OpenStudy (anonymous):

Express the vector u= (5, -1, -1) as a linear combination of the vectors v1=(2,2,4), v2=(8,5,10) and v3=(4,1,-1). That is, find c1,c2, and c3 so that u=c1v1+c2v2+c3v3 if that is possible. Hint: you may want to use what you have learned about matrix inverses to solve this problem. I got that c1=37/6, c2=7/3 and c3=-1/3 I'm pretty sure that those answers are incorrect so any help would be great!

5 years ago
OpenStudy (anonymous):

u = (5,-1,-1) v1 = (2,2,4) v2 = (8,5,10) v3 = (4,1,-1) (5,-1,-1) = a(2,2,4) + b(8,5,10) + c (4,1,-1) 5 = 2a + 8b + 4c -1 = 2a + 5b + c -1 = 4a + 10b - c --------------- - 2 = 6a + 15b 5 = 2a + 8b + 4c 15 = 6a + 24b + 12c 15 = -2 + 9b + 12c 17 = 9b + 12c 15 b = - 2 - 6 a b = - (2 + 6 a)/ 15 9 b = 17 - 12 c b = (17 - 12 c) / 9

5 years ago
OpenStudy (turingtest):

\[A=\left[\begin{matrix}2&8&4\\2&5&10\\4&1&-1\end{matrix}\right]\]\[\vec u=\left[\begin{matrix}5\\-1\\-1\end{matrix}\right]\]and let the constants a,b, and c we are looking for in the linear combination be represented by\[\vec x=\left[\begin{matrix}a\\b\\c\end{matrix}\right]\]\[A\vec x=\vec u\implies A^{-1}\vec u=\vec x\]so find \(A^{-1}\) if it exists and multiply it by \(\vec u\)

5 years ago
OpenStudy (anonymous):

you can use cramers rule to find values of a, b,c

5 years ago
OpenStudy (anonymous):

thats much quick, also if determinant =0, then it will clearly tell you about the nature of the linear combination, i.e. if it exists or not ?

5 years ago
OpenStudy (anonymous):

system of 3 equations, 3 unknowsn try cramers rule first, if that fails go for other options UNLESS explicit instructions are provided

5 years ago