I really need help with these questions, this class is linear algebra and I don\'t understand this class at all.
The answers are: T, T, T, T, F, F but why is it true and why is it false? Please explain and show work. Justify answers.
4. Suppose U,V,W B R3, A=[U V W], C=[U V W B] and X=[x y z]t.
a) If echelon form of A has a pivot in every row, then for every B, AX=B (i.e. xU+yV+zW=B) has a solution.
b) If echelon form of A has a pivot in every column, then for every B, AX=B has exactly one solution.
c) If X=[1 2 3]t is a solution of AX=B, then B is a linear combination of U,V,W.
d) If columns of A span R3, then AX=B has a solution for every B.
e) If the third row of A is not a multiple of the first row and not a multiple of the second row, then an echelon form of A has a pivot in every row.
f) It is possible that U and V span R3.
I really need help with these questions, this class is linear algebra and I don\'t understand this class at all.
The answers are: T, T, T, T, F, F but why is it true and why is it false? Please explain and show work. Justify answers.
4. Suppose U,V,W B R3, A=[U V W], C=[U V W B] and X=[x y z]t.
a) If echelon form of A has a pivot in every row, then for every B, AX=B (i.e. xU+yV+zW=B) has a solution.
b) If echelon form of A has a pivot in every column, then for every B, AX=B has exactly one solution.
c) If X=[1 2 3]t is a solution of AX=B, then B is a linear combination of U,V,W.
d) If columns of A span R3, then AX=B has a solution for every B.
e) If the third row of A is not a multiple of the first row and not a multiple of the second row, then an echelon form of A has a pivot in every row.
f) It is possible that U and V span R3.
I really need help with these questions, this class is linear algebra and I don\'t understand this class at all.
The answers are: T, T, T, T, F, F but why is it true and why is it false? Please explain and show work. Justify answers.
4. Suppose U,V,W B R3, A=[U V W], C=[U V W B] and X=[x y z]t.
a) If echelon form of A has a pivot in every row, then for every B, AX=B (i.e. xU+yV+zW=B) has a solution.
b) If echelon form of A has a pivot in every column, then for every B, AX=B has exactly one solution.
c) If X=[1 2 3]t is a solution of AX=B, then B is a linear combination of U,V,W.
d) If columns of A span R3, then AX=B has a solution for every B.
e) If the third row of A is not a multiple of the first row and not a multiple of the second row, then an echelon form of A has a pivot in every row.
f) It is possible that U and V span R3.
a) The statement is TRUE
Since there exists a pivot in each row, hence its row reduced echleon form contains an element which means the matrix A is linearly independent, hence we can write that the solution exists for this part
d) The statement is TRUE
Since if columns are independent, then rows of the matrix will also be independent, hence we can say that determinant of the matrix A will be non-zero, hence there exists the solution for the equation
e) The statement is FALSE
Reason: Let the matrices be of the form [1 2 3; 2 3 4; 3 5 7]
We can clearly see neither R2 is a multiple of R1 nor R3 is a multiple of R1, still if we do R3-R2, then it is equal to R1
hence its echleon form will not have pivot in every row
f) The statement is FALSE
Reason: Since we need atleast three vectors in R3 in order to span the complete R3 plane, where each vector is required to span one of the components in the present three components
So it doesn\'t matter what U and V you choose, there exists a vector V1 which cannot be represented in linear terms of U and V
Hence the answer is False
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