Step 1

Consider the provided question,

Given a system of five linear equations in six unknowns are all multiples of one nonzero solution.

It can be written as,

Ax=0 where \(\displaystyle{A}={5}\times{6}\) matrix.

dim(NulA)=1 because all solutions of

Ax = 0 are multiples of one nonzero solution.

Step 2

We have to explain that the system necessarily have a solution for every possible choice of constants on the right sides of the equations.

Since, A is \(\displaystyle{5}\times{6}\) matrix. n = 6

rank(A)=n-dim(NulA)=6-1=5

Image of A is 5-dimensional subspace of \(\displaystyle{R}^{{5}}\)(because A has 5 rows).

So, \(\displaystyle{C}{o}{l}{\left({A}\right)}={R}^{{5}}\)

This means that Ax=b has a solution for every b.

Consider the provided question,

Given a system of five linear equations in six unknowns are all multiples of one nonzero solution.

It can be written as,

Ax=0 where \(\displaystyle{A}={5}\times{6}\) matrix.

dim(NulA)=1 because all solutions of

Ax = 0 are multiples of one nonzero solution.

Step 2

We have to explain that the system necessarily have a solution for every possible choice of constants on the right sides of the equations.

Since, A is \(\displaystyle{5}\times{6}\) matrix. n = 6

rank(A)=n-dim(NulA)=6-1=5

Image of A is 5-dimensional subspace of \(\displaystyle{R}^{{5}}\)(because A has 5 rows).

So, \(\displaystyle{C}{o}{l}{\left({A}\right)}={R}^{{5}}\)

This means that Ax=b has a solution for every b.