\]
where \[c_{1} \ and \ c_{2}\] are constants.
From (1) have
\[\left[\begin{matrix}v & w \end{matrix}\right]\left(\begin{matrix} c_{1} \\ c_{2}\end{matrix}\right) = U \ \ \ (2)\]
or be,
\[\left[\begin{matrix}3 & 2 \\ 0 & -1 \\ -2 & -5 \end{matrix}\right]\left(\begin{matrix}c_{1} \\ c_{2}\end{matrix}\right) = \left(\begin{matrix}1 \\ -2 \\ k \end{matrix}\right) \ \ \ (3)\]
after rows operations have
\[\left[\begin{matrix}3 & 2 & 1 \\ 0 & -1 & -2 \\ 0 & 0 & k + 8 \end{matrix}\right] \ \ \ (4)\]
From (4) the condition to system can be possible, k + 8 = 0,
\[k + 8 = 0 \ \rightarrow \ k = -8\]

Answer 4

@wandeson, that is well put! Your explanation is the most robust, and of course I hinted at row operations, but tried to keep it simple for someone new to the material.

Hopefully, @cecefaith found this all helpful and will let us know if she has questions.

Answer 5

thank you, all your answers have been really helpful. you've all explained it very well. wow i didn't think it was that simple.