a. The vector v=(a,b,c,d) is a linear combination of vectors B if there exist scalars x,y,z,w such that:
v = x(1,1,1,1) + y(2,0,1,0) + z(-2,2,0,1) + w(1,-1,0,0)
This is equivalent to solving the following system of equations for x,y,z,w:
a = x + 2y - 2z + w b = x + 2z - w c = x + y d = x + z
The system has a unique solution if and only if the determinant of the coefficient matrix is nonzero. That is, if and only if:
| 1 2 -2 1 | | 1 0 2 -1 | ≠ 0 | 1 1 0 0 | | 1 0 1 0 |
Using the rule of Sarrus, we can compute the determinant as:
(1000) + (2211) + (-2*-101) + (112*-1) - (100*-1) - (2*-200) - (-2010) - (1110) = (4) + (2) - (2) - (4) - (-4) = 4
Since the determinant is nonzero, the vector v=(a,b,c,d) is a linear combination of vectors B for any values of a,b,c,d.
b. To prove that the vector w=(1,1,1,3) is a linear combination of vectors B, we need to find scalars x,y,z,w that satisfy:
w = x(1,1,1,1) + y(2,0,1,0) + z(-2,2,0,1) + w(1,-1,0,0)
This is equivalent to solving the following system of equations for x,y,z,w:
x + 2y - 2z + w = 1 x + 2z - w = 1 x + y = 1 x + z = 3
Using Gaussian elimination or matrix inversion methods, we can find that one possible solution is:
x = -3/4 y = -5/4 z = -9/4 w = -7/4
Therefore, we can write w as a linear combination of vectors B as:
w = (-3/4)(1,1,1,1) + (-5/4)(2,0,1,0) + (-9/4)(-2,2,0,1) + (-7/4)(1,-1,0,0)
Another possible solution is:
x = -5/4 y = -7/4 z = -11/4 w = -9/4
Therefore, we can also write w as a linear combination of vectors B as:
w = (-5/4)(1,1,1,1) + (-7/4)(2,0,1,0) + (-11/4)(-2,2,0,1) + (-9/4)(1,-1,0,0)
c. To find a vector v∈R^4 such that the set C={(1,1,1,1),(2,0,1,0),(−2,+20,+10),(−10,+10,+30),v} is a basis for R^4 , we need to find a vector v that is linearly independent from the other four vectors. That is,
v ≠ x(11,+11,+11,+11)+y(20,+10,+10)+z(−22,+20,+10,+10)+w(−10,+10,+30)
for any scalars x,y,z,w. This means that v cannot be a solution of the following homogeneous system of equations:
x + 2y - 2z - w = 0 x + 2z - w = 0 x + y = 0 x + z = 0
Using Gaussian elimination or matrix inversion methods again , we can find that the only solution of this system is:
x = y = z = w = 0
This means that the four vectors are already linearly independent and form a basis for R^4 by themselves. Therefore,
v can be any vector in R^4 that is not the zero vector.
For example,
v=(5,-3,-7,-9)
is a valid choice.