Suppose the solutions of a homogeneous system of four linear equations in five u

Question

Suppose the solutions of a homogeneous system of four linear equations in five unknowns are all multiples of one nonzero solution. Will the system necessarily have have a solution for every possible choice of constants on the right sides of the? equations? Explain.

The first 4 boxes are supposed to be integers.

The first box with the drop down arrow has the choices: Col A, Row A, Nul A

The second box with the drop down arrow has the choices: Col A, Row A, Nul A

The third box with the drop down arrow has the choices: may or may not be, must be

The fourth box with the drop down arrow has the choices: Col A, Row A, Nul A

The fifth box with the drop down arrow has the choices: has, does not have

The sixth box with the drop down arrow has the choices: will, will not

Suppose the solutions of a homogeneous system of four linear equations in five unknowns are all multiples of one nonzero solution. Will the system necessarily have have a solution for every possible choice of constants on the right sides of the equations? Explain. t an Consider the system as Ax - 0, where A is a 4 x 5 matrix. Because the solutions of the system are all multiples of one nonzero solution, dim Nul A-By the Rank Theorem, rank A= dim Nul A=Since dim Col A= rank A= and Consider the system as Ax=0, where A is a 4x5 matrix. Because the solutions of the system are all multiples of one nonzero solution, dim Nul A= | | . By the Rank Theorem, rank A-D-dim Nul A= Since dim Col A= rank Asi in R4 constants on the right sides of the equations and | is a subspace of R4, it follows that = R. This means that each vector b have a solution for every pesible choice of | in | and so Ax = b | a solution for all b. That is, the system I have a solution for every possible choice of

Explanation / Answer

Let us consider the system as Ax=0 where A is a 4x5 matrix of the linear system. Because the solutions of the system are all multiples of one non-zero solution, dim Nul A = 1. By the Rank-Nullity theorem, rank(A) = 5- dim Nul A = 4. Since dim Col A = rank A = 4 and row space of A is a subspace of R4 , it follows that row space of A = R4. This means that each vector b in R4 has a pre-image in R5 and so Ax = b has a solution for all b. That is the system Ax = b have a solution for every possible choice of constants on the right side of the equations.

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