LINEAR ALGEBRA THEOREMS - TEST 1
QUESTIONS AND ANSWERS
Theorem 1 (Uniqueness of the Reduced Echelon Form) - Answer-each matrix is row
equivalent to one and only one reduced echelon matrix
Theorem 2 (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 o row of the
form
[ 0 .... 0 b] with b nonzero
If a linear system is consistent, then the solution set contains either a unique solution,
when there are no free variables, or infinitely many solutions, when there is at least one
free variable
Theorem 3 - Answer-If A is an nxm a1 ..... an, and if b is in R(m), the matrix equation
Ax=b
had the same solution set as the vector equation
x(1)a(1) + .... + x(n)a(n) = b
which, in turn, has the same solution set as the system of linear equations whose
augmented matrix is
[ a(1) ... a(n) b ]
Theorem 4 - Answer-Let A be an nxm matrix. Then the following statements are
logically equivalent.
a. For each b in R(m), the equation 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 5 - 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 for, w =
p+v(h) is any solution of the homogeneous equation Ax = 0.
Theorem 7 (Characterization of Linearly Dependent Sets) - Answer-An indexed set S = {
v(1), ... , v(p) } 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 v(1) does not equal 0, then some v(j) (with j>1) is a linear combination of the
preceding vectors.
QUESTIONS AND ANSWERS
Theorem 1 (Uniqueness of the Reduced Echelon Form) - Answer-each matrix is row
equivalent to one and only one reduced echelon matrix
Theorem 2 (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 o row of the
form
[ 0 .... 0 b] with b nonzero
If a linear system is consistent, then the solution set contains either a unique solution,
when there are no free variables, or infinitely many solutions, when there is at least one
free variable
Theorem 3 - Answer-If A is an nxm a1 ..... an, and if b is in R(m), the matrix equation
Ax=b
had the same solution set as the vector equation
x(1)a(1) + .... + x(n)a(n) = b
which, in turn, has the same solution set as the system of linear equations whose
augmented matrix is
[ a(1) ... a(n) b ]
Theorem 4 - Answer-Let A be an nxm matrix. Then the following statements are
logically equivalent.
a. For each b in R(m), the equation 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 5 - 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 for, w =
p+v(h) is any solution of the homogeneous equation Ax = 0.
Theorem 7 (Characterization of Linearly Dependent Sets) - Answer-An indexed set S = {
v(1), ... , v(p) } 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 v(1) does not equal 0, then some v(j) (with j>1) is a linear combination of the
preceding vectors.
