LINEAR ALGEBRA QUIZ Q&A
consistent - Answer-a system of linear equations has either one solution or infinitely
many solutions
inconsistent - Answer-no solution
determine if a system of linear equations is consistent or inconsistent - Answer-1.
augment the matrix
2. reduce to triangular form using row operations
pivot position - Answer-a location in matrix A that corresponds to a leading 1 in the
reduced echelon form of A
parametric description of a solution set - Answer-use the free variables as the
parameters for describing a solution set
R2 - Answer-the set of all vectors with two entries
R3 - Answer-vectors in R3 are 3x1 column matrices with three entries
linear combination (y) - Answer-given vectors v1, v2, ..., vp in Rn and scalars c1, c2, ...,
cp, the vector y is called a linear combination and is defined by y = c1v1 + ... + cpvp
determine wether b is a linear combination of a1 and a2 - Answer-1. augment the matrix
to [a1 a2 b]
2. if solution is consistent, b is a linear combination of a1 and a2
span - Answer-the span of {v1,...,vp} is a set of all linear combinations of v1...vp
must contain the zero vector
determine if b is in span{v1,...,vp} - Answer-1. determine whether the vector equation
x1v1 + x2v2 + ... + xpvp = b
2. equivalently, determine whether the augmented matrix [v1 ... vp b] has a solution
matrix multiplication - Answer-Ax is defined only if the number of columns of A equals
the number of entries in x
matrix equation - Answer-Ax = b
vector equation - Answer-x1a1 +x2a2 + ... + xnan = b
solution of Ax = b - Answer-exists if and only if b is a linear combination of the columns
of A
, coefficient matrix theorem - Answer-let A be an mxn matrix
1. for each b in Rm, Ax = b has a solution
2. each b in Rm is a linear combination of the columns of A
3. the columns of A span Rm
4. A has a pivot position in every row
homogenous linear system - Answer-1. can be written in the form Ax = 0 where A is an
mxn matrix and 0 is the zero vector in Rm
2. always has at least one solution
3. Ax = 0 has a nontrivial solution if and only if the equation has at least one free
variable
trivial solution - Answer-the zero solution (for Ax = 0, x = 0)
nontrivial solution - Answer-a nonzero vector x that satisfies Ax = 0
determine if the homogenous system has a nontrivial solution - Answer-1. let A be the
matrix of coefficients of the system
2. row reduce the augmented matrix [A 0] to echelon form
3. determine if a free variable exists
4. to describe the solution set, continue row reduction to reduced echelon form
parametric vector equation - Answer-x = su + tv (s,t in R)
solutions of nonhomogenous systems - Answer-the general solution (many solutions)
can be written in parametric vector form as one vector plus an arbitrary linear
combination of vectors that satisfy the corresponding homogenous system
describe all solutions of Ax = b - Answer-1. perform row operations on [A b]
2. express the solution in terms of the free variables
linearly independent - Answer-1. a set of vectors {v1,...,vp} in Rn is said to be linearly
undefended if the vector equation x1v1 + x2v2 + ... + xpvp = 0 has only the trivial
solution (x=0)
2. pivot in every column
3. one-to-one
linearly dependent - Answer-1. the set of vectors {v1,...,vp} is linearly dependent if there
exist weights c1,...,cp (not all zero) such that c1v1 + c2v2 + ... + cpp = 0
2. a nontrivial solution exists (at least one free variable)
determine if the set {v1, v2, v3} is linearly independent - Answer-1. A = [v1 v2 v3]
2. augment the matrix to [A 0] and row reduce
3. if there is at least one free variable, the set is NOT linearly independent
consistent - Answer-a system of linear equations has either one solution or infinitely
many solutions
inconsistent - Answer-no solution
determine if a system of linear equations is consistent or inconsistent - Answer-1.
augment the matrix
2. reduce to triangular form using row operations
pivot position - Answer-a location in matrix A that corresponds to a leading 1 in the
reduced echelon form of A
parametric description of a solution set - Answer-use the free variables as the
parameters for describing a solution set
R2 - Answer-the set of all vectors with two entries
R3 - Answer-vectors in R3 are 3x1 column matrices with three entries
linear combination (y) - Answer-given vectors v1, v2, ..., vp in Rn and scalars c1, c2, ...,
cp, the vector y is called a linear combination and is defined by y = c1v1 + ... + cpvp
determine wether b is a linear combination of a1 and a2 - Answer-1. augment the matrix
to [a1 a2 b]
2. if solution is consistent, b is a linear combination of a1 and a2
span - Answer-the span of {v1,...,vp} is a set of all linear combinations of v1...vp
must contain the zero vector
determine if b is in span{v1,...,vp} - Answer-1. determine whether the vector equation
x1v1 + x2v2 + ... + xpvp = b
2. equivalently, determine whether the augmented matrix [v1 ... vp b] has a solution
matrix multiplication - Answer-Ax is defined only if the number of columns of A equals
the number of entries in x
matrix equation - Answer-Ax = b
vector equation - Answer-x1a1 +x2a2 + ... + xnan = b
solution of Ax = b - Answer-exists if and only if b is a linear combination of the columns
of A
, coefficient matrix theorem - Answer-let A be an mxn matrix
1. for each b in Rm, Ax = b has a solution
2. each b in Rm is a linear combination of the columns of A
3. the columns of A span Rm
4. A has a pivot position in every row
homogenous linear system - Answer-1. can be written in the form Ax = 0 where A is an
mxn matrix and 0 is the zero vector in Rm
2. always has at least one solution
3. Ax = 0 has a nontrivial solution if and only if the equation has at least one free
variable
trivial solution - Answer-the zero solution (for Ax = 0, x = 0)
nontrivial solution - Answer-a nonzero vector x that satisfies Ax = 0
determine if the homogenous system has a nontrivial solution - Answer-1. let A be the
matrix of coefficients of the system
2. row reduce the augmented matrix [A 0] to echelon form
3. determine if a free variable exists
4. to describe the solution set, continue row reduction to reduced echelon form
parametric vector equation - Answer-x = su + tv (s,t in R)
solutions of nonhomogenous systems - Answer-the general solution (many solutions)
can be written in parametric vector form as one vector plus an arbitrary linear
combination of vectors that satisfy the corresponding homogenous system
describe all solutions of Ax = b - Answer-1. perform row operations on [A b]
2. express the solution in terms of the free variables
linearly independent - Answer-1. a set of vectors {v1,...,vp} in Rn is said to be linearly
undefended if the vector equation x1v1 + x2v2 + ... + xpvp = 0 has only the trivial
solution (x=0)
2. pivot in every column
3. one-to-one
linearly dependent - Answer-1. the set of vectors {v1,...,vp} is linearly dependent if there
exist weights c1,...,cp (not all zero) such that c1v1 + c2v2 + ... + cpp = 0
2. a nontrivial solution exists (at least one free variable)
determine if the set {v1, v2, v3} is linearly independent - Answer-1. A = [v1 v2 v3]
2. augment the matrix to [A 0] and row reduce
3. if there is at least one free variable, the set is NOT linearly independent
