Linear Algebra exam 1 True/false

A set B = {v1, v2, ..., vp} of vectors in R is always linearly independent when p < n.

FALSE, The set B = {v1, v2} of two vectors in R3 is linearly dependent when v2 is a scalar multiple of v1 .

If u, v, and w are nonzero vectors in R2 and u is not a multiple of v, is w a linear combination of u and v?

ALWAYS, When u, v are nonzero vectors and u is not a multiple of v, they are linearly independent. But then Span{u, v} = R2, so every vector w in R2 is a linear combination of u, v. Consequently, if u, v, and w are nonzero vectors in R2 and u is not a multiple of v, then w is a linear combination of u, v.

If a system of linear equations has no free variables, then it has a unique solution

FALSE, A linear system must have no solution, unique solutions, or infinitely many solutions. Having no free variables indicates that the system does not hae infinitely many solutions, but it does not determine which of the other two cases it might be.

When u, v are nonzero vectors, then Span{u, v} contains only the line through u and the origin, and the line through v and the origin.

FALSE, By definition, Span{u, v} consists of all linear combinations y = c1u+c2v of linear combinations of u, v where c1, c2 are scalars. On the other hand, the line through u and the origin consists of all scalar multiples su of u, while the line through v and the origin consists of all scalar multiples tv of v. Now Span{u, v} contains su and tv for all s, t, so Span{u, v} contains the lines through u and v. But Span{u, v} also contains u + v which does not lie on either line.

Every matrix is row equivalent to a unique matrix in echelon form

FALSE, Every matrix is row equivalent to a unique matrix in REDUCED ECHELON FORM, however the echelon form of a matrix need not be unique.

If a 4 × 3 coefficient matrix for a system has three pivot columns, then the associated system is always consistent.

FALSE, If the coefficient matrix of a system is 4 × 3, then the augmented matrix is 4 × 4, and could have four pivot columns. So even if the coefficient matrix for a system has three pivot columns, the rightmost column of the augmented matrix could be a pivot column, in which case the system would be inconsistent.

If A is an m × n matrix and the equation Ax = b is consistent for some b in Rm, then the columns of A span Rm.

FALSE, When A is m×n, then the columns of A span Rm if and only if the equation Ax = b is consistent for every b in Rm. It is not enough to say the equation is con- sistent for some b in Rm

The homogeneous equation Ax = 0 has the trivial solution x = 0 if and only if the equation has at least one free variable.

FASLE, Since A0 = 0, the homogeneous equation Ax = 0 always has the trivial solution. The equation has a nontrivial solution if and only if it has at least one free variable.

The sum of the vector u-v and the vector v is the vector v

False, By vector addition (u-v)+v=u-v+v = u=/v

Four points co plainer?

Find a plane with three of the points using the system of equation z=ax+by, find a and b then test the 4th point.

Collinear points?

Pick two points, use y=mx+b and test the third point

If x is a nontrivial solution of Ax = 0, is every entry in x nonzero?

SOMETIMES, nontrivial solutions of Ax = 0 may contain zero entries.

If one row in an echelon form of an augmented matrix is [00050] when is the associated linear system inconsistent

SOMETIMES, the row shown corresponds to the equation 5x4=0 in the associated linear system. given only one row in a system can not determine if the equation is inconsistent. you can have examples of both with different values in the other rows.

If an augmented matrix [A b] is trans- formed into [C d] by elementary row opera- tions, then the linear systems they represent have exactly the same solution set.

TRUE , When [A b] is transformed into [C d] by el- ementary row operations, they are row equiv- alent by definition. But the linear systems corresponding to row equivalent augmented matrices have exactly the same solution set.

A consistent System of Linear equations has one or more solutions

TRUE, By def'n a consistent system has at least one solution

The columns of a matrix A are linearly independent when the equation Ax=0 has only the solution x=0

TRUE, If A = [a1 a2 ··· an], then columns {a1, a2, . . . , an} of A are linearly indepen- dent when x1a1 + x2a2 + . . . + xnan = 0=⇒ x1=x2=...=xn=0. But by matrix-vector multiplication, Ax = x1a1+x2a2+...+xnan. So the columns of A are linearly independent when Ax=0 =⇒ x=0, i.e., when Ax = 0 has only the solution x = 0. Consequently, the statement is

If matrices A and B are row equivalent, they have the same reduced echelon form.

TRUE, If A is row equivalent to B, then A can be transformed by elementary row operations first into B and then further transformed into the reduced echelon form U of B. Since the reduced echelon form of A is unique, it must be U.

A vector b is a linear combo of the columns of a matrix A iff the equation Ax=b has at least one solution

TRUE, If A is the mxn matrix A=[a1 a2 ...an] having columns a1,a2,...an in Rm, and if b is a vector in Rm, the matrix equation Ax=b has the same solution set as the vector eqution x1a1+x2a2+...+xnan=b, but htis vector equation is a linear combo of the columns of A so the matrix equation Ax=b has a solution if anf only b is a linear combo of the columns of A

If u and v are linearly independent and w is in Span{u, v}, then the set {u, v, w} is linearly dependent.

TRUE, If w is in Span{u, v}, then w can be writ- ten as a linear combination w = c1u+c2v of u and v. But then {u, v, w} is a linearly dependent set.

Any linear combination of vectors can al- ways be written in the form Ax for a suitable matrix A and vector x

TRUE, Then A is an m×n matrix and x is a vector in Rn such that Ax = c1a1 +c2a2 +...+cnan = v.

A basic variable in a linear system is a variable that corresponds to a pivot column in the coefficient matrix

TRUE, When A is the coefficient matrix of a system of linear equations in n varables, then the variables corresponding to pivot columns in A are called BASIC VARIABLES, while the other variables are called free variables.

A basic variable in a linear system is a variable that corresponds to a pivot column in the coefficient matrix.

TRUE, When A is the coefficient matrix of a system of linear equations in n variables, then the variables corresponding to pivot columns in A are called basic variables, while the other variables are called free variables.

For every Matrix equation Ax=b there corresponds a vector equation having the same solution set

TRUE, When A=[a1,a2,...,an] is an mxn matrix and b is a vector in Rm, then Ax=[a1,a2,...,an][x1,x2,...xn](x is veticle)=x1a1+x2a2+...+xnan Thus the matrix equation Ax=b has the same solution set as the vector equation x1a1+s2a2+...+xnan=b

Any list of five real numbers is a vector in R5

TRUE, if n is a pos integer, Rn denotes the collection of all lists of n real numbers. Hence R5 is the collection of all lists of five numbers.

The equation Ax=b is homogeneous if the zero vector is a solution

TRUE, if the zero vector x=0 is a solution o Ax=b the b=Ax=A0=0 thus the equation becomes Ax=0 which is a homogeneous equation

A pivot column in the augmented matrix for a linear system corresponds to a basic variable in a linear system.

The last column in the augmented matrix could be a pivot column. But this column corresponds to the right hand values, not a variable. In fact, when this last column is a pivot column, the associated system of linear equations in inconsistent.

When u=(vector) v=(vector) then the vectors in Span{u, v} lie on a line through the origin.

The vectors u and v must be multiples of each other, to be linearly dependent. then they would lie on the same line in R1

An example of a linear combo of vectors v1 and v2 is the vector -2v1

True, y=c1v1+c2v2, c1=-2v1 and c2=0

The set of vectors {v1, v2, v3} is a spanning set for R3 when (given 3 vectors in problem)

We have to check if each x in R3 can be written as a linear combination x = c1v1 +c2v2 +c3v3, Ax=[2 4 2 x1] for all x3, do RREF and see what the last row looks like, if it =0 then its inconsistent

a matrix solution is consistent when

every row has a pivot column