Respuesta :
Answer:
a) Yes, it is a linear combination.
b) Impossible to write as a linear combination.
Step-by-step explanation:
Recall that given vectors u,v,w we say that w is a linear combination of u and v if there exists real numbers a,b such that
[tex]w=au+bv[/tex]
a) [tex] u = (2,19,-16, -4)[/tex]. So, we have the following
[tex] (2,19,-16, -4)=a(6,-7,8,6)+b(4,6,-4,1)[/tex]. Which give us the following equations
[tex]6a+4b = 2[/tex]
[tex]-7a+6b = 19[/tex]
[tex]8a-4b = -16[/tex]
[tex]6a+b =-4[/tex]
Note that if we add the first and the third equation, we get that [tex]14a = -14[/tex] which implies that a=-1. In the first equation, if a=-1, then [tex]4b=2+6[/tex] which implies that b=2. We must check that when (a,b) =(-1,2) the four equations are still valid.
So
[tex]6(-1)+4(2) = -6+8 = 2[/tex]
[tex]-7(-1)+6(2) = 7+12 = 19[/tex]
[tex]8(-1)-4(2) = -8-8 = -16[/tex]
[tex]6(-1)+(2) =-6+2 = -4[/tex]
Since all equations are met, we have written the desired vector u as the linear combination of the initial vectors.
b) Repeating the same analysis, we get
[tex](432 , 1134 , −18, 132)=a(6,-7,8,6)+b(4,6,-4,1)[/tex]
[tex]6a+4b = 432[/tex]
[tex]-7a+6b = 1134[/tex]
[tex]8a-4b = -18[/tex]
[tex]6a+b =132[/tex]
adding the first and third equation we get [tex]14a = 414[/tex] so a = 207/7. Replacing this value will give us that b=891/14.
However,
note that
[tex]-7\frac{207}{7}+6\frac{891}{14} = \frac{1124}{7}\neq 1134[/tex]. Then, it is impossible to write the linear combination.
