Exhibit a basis for the plane spanned by the three vectors := [1 2 -1] := [-2 0
ID: 3075752 • Letter: E
Question
Exhibit a basis for the plane spanned by the three vectors := [1 2 -1] := [-2 0 3] := [7 6 -9] What is the dimension of this plane? Find a third vector in R3 to augment to the two vectors in your basis from (1), to create a basis of three vectors for R3. Exhibit a basis for the solution space to the homogeneous matrix equation What is the dimension of this subspace of R4? Find two more vectors in R4 to augment with the two vectors in your basis from part(3), to create a basis for R4. Exhibit a basis for the span of the four columns of the matrix in (3), made out of two of the original columns. What is the dimension of this subspace of R3? Thank you so much for all the help! This homework assignment has been incredibly vague!Explanation / Answer
may this example help you First, which of the following are bases for R^3? a) (2,-3,1), (4,1,1), (0,-7,1) b) (1,6,4), (2,4,-1), (-1,2,5) solution-For the first question, part (a), the vectors are not bases if you can express one of them in terms of the other two, i.e. they are linearly dependent. Note: p[ 2] + q[4] + r[ 0] = 0 [-3] [1] [-7] [ 1] [1] [ 1] This gives 3 homogeneous equations in 3 unknowns, p, q and r, and if they are not consistent, the vectors are independent. If they are consistent one vector is a linear combination of the other two and so the vectors could not form a basis for R^3. The condition for valid solutions of p, q, r other than the trivial ones p = q = r = 0 is that the determinant shown below equals zero. |2 4 0| M = |-3 1 -7| = 0 (if true then vectors are linearly dependent) |1 1 1| Since, Det(M) = 2 - 28 + 0 - 0 + 14 + 12 = 0, the vectors are linearly dependent and DO NOT form a basis for R^3. For part (b), by the same reasoning as before, we must check the value of the determinant shown below: |1 6 4| M = |2 4 -1| = 20 + 6 + 16 + 16 + 2 - 60 = 0 |-1 2 5| again the vectors are linearly dependent and DO NOT form a basis for R^3. For your second question, notice the 3 equations are all the same equation. If these are coordinates then X1 - 2X2 + X3 = 0 is the equation of a plane and so the dimension is 2. Any two lines chosen at random in this plane can be used as base vectors. For example X1 = 2, X2 = 1 and X3 = 1 will satisfy the equation, so one base vector is (2, 1, 1). A second is (1, 1, 2), but there are an infinity of pairs (as long as they are not parallel) that can be used as base vectors. Finally, for the third problem, any vector not in the plane of v1 and v2 could form the third vector of the basis. For example the vector perpendicular to both these is: |i j k| |1 -1 0| = i(2) - j(-2) + k(4) = i + j + 2k |3 1 -2| So the third vector could be (1, 1, 2).
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