1.
To determine whether the given vectors are a basis for mathbb R^3, we need to check whether they span mathbb R^3 and whether they are linearly independent.
To check whether the vectors span mathbb R^3, we can form a matrix with the vectors as columns and row reduce it to see if we get a matrix in reduced row echelon form with three nonzero rows:
[1 3 -3]
[0 2 -5]
[-2 -4 1]
Row reducing this matrix, we get:
[1 0 0]
[0 1 0]
[0 0 1]
Since we have three nonzero rows, the vectors span mathbb R^3.
To check whether the vectors are linearly independent, we can set up the equation:
a[1] * [1] + a[2] * [3] + a[3] * [-3] = [0]
[0] [2] [-4] [0]
[-2] [-4] [1] [0]
where a[1], a[2], and a[3] are constants. We can solve this system of equations using row reduction:
[1 3 -3 0]
[0 2 -5 0]
[-2 -4 1 0]
R2 = R2 - R1 * 3/2
R3 = R3 + R1 * 2
[1 3 -3 0]
[0 2 -5 0]
[0 2 -5 0]
R3 = R3 - R2
[1 3 -3 0]
[0 2 -5 0]
[0 0 0 0]
Since we have a row of zeros, we can see that the vectors are linearly dependent.
Therefore, the given vectors do not form a basis for mathbb R^3.
2.
To find a basis for the nullspace of the given matrix, we need to solve the system of equations:
[1 0 -3 2][x1] [0]
[0 1 -5 4][x2] = [0]
[3 -2 1 -2][x3] [0]
We can set up the corresponding augmented matrix and row reduce it:
[1 0 -3 2 | 0]
[0 1 -5 4 | 0]
[3 -2 1 -2 | 0]
R3 = R3 - 3R1
R3 = R3 + 2R2
[1 0 -3 2 | 0]
[0 1 -5 4 | 0]
[0 -2 8 -2 | 0]
R3 = R3 + 2R2
[1 0 -3 2 | 0]
[0 1 -5 4 | 0]
[0 0 -2 6 | 0]
R3 = -R3/2
[1 0 -3 2 | 0]
[0 1 -5 4 | 0]
[0 0 1 -3 | 0]
R1 = R1 + 3R3
R2 = R2 + 5R3
[1 0 0 -7 | 0]
[0 1 0 11 | 0]
[0 0 1 -3 | 0]
Therefore, the solution to the system of equations is x1 = 7x4, x2 = -11x4, x3 = 3x4, where x4 is a free variable.
A basis for the nullspace of the matrix is the vector:
[7]
[-11]
[3]
[1]
3.
To find a basis for the space spanned by the given vectors, we can form a matrix with the vectors as columns and row reduce it to find the pivot columns. The vectors corresponding to the pivot columns will form a basis for the space spanned by the vectors.
[1 0 -3 2 0]
[0 1 2 -3 0]
[-3 -8 1 6 0]
[2 7 -8 9 0]
Row reducing this matrix, we get:
[1 0 0 1 0]
[0 1 0 2 0]
[0 0 1 -1 0]
[0 0 0 0 0]
The pivot columns are the first three columns, so a basis for the space spanned by the vectors is:
[[1], [0], [-3], [2]]
[[0], [1], [2], [-3]]
[[-3], [-8], [1], [6]]
4.
To find a linear dependence among the given polynomials, we need to find constants c1, c2, and c3, not all zero, such that:
c1 * p1(t) + c2 * p2(t) + c3 * p3(t) = 0
Substituting the given polynomials, we get:
c1 * (1 + t) + c2 * (1 - t) + c3 * 2 = 0
Simplifying, we get:
(c1 + c2) + t(c1 - c2) + 2c3 = 0
This equation must hold for all values of t. Therefore, we can equate the coefficients of t and the constant term to get a system of equations:
c1 - c2 = 0
2c3 = 0
From the second equation, we get c3 = 0, which means that c1 + c2 = 0 from the first equation. Therefore, we have a non-trivial linear dependence:
c1 * p1(t) + c2 * p2(t) = 0
Taking c1 = c2 = 1, we get:
p1(t) + p2(t) = 0
Therefore, a non-trivial linear dependence among the given polynomials is p1(t) + p2(t) = 0.
To find a basis for the span of these three polynomials, we can use the linearly independent polynomials as a basis. Since p1(t) and p2(t) are linearly independent, we can use them as a basis. Therefore, a basis for the span of the given polynomials is:
{1 + t, 1 - t}