Answer :
The vectors ⎣⎡11⎦⎤ and ⎣⎡11⎦⎤ are independent for b ≠ 1. the vector ⎣⎡021⎦⎤ is not contained in the span of the given vectors.
To show that the vector ⎣⎡021⎦⎤ is not contained in the span of the vectors ⎣⎡111⎦⎤, ⎣⎡001⎦⎤, and ⎣⎡221⎦⎤, we need to demonstrate that there are no constants c1, c2, and c3 such that:
c1⎣⎡111⎦⎤ + c2⎣⎡001⎦⎤ + c3⎣⎡221⎦⎤ = ⎣⎡021⎦⎤
Setting up the equation:
c1(1, 1, 1) + c2(0, 0, 1) + c3(2, 2, 1) = (0, 2, 1)
This leads to a system of equations:
c1 + 2c3 = 0 ...(1)
c1 + 2c3 = 2 ...(2)
c1 + c3 = 1 ...(3)
From equations (1) and (2), we can see that the system is inconsistent since the left sides are identical while the right sides differ. This means that there are no values of c1, c2, and c3 that satisfy the system of equations, and thus the vector ⎣⎡021⎦⎤ is not contained in the span of the given vectors.
4. For the vectors ⎣⎡11⎦⎤ and ⎣⎡1b⎦⎤ to be independent, we need to determine the values of b that make the two vectors linearly independent.
Setting up the equation for linear dependence:
c1⎣⎡11⎦⎤ + c2⎣⎡1b⎦⎤ = ⎣⎡0⎦⎤
Expanding the equation:
c1(1, 1) + c2(1, b) = (0, 0)
This leads to a system of equations:
c1 + c2 = 0 ...(4)
c1 + bc2 = 0 ...(5)
To find the values of b that make the vectors linearly independent, we need to find the values that make the system of equations inconsistent.
From equations (4) and (5), we can see that for the system to be inconsistent (i.e., no solution), the determinant of the coefficient matrix must be zero:
|1 1|
|1 b| = 0
The determinant equation is: (1 * b) - (1 * 1) = 0
b - 1 = 0
b = 1
Therefore, the vectors ⎣⎡11⎦⎤ and ⎣⎡11⎦⎤ are independent for b ≠ 1.
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