LINEAR ALGEBRA THEOREMS: EXAM
#1 QUESTIONS WITH CORRECT
ANSWERS
Uniqueness of the Reduced Echelon Form - Answer-Each matrix is row equivalent to
one and only one reduced echelon matrix.
Existence and Uniqueness Theorem - Answer-A linear system is consistent if and only if
the rightmost column of the augmented matrix is not a pivot column-that is, if and only if
an echelon form of the augmented matrix has no row of the form [0 . . . 0 b] with b
nonzero
If a linear system is consistent, then the solution set contains either (i) a unique solution,
when there are no free variables, or (ii) infinitely many solutions, when there is at least
one free variable.
Theorem 3 - Answer-If A is an mxn matrix, with columns a1,. . ., an, and if b is in R^m,
the matrix equation Ax=b has the same solution set as the vector equation
x1a1+x2a2+...+xnan=b which in turn, has the same solution set as the system of linear
equations whose augmented matrix is [a1 a2 . . . an b]
Theorem 4 - Answer-Let A be an mxn matrix. Then the following statements are
logically equivalent. That is, for a particular A, either they are al true statements or they
are all false.
a. for each b in R^m, the equation of Ax=b has a solution.
b. Each b in R^m is a linear combination of the columns of A
c. The columns of A span R^m
d. A has a pivot position in every row.
Theorem 5 - Answer-If A is an mxn matrix, u and v are vectors in R^n, and c is a scalar,
then:
a. A(u+v) =Au+Av
b. A(cu)=c(Au)
Theorem 6 - Answer-Suppose the equation Ax=b is consistent for some given b, and let
p be a solution. Then the solution set of Ax=b is the set of all vectors of the form w=p+v,
where v is any solution of the homogeneous equation Ax=0.
Characterization of Linearly Dependent Sets - Answer-An indexed set S= {v1, . . . .,vp}
of two or more vectors is linearly dependent if and only if at least one of the vectors in S
is a linear combination of the others. In fact, if S is linearly dependent and v1 does not
equal 0, then some v is a linear combination of the preceding vectors, v1, . . . ., v(j-1)
#1 QUESTIONS WITH CORRECT
ANSWERS
Uniqueness of the Reduced Echelon Form - Answer-Each matrix is row equivalent to
one and only one reduced echelon matrix.
Existence and Uniqueness Theorem - Answer-A linear system is consistent if and only if
the rightmost column of the augmented matrix is not a pivot column-that is, if and only if
an echelon form of the augmented matrix has no row of the form [0 . . . 0 b] with b
nonzero
If a linear system is consistent, then the solution set contains either (i) a unique solution,
when there are no free variables, or (ii) infinitely many solutions, when there is at least
one free variable.
Theorem 3 - Answer-If A is an mxn matrix, with columns a1,. . ., an, and if b is in R^m,
the matrix equation Ax=b has the same solution set as the vector equation
x1a1+x2a2+...+xnan=b which in turn, has the same solution set as the system of linear
equations whose augmented matrix is [a1 a2 . . . an b]
Theorem 4 - Answer-Let A be an mxn matrix. Then the following statements are
logically equivalent. That is, for a particular A, either they are al true statements or they
are all false.
a. for each b in R^m, the equation of Ax=b has a solution.
b. Each b in R^m is a linear combination of the columns of A
c. The columns of A span R^m
d. A has a pivot position in every row.
Theorem 5 - Answer-If A is an mxn matrix, u and v are vectors in R^n, and c is a scalar,
then:
a. A(u+v) =Au+Av
b. A(cu)=c(Au)
Theorem 6 - Answer-Suppose the equation Ax=b is consistent for some given b, and let
p be a solution. Then the solution set of Ax=b is the set of all vectors of the form w=p+v,
where v is any solution of the homogeneous equation Ax=0.
Characterization of Linearly Dependent Sets - Answer-An indexed set S= {v1, . . . .,vp}
of two or more vectors is linearly dependent if and only if at least one of the vectors in S
is a linear combination of the others. In fact, if S is linearly dependent and v1 does not
equal 0, then some v is a linear combination of the preceding vectors, v1, . . . ., v(j-1)
