1 2 0 1 D-0 0 1 1 ) Using the matrix A, and its reduced row echelon matrix D, fi
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Question
1 2 0 1 D-0 0 1 1 ) Using the matrix A, and its reduced row echelon matrix D, find a basis for the subspace generated by (C1,-20.3). (2,-4,0,6).C-1,12.0), (0,-1,2,3) and for those vectors that are not part of the basis, write them as a linear combination of the basis.(3 points) Using the matrix A, and its reduced row echelon matrix D, find a basis, and the dimension for solution of the following homogenous system (3 points) (li) -2x1-4x2 +x-x0 2x3+2x 33t6 x2 + 3x4 =0 Q.3 Consider the following two basis for R2 the basis B ((1,0), (1,1)), and the basisExplanation / Answer
(i)
The set S = {v1, v2, v3, v4} of vectors in R4 is linearly independent if the only solution of
(*) c1v1 + c2v2 + c3v3 + c4v4 = 0
is c1, c2, c3, c4 = 0.
In this case, the set S forms a basis for span S.
Otherwise (i.e., if a solution with at least some nonzero values exists), S is linearly dependent.
If this is the case, a subset of S can be found that forms a basis for span S.
With our vectors v1, v2, v3, v4, (*) becomes:
Rearranging the left hand side yields
The matrix equation above is equivalent to the following homogeneous system of equations
We now transform the coefficient matrix of the homogeneous system above to the reduced row echelon form to determine whether the system has
which corresponds to the system
Those columns in the matrix that do not contain leading entries, correspond to unknowns that will be arbitrary. The system has infinitely many solutions:
Since the variables c2, c4 are arbitrary, then each of the vectors v2, v4 can be expressed as a linear combination of vectors in the set T = {v1, v3}. For example, set c2 = 1, c4 = 0, and use the equation (*) to express v2 as a linear combination of the remaining vectors in the set S:
v2 = 2v1 +0v3
Since the set T = {v1, v3} is linearly independent and it spans span S, then the set
forms a basis for span S and here Rank of the matrix = 2
(ii)
The set S = {v1, v2, v3, v4} of vectors in R4 is linearly independent if the only solution of
(*) c1v1 + c2v2 + c3v3 + c4v4 = 0
is c1, c2, c3, c4 = 0.
In this case, the set S forms a basis for span S.
Otherwise (i.e., if a solution with at least some nonzero values exists), S is linearly dependent.
If this is the case, a subset of S can be found that forms a basis for span S.
With our vectors v1, v2, v3, v4, (*) becomes:
Rearranging the left hand side yields
The matrix equation above is equivalent to the following homogeneous system of equations
We now transform the coefficient matrix of the homogeneous system above to the reduced row echelon form to determine whether the system has
The reduced row echelon form of the coefficient matrix of the homogeneous system (**) is
which corresponds to the system
Those columns in the matrix that do not contain leading entries, correspond to unknowns that will be arbitrary. The system has infinitely many solutions:
Since the variables c3, c4 are arbitrary, then each of the vectors v3, v4 can be expressed as a linear combination of vectors in the set T = {v1, v2}. For example, set c3 = 1, c4 = 0, and use the equation (*) to express v3 as a linear combination of the remaining vectors in the set S:
v3 = -4v1-2v2
Since the set T = {v1, v2} is linearly independent and it spans span S, then the set
forms a basis for span S.
c1 1 -2 0 3 + c2 2 -4 0 6 + c3 -1 1 2 0 + c4 0 -1 2 3 = 0 0 0 0Related Questions
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