(1 point) True False Problem Mark each statement as (necessarily) true or false.
ID: 3283495 • Letter: #
Question
(1 point) True False Problem Mark each statement as (necessarily) true or false. a. The columns of matrix A are linearly independent if the equation Ax = 0 has the trivial solution. True b. Two vectors are linearly dependent if and only if they are colinear. True c. If S is a set of linearly dependent vectors, then every vector in S can be written as a linear combination of the other vectors in S. True d. If a set S of vectors contains fewer vectors than there are entries in the vectors, then the set must be linearly independent. False e. The columns of a matrix with dimensions m X n, where mExplanation / Answer
let A is m*n matrix .
a) Since columns are linearly independent then rank of column space =n
and rank(A)=n , we know if for matrix A of order m*n with rank n then having unique solution which is zero solution for AX=0 . hence trivial solution .Hence "TRUE"
b) only two vectors X,Y are linearly independent iff one vector is equal to the scalar multiple of other vector . That is X=cY which implies they are colinear . hence "TRUE"
c) NO . it is wrong . The theorem is that there does not exist at least one vector which is linear combinations of others. Because if possible then the set will be linearly dependent . as for example {(1,0,0),(0,1,0),(0,0,1)} is linearly independent and no vector can write as the linear combinations of others So "FALSE"
d) NO . It is also wrong . as for example : {(1,2,3,4),(2,4,6,8)} there are 2 vectors < entries =4 . but (2,4,6,8)=2*(a,2,3,4) . So dependent . statement is "FALSE"
e) Yes it is right . beacuse m<n . so total rank (A)<=m. So A will have at most m limearly independent vectors . Now as n>m so it is not possible to have a linearly independent of n vector . because if possible then rank of A will be n >m , not pssible . Hence statement is "TRUE"
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