1.5 solution sets of linear equations

homogeneous

A system of linear equations is said to be homogeneous if it can be written in the form Ax = 0, where A is the mxn matrix and 0 is the zero vector in Rm. A homogeneous equation is always consistent. The homogeneous equation Ax = 0 has a nontrivial solution if and only if the equation has at least one free variable.

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

A system of linear equations is said to be homogeneous if it can be written in the form Ax=0​, where A is an m×n matrix and 0 is the zero vector in ℝm. If the zero vector is a​ solution, then b=Ax=A0=0.

The equation x = p + tv describes a line through v parallel to p

False, The effect of adding p to v is to move p in a direction parallel to the line through v and 0. So the equation x = p + tv describes a line through p parallel to v.

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

False, a nontrivial solution of Ax = 0 is any nonzero vector x that satisfies the equation. A nontrivial solution can have some zero entries so long as not all of its entries are zero

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

False, the equation Ax = 0 ALWAYS has the trivial solution

The solution set of Ax = b is the set of all vectors of the form w = p + v_h, where v_h is any solution of the equation Ax = 0

False, the solution set can be empty. The statement is only true when there exists a vector p such that Ap = b

The solution set of Ax = b is obtained by translating the solution set of Ax = 0

False, this only applies to a consistent system

Suppose A is a 3x3 matrix and y is a vector R3 such that the equation Ax = y doesn't have a solution. Does there exist a vector z in R3 such that the equation Ax = z has a unique solution?

If Ax = z had a unique solution then there are no free variables because each row would have a pivot. Since Ax = y has no solution, there is a zero row in the row reduced echelon form of the coefficient matrix A. [ 0 0 0 1 ] One of the columns of matrix A doesn't have a pivot. If Ax = z has a unique solution there aren't free columns in the row reduced matrix A because otherwise [ A z ] would have free variables. Since matrix A must have a free column, this is a contradiction. So there isn't a vector z E R3 such that the equation Ax = z has a unique solution

If b≠0​, can the solution set of Ax=b be a plane through the​ origin?

No. If the solution set of Ax=b contained the​ origin, then 0 would satisfy A0=b​, which is not true since b is not the zero vector.

Suppose Ax=b has a solution. Explain why the solution is unique precisely when Ax=0 has only the trivial solution.

Since Ax=b is​ consistent, its solution set is obtained by translating the solution set of Ax=0. So the solution set of Ax=b is a single vector if and only if the solution set of Ax=0 is a single​ vector, and that happens if and only if Ax=0 has only the trivial solution.

trivial solution

The zero solution

The equation x = x_2u+ ... + x_3v, with x_2 and x_3 free (and neither u nor v a multiple of the other), describes a plane through the origin

True, if the zero vector is a solution then b = Ax = A0 = 0

A homogeneous equation is always consistent.

True. A homogenous equation can be written in the form Ax=0​, where A is an m×n matrix and 0 is the zero vector in ℝm. Such a system Ax=0 always has at least one solution, namely, x=0. Thus a homogenous equation is always consistent.

The effect of adding p to a vector is to move the vector in a direction parallel to p

True. Given v and p in ℝ2 or ℝ3​, the effect of adding p to v is to move v in a direction parallel to the line through p and 0.

Suppose A is the 3x3 zero matrix (with all zero entries). Describe the solution set of the equation Ax = 0

When A is the 3x3 zero matrix, every x in R3 satisfies Ax = 0. So the solution set is all vectors in R3

nontrivial solution

a nonzero vector x that satisfies Ax = 0