Let vector 'v' = (1,1) and vector 'w' = (1,-1).
a)Find real numbers t1 and t2 such that t1(v) + t2(w) = (13,-12).
b)Find real numbers t1 and t2 such that t1(v) + t2(w) = (x1,x2)for any x1,x2 that is an element of R.
t1(1,1) + t2(1,-1) = (13,-12)
(t1,t1) + (t2,-t2) = (13,-12)
(t1+t2 , t1 -t2) = (12, -12)
so
t1+t2 = 13
t1-t2 = -12
add them
2t1 = 1
t1 = 1/2
subtract them
2t2 = 25
t2 = 25/2
b) using he results of above
replace 13 with x1 and -12 with x2 and repeat my steps
To find real numbers t1 and t2 that satisfy the given equations, we can use the concept of linear combinations.
a) Find t1 and t2 such that t1(v) + t2(w) = (13, -12):
To solve this equation, we need to multiply each vector by its corresponding coefficient, t1 and t2, and then add them together to obtain (13, -12).
t1(v) + t2(w) = t1(1, 1) + t2(1, -1) = (t1 + t2, t1 - t2)
Equating the corresponding components of both sides, we get the following system of equations:
t1 + t2 = 13
t1 - t2 = -12
We can solve this system of equations by adding the two equations together:
(t1 + t2) + (t1 - t2) = 13 + (-12)
2t1 = 1
Dividing both sides by 2, we obtain:
t1 = 1/2
Substituting t1 = 1/2 into one of the original equations, we can find t2:
(1/2) + t2 = 13
t2 = 13 - 1/2
t2 = 25/2
Therefore, real numbers t1 = 1/2 and t2 = 25/2 satisfy the equation t1(v) + t2(w) = (13, -12).
b) Find t1 and t2 such that t1(v) + t2(w) = (x1, x2) for any x1, x2 ∈ ℝ:
In order to find real numbers t1 and t2 for any given x1 and x2, we can set up a similar system of equations.
t1 + t2 = x1
t1 - t2 = x2
We can solve this system of equations using the same approach as in part a:
Adding the two equations:
(t1 + t2) + (t1 - t2) = x1 + x2
2t1 = x1 + x2
Dividing both sides by 2:
t1 = (x1 + x2)/2
Substituting the expression for t1 into one of the original equations, we can solve for t2:
(x1 + x2)/2 + t2 = x1
t2 = x1 - (x1 + x2)/2
t2 = (2x1 - (x1 + x2))/2
t2 = (x1 - x2)/2
Therefore, for any x1 and x2 ∈ ℝ, real numbers t1 = (x1 + x2)/2 and t2 = (x1 - x2)/2 satisfy the equation t1(v) + t2(w) = (x1, x2).