Download Chapter 8 - James Bac Dang

CHAPTER 8
MATRICES AND
DETERMINANTS
8.1 Matrix Solutions to Linear
Systems
• Objectives
– Write the augmented matrix for a linear
system
– Perform matrix row operations
– Use matrices & Gaussian elimination to solve
systems
– Use matrices & Gauss-Jordan elimination to
solve systems
Solving a system of 3 equations
with 3 variables
• Graphically, you are attempting to find
where 3 planes intersect.
• If you find 3 numeric values for (x,y,z), this
indicates the 3 planes intersect at that
point.
Representing a system of
equations in a matrix
• If a linear system of 3 equations involved 3
variables, each column represents the different
variables & constant, and each row represents a
separate equation.
• Example: Write the following system as a matrix
2x + 3y – 3z = 7
5x + y – 4z = 2
4x + 2y - z = 6
2 3 3 7
5 1 4 2
4 2 1 6
Solving linear systems using
Gaussian elimination
• Use techniques learned previously to
solve equations (addition & substitution) to
solve the system
• Variables are eliminated, but each column
represents a different variable
• Perform addition &/or multiplication to
simplify rows. Have one row contain 2
zeros, a one, and a constant. This allows
you to solve for one variable.
• Work up the matrix and solve for the
remaining variables.
A matrix should look like this after
Gaussian elimination is applied
1
0
0
y1
1
0
z1
z2
1
k1
k2
k3
Examples 3 & 4 (p. 541-543)
• These examples outline a stepwise
approach to use Gaussian elimination.
• One row (often the bottom row) will
contain (0 0 0 1 constant) using this
method.
• After the last variable is solved in the
bottom equation, substitute in for that
variable in the remaining equations.
• GO THROUGH EXAMPLES CAREFULLY
Question: Must the last row
contain (0 0 0 1 constant)?
• Yes, in Gaussian elimination, but there are other
options.
• Look again at example 3, page 539.
• If you multiply the first row by (-1) and add it to
the 2nd row, the result is (-2 0 0 -12)
• What does this mean? -2x = -12, x=6
• By making an informed decision as to what
variables to eliminate, we solved for a variable
much more quickly!
• Next, multiply row 1 by (-1) & add to 3rd row: (-2
2 0 6). Recall x=-6, therefore this becomes
-2(6)+2y=6, y=3
Now knowing x & y, solve for z in any row (row 2?)
6 + 3 + 2z = 19, z=5
Solution: (6,3,5)
If this method is quicker, why would
we use Gaussian elimination?
• Using a matrix to solve a system by
Gaussian elimination provides a standard,
programmable approach.
• When computer programs (may be
contained in calculators) solve systems.
This is the method utilized!
8.2 Inconsistent & Dependent
Systems & Their Applications
• Objectives
– Apply Gaussian elimination to systems
without unique solutions
– Apply Gaussian elimination to systems with
more variables than equations
– Solve problems involving systems without
unique solutions.
How would you know, with
Gaussian elimination, that there are
no solutions to your system?
• When reducing your matrix (attempting to
have rows contain only 0’s, 1’s & the
constant) a row becomes 0 0 0 0 k
• What does that mean? Can you have 0
times anything equal to a non-zero
constant? NO! No solution!
• Inconsistent system – no solution
Graphically, what is happening with
an inconsistent system?
• Recall, with 3 variables, the equation
represents a plane, therefore we are
considering the intersection of 3 planes.
• If a system is inconsistent, 2 or more of
the planes may be parallel OR 2 planes
could intersect forming 1 line and a
different pair of planes intersect at a
different line, therefore there is nothing in
common to all three planes.
Could there be more than one
solution?
• Yes! If the planes intersect to form a line,
rather than a point, there would be
infinitely many solutions. All pts. lying on
the line would be solutions.
• You can’t state infinitely many points,
so you state the general form of all
points on the line, in terms of one of
the variables.
What if your system has 3 variables
but only 2 equations?
• Graphically, this is the intersection of 2
planes.
• 2 planes cannot intersect in 1 point, rather
they intersect in 1 line. (or are parallel,
thus no solution)
• The solution is all points on that line.
• The ordered triple is represented as one of
the variables (usually z) and the other 2 as
functions of that variable: ex: (z+2,3z,z)
Dependent system
• Notice when there were infinitely many
solutions, two variables were stated in
terms of the 3rd. In other words, the x & y
values are dependent on the value
selected for z.
• If there are infinitely many solutions, the
system is considered to be dependent.
8.3 Matrix Operations & Their
Applications
• Objectives
– Use matrix notation
– Understand what is meant by equal matrices
– Perform scalar multiplication
– Solve matrix equations
– Multiply matrices
– Describe applied situations with matrix
operations
What is a matrix?
• A set of numbers in rows & columns
• m x n describes the dimensions of the matrix (m
rows & n columns)
• The matrix is contained within brackets.
• Example of a 3 x 3 matrix:
4 2 5
1  4 3


0  3 8
Add & Subtract matrices
• Only if they are of the same dimensions
• Add (or subtract) position by position (i.e.
the term in the 3rd row 2nd column of the 1st
matrix + the term in the 3rd row 2nd column
of the 2nd matrix)
Scalar Multiplication
• When the multiplier is on the outside of the
matrix, every term in the matrix is multiplied by
that constant.
1  4  4  16
4 



5  2 20  8 
• You can combine multiplication and addition of
matrices (PRS on next slide)
Find 3A + B if
3
1) 
6
9
2) 
18
3
3) 
6
5
4) 
10
 2

4
 6

12 
3

9
0

8
1 1
2  3
A
,B  


2 3
4 1 
Multiplication of matrices
• When multiplying matrices, order matters!
• The first row of the 1st matrix is multiplied
(term by term) by the 1st column of the 2nd
matrix. The sum of these products is the
new term for the element in the 1st row, 1st
column of the new product matrix.
• The # columns in 1st matrix MUST equal #
rows in 2nd matrix (otherwise terms won’t
match up in the term by term
multiplication)
Multiply:
•
•
•
•
•
•
•
•
•
 3 4  3 1  1 2
 2 5  4  2 1 0

 

1)1st row of 1st by 1st column of 2nd (3(3)+4(4)=25)
2) 1st row of 1st by 2nd column of 2nd (3(1)+4(-2)=-5)
3)1st row of 1st by 3rd column of 2nd (3(-1)+4(1)=1)
4)1st row of 1st by 4th column of 2nd (3(2)+4(0)=6)
5)2nd row of 1st by 1st column of 2nd (-2(3)+5(4)=14)
6)2nd row of 1st by 2nd column of 2nd (-2(1)+5(-2)=-12)
7)2nd row of 1st by 3rd column of 2nd (-2(-1)+5(1)=7)
8)2nd row of 1st by 4th column of 2nd (-2(2)+5(0)=-4)
(continue on next slide)
New matrix is created
• Multiplying 1st row by 1st column fills position 11,
multiplying 1st row by 2nd column fills position 12,
etc
• Result: 25
5 1 6 
14  12 7  4


• Compare this matrix with the results from the
previous slide
8.4 Multiplicative Inverses of
Matrices & Matrix Equations
• Objectives
– Find the multiplicative inverse of a square
matrix
– Use inverses to solve matrix equations
– Encode & decode messages
What is the multiplicative inverse of
a matrix?
• It’s the matrix that you must multiply another
matrix by to result in the identity matrix.
• What is the identity matrix? It’s the matrix that
you would multiply another matrix by that would
not change the value of the original matrix.
• The identity matrix of a 3x3 matrix is:
1 0 0
0 1 0 


0 0 1
How do we find the multiplicative
inverse of a matrix?
• Example 2, p. 574, outlines the steps. In general,
if it’s a 2x2 matrix, you’re finding the value of 4
unknowns using 2systems of equations, each with
2 variables.
• Example on next slide for finding a multiplicative
inverse of a matrix ( A1 ), which means the
multiplicative inverse, NOT 1/A.
Find the multiplicative
inverse of
5 1 
 2 2


5 1 a b  1 0
 2 2    c d   0 1 

 
 

5a  c  1
 8a  2, a  1 / 4, c  1 / 4
2a  2c  0

8
b

1
,
b


1
/
8
,
d

9
/
40
5b  d  0
2b  2d  1
• Multiplicative inverse is:  1 / 4  1 / 8
 1 / 4 9 / 40


Quick method for finding the
multiplicative inverse of a 2x2
matrix
a b  1
 d  b
1
,A 

• If A  


c
d

c
a
ad

bc




What if the matrix is greater than a
2x2?
• Form an augmented matrix A I  where A is
the original matrix & I is the identity matrix.
• Perform elimination and substitution until the
original matrix (A) appears as the identity matrix.
• The resulting matrix left where the identity matrix
began is the multiplicative inverse matrix. ( A1 )
8.5 Determinants & Cramer’s Rule
• Objectives
– Evaluate a 2nd-order determinant
– Solve a system of linear equations in 2
variables using Cramer’s rule
– Evaluate a 3rd-order determinant
– Solve a system of linear equations in 3
variables using Cramer’s rule
– Use determinants to identify inconsistent &
dependent systems
– Evaluate higher-order determinants
Determinant of a 2x2 matrix
• If A is a matrix, the determinant is
A
3 2
 3  2
A
,A
 3(5)  (2)4  23

4 5
4 5 
When are determinants useful?
• They can be used to solve a system of equations
• Cramer’s Rule
a1 x  b1 y  c1
a2 x  b2 y  c2
c1 b1
c2 b2
x
,y
a1 b1
a2 b2
a1
a2
a1
a2
c1
c2
b1
b2
Finding a determinant of a 3x3
matrix
• More complicated, but it can be done!
a1
a2
a3
b1
b2
b3
c1
b2
c2  a1 
b3
c3
c2
b1
 a2 
c3
b3
c1
b1
 a3 
c3
b2
c1
c2
• It’s often easier to pick your “home row/column”
(the one with the multipliers) to be a row/column
that has one or more zeros in it.
Determinants can be used to solve a
linear system in 3 variables
CRAMER’S RULE
Dy
Dx
Dz
x
,y
,z 
D
D
D
What is D and Dx , D y , Dz
• D is the determinant that results from the
coefficients of all variables.
• Dx is the determinant that results when each x
coefficient is replaced with the given constants.
• D y is the determinant that results when each y
coefficient is replaced with the given constants.
• Dz is the determinant that results when the z
coefficients are replaced with the given constants.
Find z, given
2x + y = 7
-x + 3y + z = 5
3x + 2y – 4z = 10
2
1
7
1)
2
1
3
2
1
3
( 2)
2
1
3
1 0
3 1
5 10
1 0
3 1
2 4
1 0
3 1
2 4
1 7
3 5
2 10
2
1
3
3)
2
1
3
2
1
3
( 4)
2
1
3
1 7
3 5
2 10
1 0
3 1
2 4
1 7
3 5
2 10
1 7
3 1
2 4