# Adv. Linear Algebra: Review Topics Chapter 1

Here is the link to my review topics for exam 1.

http://math.uh.edu/~bgb/Courses/Math4377/Math4377-Ex1-Topics.pdf

1.2

Does the solution set to a linear system change under elementary row operations?

The solution set does not change under elementary row operations.

What are independent/free variables? How can we tell there are free variables by looking at the row-reduced echelon form?

I would give an example:

1 0 0 0
0 1 0 0
0 0 0 0

x1=x2=independent, x3=x4=free=linear combination of alpha1 (0 1 0 0) & alpha2 (0 0 0 0)

1.3

How can we rewrite a linear system Ax = b in vector form?

Ax=b

A=

A11 ... A1n
.
.
.
Am1 ... Amn

x=

x1
.
.
.
xn

b=

b1
.
.
.
bm

Can we solve the system if b can be written as a linear combination of the column vectors of A?

Yes, I would just put it in augmented form and apply elementary row reductions. I would find that some columns have no pivot variables, thus the variable in that column is free, which is the consequence of being a linear combination of other columns.

1.4

How do the solutions to an inhomogeneous system relate to the solutions of the corresponding homogeneous one?

Homogeneous systems are equal to zero a1x1+...anxn=0, thus it is linear independent in which the a1=...=an=0 or it contains only the trivial solution, all xn=0.

Inhomogeneous systems are not equal to zero, thus it's solutions will not all be zero.

THANKS!!!

The solution set does not change under elementary row operations.
Why?

What are independent/free variables? How can we tell there are free variables by looking at the row-reduced echelon form?

I would give an example:

1 0 0 0
0 1 0 0
0 0 0 0

x1=x2=independent, x3=x4=free=linear combination of alpha1 (0 1 0 0) & alpha2 (0 0 0 0)
Hmm... that's not a good example.

How can we rewrite a linear system Ax = b in vector form?

Ax=b

A=

A11 ... A1n
.
.
.
Am1 ... Amn

x=

x1
.
.
.
xn

b=

b1
.
.
.
bm
Sorry, I don't understand this.

Can we solve the system if b can be written as a linear combination of the column vectors of A?

Yes, I would just put it in augmented form and apply elementary row reductions. I would find that some columns have no pivot variables, thus the variable in that column is free, which is the consequence of being a linear combination of other columns.
It is not necessarily true that some of the variables will be free.

How do the solutions to an inhomogeneous system relate to the solutions of the corresponding homogeneous one?

Homogeneous systems are equal to zero a1x1+...anxn=0, thus it is linear independent in which the a1=...=an=0 or it contains only the trivial solution, all xn=0.

Inhomogeneous systems are not equal to zero, thus it's solutions will not all be zero.
That doesn't explain how they are related.

Defennder
Homework Helper
How can we rewrite a linear system Ax = b in vector form?

Ax=b

A=

A11 ... A1n
.
.
.
Am1 ... Amn

x=

x1
.
.
.
xn

b=

b1
.
.
.
bm
That looks ok, if they would accept matrices as part of the answer. Otherwise you're better off writing the matrix A as a linear combination of all its column vectors with the entries of column vector X as coefficients.

How do the solutions to an inhomogeneous system relate to the solutions of the corresponding homogeneous one?

Homogeneous systems are equal to zero a1x1+...anxn=0, thus it is linear independent in which the a1=...=an=0 or it contains only the trivial solution, all xn=0.

Inhomogeneous systems are not equal to zero, thus it's solutions will not all be zero.

THANKS!!!

THis not only doesn't explain the relationship between the solutions of an inhomogeneous system of equations to those of corresponding homogeneous, but it is even not completely true. Because you said that a1=...=an=0 but this is true only when the system has a unique solution, but there are cases where the system has infinitely many solutions, so not all of a1, a2,....,an need to be zero., and thus the vector columns of the corresponding coefficient matrix need not necessarily be independent, like i just pointed out.

The relationship between the solutions of an inhomogeneous system of equations to thos of a homogeneous is that each of them differs by a constant term $$b_i$$ i=1,2,....,m

That is if the sols of the hom. syst. of eq, are say, $$x_i=kx_j$$ then those of the inhomog. roughly speaking will be,

$$x_i=kx_j+b_i; : b_i=constant$$

Why?

.

Because there is a theorem that states exactly this, and that we all, i assume, have proved it!