What sets are linearly independant2 -74 -8-6 -121 pi 02 root(2) 04 sin2 03 e 02 4 1-1 1 0 5 0 0
2 4 1-1 1 0 5 0 0(pivot in every row when reduced)
We can determine if the given set of vectors is linearly independent or not by checking if their linear combination results in a non-trivial solution to the equation:
c1v1 + c2v2 + c3v3 + … + cnvn = 0
where c1, c2, c3, …, cn are scalars and v1, v2, v3, …, vn are the given vectors.
If the only solution to this equation is c1 = c2 = c3 = … = cn = 0, then the vectors are linearly independent. Otherwise, they are linearly dependent.
In this problem, we have the following set of vectors:
v1 = (2, -74, -8)
v2 = (-6, -121, pi)
v3 = (0, 2, sqrt(2))
v4 = (0, 4sin(2), 0)
v5 = (3e, 0, 2)
v6 = (4, 1, -1)
v7 = (1, 0, 5)
v8 = (0, 0, 0)
Notice that v8 is a zero vector and any set containing a zero vector is linearly dependent.
Let’s create an augmented matrix using the coefficients of this equation:
[ 2 -6 0 0 3 4 1 0 | 0 ]
[-74-121 2 4 0 1 0 0 | 0 ]
[ -8 pi sqrt(2) 0 2 -1 5 0 | 0 ]
Now, we’ll perform row operations to reduce it to row echelon form.
R2 + 37R1 -> R2 (to eliminate coeffcients in column 1)
R3 + 4R1 -> R3 (to eliminate coefficients in column 1)
[ 2 -6 0 0 3 4 1 0 | 0 ]
[ 0 -299 2 4 111 145 37 0 | 0 ]
[ 0 -pi sqrt(2) 0 14 -5 9 0 | 0 ]
R3 + (pi/299)R2 -> R3 (to eliminate coefficients in column 2)
[ 2 -6 0 0 3 4 1 0 | 0 ]
[ 0 -299 2 4 111 145 37 0 | 0 ]
[ 0 0 -sqrt(2) 4pi/299 42pi/299 99pi/299 9-pi/299 0 | 0 ]
The reduced row echelon form of this matrix shows that the only solution to this equation is c1 = c2 = c3 = … = cn = 0.
Therefore, the given set of vectors is linearly independent except for the zero vector.
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