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Summary Linear Algebra 2DM80
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Summary of Linear Algebra course Biostatistics and Linear Algebra for first year medical sciences and technology or biomedical engineering.
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Linear Algebra and it’sapplications
Jaar 1 Kwartiel 3
1.1 systems of linear equations
A linear equation in the variables x1, …, xn is an equation that can be written
in the form:
a 1 x 1+a 2 x 2+…+ anxn=b
A system of linear equations or a linear system is a collection of one or more
linear equations involving the same variables – say, x1, …, xn.
A solution of the system is a list (s1, s2, … sn) of numbers that make each
equation a true statement when the values s1, …, sn are substituted for x1, …,
xn. The set of all possible solutions is called the solution set of the linear
system. Two linear systems are called equivalent if they have the same solution
set.
A system of linear equations has
1. no solution
2. exactly one solution
3. infinitely many solutions.
A system of linear equations is said to be consistent if it has either one solution
or infinitely many solutions; a system is inconsistent if it has no solution.
Linear system:
x 1−2 x 2+ x 3=0
2 x 2−8 x 3=8
5 x 1−5 x 3=10
Coefficient matrix:
[ ]
1 −2 1
0 2 −8
5 0 −5
Augmented matrix:
[ ]
1 −2 1 0
0 2 −8 8
5 0 −5 10
If m and n are positive integers, an m x n matrix is a rectangular array of
numbers with m rows and n columns.
The basic strategy to solve a linear system is to replace on system with an
equivalent system that is easier to solve. Three basic operations are used to
simplify a linear system: replace one equation by the sum of itself and a multiple
of another equation, interchange two equations, and multiply all the terms in an
equation by a nonzero constant.
If the augmented matrices of two linear systems are row equivalent, then the two
systems have the same solution set.
Two fundamental questions about a linear system:
1
, 1. Is the system consistent; that is, does at least one solution exist?
2. If a solution exists, is it the only one; that is, is the solution unique?
1.2 row reduction and echelon forms
A rectangular matrix is in echelon form (or row echelon form) it has it has the
following three properties:
1. All nonzero rows are above any rows of all zeros.
2. Each leading entry of a row is in a column to the right of the leading entry
of the row above it.
3. All entries in a column below a leading entry are zeros.
If a matrix is in echelon form satisfies the following additional conditions, then it
is in reduced echelon form (or reduced row echelon form).
4. The leading entry in each nonzero row is 1.
5. Each leading 1 is the only nonzero entry in its column.
Reduced echelon matrix:
[ ]
1 0 0 29
0 1 0 16
0 0 13
Uniqueness of the reduced echelon form:
Each matrix is row equivalent to one and only on reduced echelon
matrix.
Since the reduced echelon form is unique, the leading entries are always in the
same positions in any echelon form obtained from a given matrix.
A pivot position in a matrix A is a location in A that corresponds to a leading 1 in
the reduced echelon form of A. A pivot column is a column of A that contains a
pivot position.
A pivot is a nonzero number in a pivot position that is used as needed to create
zeros via row operations.
The row reduction algorithm
1. Begin with the leftmost nonzero column. This is a pivot column. The pivot
position is at the top.
2. Select a nonzero entry in the pivot column as a pivot. If necessary,
interchange rows to move this entry into the pivot position.
3. Use row replacement operations to create zeros in all positions below the
pivot.
4. Cover or ignore the row containing the pivot position and cover all rows, if
any, above it. Apply steps 1-3 to the submatrix that remains. Repeat the
process until there are no more nonzero rows to modify.
5. Beginning with the rightmost pivot and working upward and to the left,
create zeros above each pivot. If a pivot is not a 1, make it 1 by a scaling
operation.
The variables corresponding to pivot columns in the matrix are called basic
variables. The other variables are called free variables.
[ ]{ }
1 0 −5 1 x 1=1+5 x 3
0 1 1 4 x 2=4−x 3 parametric description
0 0 0 0 x 3 is free
2
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