12.1 Systems of Linear Equations: Substitution and Elimination

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12.1 Systems of Linear Equations Substitution
and Elimination
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A system of equations is a collection of two or
more equations, each containing one or more
variables.
A solution of a system of equations consists of
values for the variables that reduce each
equation of the system to a true statement.
To solve a system of equations means to find all
solutions of the system.
When a system of equations has at least one
solution, it is said to be consistent otherwise
it is called inconsistent.
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An equation in n variables is said to be linear
if it is equivalent to an equation of the form
where
are n distinct variables,
are constants, and at least one of the
as is not zero.
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If each equation in a system of equations is
linear, then we have a system of linear equations.
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Two Linear Equations Containing Two Variables
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If the graph of the lines in a system of two
linear equations in two variables intersect, then
the system of equations has one solution, given
by the point of intersection. The system is
consistent and the equations are independent.
y
Solution
x
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If the graph of the lines in a system of two
linear equations in two variables are parallel,
then the system of equations has no solution,
because the lines never intersect. The system is
inconsistent.
y
x
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If the graph of the lines in a system of two
linear equations in two variables are coincident,
then the system of equations has infinitely many
solutions, represented by the totality of points
on the line. The system is consistent and
dependent.
y
x
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Two Algebraic Methods for Solving a System
1. Method of substitution 2. Method of
elimination
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Use Method of Substitution to solve
(1)
(2)
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Use graphing utility to solve the previous system
of equations
(1) (2)
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A system of three linear equations containing
three variables has either

• Exactly one solution (consistent system with
  independent equations).
• No solution (inconsistent system).
• Infinitely many solutions (consistent system with
  dependent equations).

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(1)
Solve
(2)
(3)
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12.2 Systems of Linear Equations Matrices
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A matrix is defined as a rectangular array of
numbers,
Column j
Column n
Column 1
Column 2
Row 1
Row 2
Row i





Row 4
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Augmented Matrix
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Row Operations on an Augmented Matrix
1. Interchange any two rows.
2. Replace a row by a nonzero multiple of that
row.
3. Replace a row by the sum of that row and a
constant multiple of some other row.
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Solve
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Matrix Method for Solving a System of Linear
Equations (Row Echelon Form)

• Write the augment matrix of the system.
• Perform operations that place the number 1 in row
  1 column1.
• Perform operations that leave the entry 1 in row
  1 and column1 unchanged, while causing 0s appear
  below it in column1.

• Perform operations that place 1 in row 2 column
  2, but leave the entries in columns to the left
  unchanged. The perform operations to place 0s
  below it.(If rows that contain only 0s are
  obtained place them to the bottom of the matrix.)
• Repeat previous step.

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Solve
using a graphing utility.
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Solve using a graphing utility
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12.3 Systems of Linear EquationsDeterminants
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If a, b, c, and d are four real numbers, the
symbol
is called a 2 by 2 determinant. Its value is
ad-bc that is
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Theorem Cramers Rule
The solution to the system of equations
is given by
D ad-bc
provided that
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A 3 by 3 determinant is symbolized by
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12.4Matrix Algebra
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Consider the two matrices
(a) Find AB
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(b) Find A-B.
(c) Find 3A
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Let A denote an m by r matrix and let B denote an
r by n matrix. The product AB is defined as the
m by n matrix whose entry in row i, column j is
the product of the ith row of A and the jth
column of B.
The definition of the product AB of matrices A
and B, in this order requires that the number of
columns in A equal the number of rows of B.
Otherwise the product is undefined.
A m by r
B r by n
Must be same for AB to exist.
AB is m by n.
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Find the product AB of

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Associative Property
A(BC)(AB)C
Distributive Property
A(BC)ABAC
Commutative Property
Matrix multiplication is not commutative.
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Identity Property
If A is an m by n matrix, then
Im A A and AIn A
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Let A be a square n by n matrix. If there exists
an n by n matrix A-1, read A inverse, for
which
Then A-1 is called the inverse of the matrix A.
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Find the inverse of
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Procedure for Finding the Inverse of a
Nonsingular Matrix

• Form the matrix AIn.
• Transform the matrix AIn into reduced row
  echelon form.
• The reduced row echelon of AIn form will
  contain the identity matrix In on the left of
  the vertical bar the n by n matrix on the right
  of the vertical bar is the inverse of A.

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12.6 Systems of Nonlinear Equations
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Two algebraic methods for solving systems of
equations 1. Substitution 2. Elimination
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Solve by substitution
(1)
(2)
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12.7 Systems of Inequalities
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Graph the linear inequality 2x - y - 8 gt0.
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2x - y -8 gt 0
Test point (1, 3)
Does not belong to graph
2(1) - 3 - 8 -9 lt 0
Test point (5, 1)
Belongs to graph
2(5) - 1 - 8 1 gt 0
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Steps for Graphing an Inequality by Hand

• Replace the inequality symbol by an equal sign
  and graph the resulting equation. If the
  inequality is strict, use dashes otherwise use
  solid mark. This graph separates the xy-plane
  into two regions.
• Select a test point P in one of the regions.
• If the coordinates of P satisfy the inequality,
  then so do all the points in that region. Shade
  that region.
• If the coordinates of P do not satisfy the
  inequality, then none of the points in that
  region do. Shade the opposite region.

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Steps for Graphing an Inequality Using a Graphing
Utility

• Replace the inequality symbol by an equal sign
  and graph the resulting equation. This graph
  separates the xy-plane into two regions.
• Select a test point P in one of the regions.
• Use graphing utility to check if P satisfy the
  inequality. If it does then so do all the points
  in that region. Use graphing utility to shade
  that region.
• If the coordinates of P do not satisfy the
  inequality, then none of the points in that
  region do. Shade the opposite region.

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Graph
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y x - 6
y -2x 5
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3
1
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