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Vector Equations Matrix Equations Homogenous (Matrix Equation…
Vector Equations
Matrix Equations
Homogenous
Matrix Equation
Representation
In matrix equation form this would just be Ax=b
A represents an m x n matrix
x is a column matrix of x variables
b is also a column matrix but of real numbers?
This is only the case where the columns of A equal the number of rows for x
Essentially each x is a weight and each weight is designed for a vector
\(x_1\) is for the first vector which happens to be the first column
\(x_2\) is for the second vector which happens to be the second column
So on so forth
Can be read as column of A times the row of x
Example
Solution
A solution for Ax=b exists if and only if b is a linear combinations of the columns of A
In other words, a solution exists only when
\(x_1a_1+x_2a_2...x_na_n=b \)
Is it consistent
Example
Reducing it we get
Simplifying the last column of the last row we get
\(b_1-\frac{1}{2}+b_3=0\)
This represents the span of \(a_1,a_2,a_3\)
And using this information we can see that A is not consistent for ALL of \(b_1,b_2,b_3\) since we can get a form of \(a \neq 0\) where a is nonzero
Remember the most right hand column can not be a pivot column
Properties
Significance
Now we have 3 ways to view a system of linear equations:
Matrix equation
Vector equation
System of linear equtions
The way to solve them is the same;
row reduce the augmented matrix(6)
Homogeneous Linear System
Form
A system is considered homogeneous if it takes the form Ax=0
Solution
Always has at least one solution which is trivial x=0
A homogeneous system only has a non trivial solution if and only if there's a free variable
Why must there be a free variable?
If a system is consistent it can be unique if there's no free variables or there can be an infinite number of solutions if there are free variables
Free variable
\( \begin{pmatrix} 1 & 0 & -\frac{4}{3}&0\\ 0&1&0&0\\ 0&0&0&0\\ \end{pmatrix} \)
\(x_1-\frac{4}{3}x_3=0\\ x_2=0\\ x_3=\text{free variable}\)
No we can find cases where \(x_1=\frac{4}{3}x_3\) which means we'd have a case where A multiplied to
x
gets us the zero vector where not all the xs are 0s
No free variables
\(x_1=0\\ x_2=0\\ x_3=0\)
This is the zero vector which counts as the trivial case which is always a solution
\( \begin{pmatrix} 3 & 5 & -4&0\\ -3&-2&4&0\\ 6&1&-8&0\\ \end{pmatrix} \)
Reducing it we get
\( \begin{pmatrix} 1 & 0 & -\frac{4}{3}&0\\ 0&1&0&0\\ 0&0&0&0\\ \end{pmatrix} \)
This means
\(x_1-\frac{4}{3}x_3=0\\ x_2=0\\ x_3=\text{free variable}\)
Putting it into parametric vector form
1 more item...
Geometrically
We can view the solutions as spans
\(10x_1-3x_2-2x_3=0\)
x=\( \begin{pmatrix} x_1\\ x_2\\ x_3 \\ \end{pmatrix} \)=\( \begin{pmatrix} .3x_2+.2x_3\\ x_2\\ -x_3 \\ \end{pmatrix} \)=\( \begin{pmatrix} .3x_2+.2x_3\\ x_2+0\\ 0+x_3 \\ \end{pmatrix} \)=\( \begin{pmatrix} .3x_2\\ x_2\\ 0 \\ \end{pmatrix} \)+\( \begin{pmatrix} .2x_3\\ 0\\ x_3 \\ \end{pmatrix} \)
=\(x_2 \begin{pmatrix} .3\\ 1\\ 0 \\ \end{pmatrix} \)+\( x_3\begin{pmatrix} .2\\ 0\\ 1 \\ \end{pmatrix} \)
So we can essentially view this as two vectors being scaled by anything, which is close to what a span is
Put basic variables in terms of free variables then split up the variables. Finally factor out the free variables
Solving it the way we did is also known as
Parametric vector form
It takes the form
x
=s
u
+t
v
Where s and t are all real numbers
Non homogeneous
These solutions can be viewed as a transformation to the homogeneous solution
We just add vector p to the solution set of the homogeneous matrix equation
In 3D
For Homogenous equations
One free variable is a line
Two free variables is a plane
Not entirely sure why
Non-homogenous systems in terms of homogeneous ones
We can view a non-homogeneous system as a vector plus arbitrary linear combinations that satisfy the homogeneous system
An example
#
\( \begin{pmatrix} 3 & 5 & -4&7\\ -3&-2&4&-1\\ 6&1&-8&-4\\ \end{pmatrix} \)
Reducing it we get
\( \begin{pmatrix} 1 & 0 & -\frac{4}{3}&-1\\ 0&1&0&2\\ 0&0&0&0\\ \end{pmatrix} \)
Putting it into parametric vector form we get
x=\( \begin{pmatrix} x_1\\ x_2\\ x_3 \\ \end{pmatrix} \)=\( \begin{pmatrix} -1+\frac{4}{3}x_3\\ 2\\ x_3 \\ \end{pmatrix} \)=\( \begin{pmatrix} -1+\frac{4}{3}x_3\\ 2+0\\ 0+x_3 \\ \end{pmatrix} \)=\( \begin{pmatrix} -1\\ 2\\ 0 \\ \end{pmatrix} \)+\( \begin{pmatrix} \frac{4}{3}x_3\\ 0\\ x_3 \\ \end{pmatrix} \)
=\( \begin{pmatrix} -1\\ 2\\ 0 \\ \end{pmatrix} \)+\( x_3\begin{pmatrix} \frac{4}{3}\\ 0\\ 1 \\ \end{pmatrix} \)
The key thing to note is that the first vector is from the non-homogeneous and the vector with the free variable factored out is from the homogeneous version
#
Theorem
Simply put means if the matrix equation is consistent then our solution set is the same as the homogeneous equation plus a vector
Also written as
x
=
p
+t
v
x
is the result
p
is one particular solution of A
x
=
b
t is a scalar
t
v
is the solution set from the homogeneous equation
How to write parametric vector form
Don't
Examples
The reason why b is true is because there's a pivot in every column for the coefficient matrix
This means we'll never get the case 0 0 0 b where b is nonzero since there's always a number and takes the form
0 0 X b where b could be a nonzero and X is a nonzero
If the coeffiecient matrix had a row of all zeros then there would be a case where
0 0 0 b where b is nonzero
Vector Equations
A matrix with only one column is known as a column vector and as a
vector
Can be any real number
Terminology
\(R^2\),\(R^3\)..\(R^n\)
R represents the real numbers that appear in the vector
The number represents the amount of rows, or you can view them as how many numbers there are in the column
\(R^2\)
\( \begin{pmatrix} 1 \\ .4 \\ \end{pmatrix} \)
\(R^3\)
\( \begin{pmatrix} 1 \\ .4\\ .7 \\ \end{pmatrix} \)
Operations
Vector addition
Tip to tail
Imagine you're walking. A vector tells you how much to go and in what direction
So when we want to add two vectors we follow one vector's order first, order doesn't matter, then the other one
Geometrically this is essentially a parallelogram
Another name is tip to tail
You place the second vector's tail on the first vector's tip
Then the sum would be a line that goes from where you started to where you end
ORDER DOESN'T MATTER
Tip:Arrow head
Tail:End of vector
Scalar multiplication
This can be viewed as stretching or shrinking a vector
Let's say
u
is a vector
5
u
means we stretch
u
out 5 times larger than it was
1/2
u
means we shrink it to half the size
The multiple can be any real number including zero
Vector subtraction
Tail to tail
We can view subtraction as adding
u
-
v
is the same thing as
u
+
-v
-
v
just means same magnitude but opposite direction
u+v
Tail to tail method just means place the vectors tail to tail and the answer will be the line from the tip of the one being subtracted to the tip of the other vector
In our case from the tip of
v
to the tip of
u
Vector into a Matrix
We can set multiple vectors into a matrix by making each vector a column
Turns out the solutions sets are the same
So when we solve for the echelon forms our answers represent the "weights" or \(x_1,x_2...x_n\)
Theorems
To determine whether the columns of A span \(R^4\),determine whether A has a pivot position in every row.
This just means the columns which can be viewed as a vector are on Span {\(a_1\),\(a_2\),...\(a_n\)}
