First review:
linear systems
Solve
the following linear system of equations (3 methods shown):

method 1: elimination
and backward substitution:


3x + 4y = 20
2x  10y = 8


first
solve equation 1 for x and get

x
= (20  4y) / 3





substitute this into the
second equation and solve for y

2 ( 20
– 4y )/ 3 
10y = 8
40 – 8y  30y = 24
y = 16 / 38 = 8 / 19


substitute this value for
into the first equation:

3x + 4 (8
/ 19) = 20
57x + 32 = 380
x = 348 / 57 = 116 / 19





check: 3(
116/19 ) + 4(8/19)
= (348+32)/19 = 380/19 = 20.

(either equation can be used)









method 2: cancel terms
by multiplying constants and adding equations


2{ 3x + 4y = 20}


multiple the first equation
by 2, and the second by 3, then add

3{2x  10y = 8}





after multiplication

6x + 8y = 40



6x +30y = 24


now add

0x +38y = 16



38y = 16



y = 16/38 = 8/19


Now substitute this result
back into either equation and solve for x.



Instead let's use the
multiplication technique to solve for x

5{ 3x + 4y = 20}
2{2x  10y = 8}


multiply the equations as
shown to eliminate y

15x
+ 20y = 100
4x 20y = 16


now add the equations

19x +0 = 116


solve for x

x = 116/19






method 3:
make coefficients along the diagonal 1 and all others zero.


Original equations

3x + 4y = 20



2x 10y = 8




1

divide 1st
equation by 3

1x + 4/3 y = 20/3



2x  10y = 8

2

multiply 1st
equation by 2 then add to second equation (to make that
coefficient 0)

2x 8/3y = 40/3



0x 38/3 y = 16/3




3

rewrite the
equations

1x + 4/3y = 20/3



38/3y = 16/3

4

multiply the 2nd
equation by 3/38

1 y = 16/38 = 8/19




5

rewrite the
equations

x + 4/3y = 20/3


the diagonals are
both 1 now

1y = 8/19

6

multiply the 2nd
equation by 4/3 and add the the first to eliminate the y term in
the first equation

4/3y = 32/57



1x + 0y = 348/57 =
116/19




7

rewrite the
equations

x = 116/19



y = 8/19


Notice, that once
we get coefficients equal to one along the diagonal and zero
elsewhere we have the solution to the system.


Notice
that the denominator is the same for x and y, 19. This is always the
case. Also there is a relationship between the x and y coefficients
that will yield the numerators of the answers. To see this we'll
represent this system with a matrix. Using a matrix and its
determinate can make solving systems of linear equations faster then
the previous two methods. The formulas obtained are for
explanation only! We will use visualization instead as explained
later.
First,
let's talk about matrices: (mā'  trə  sēs)
A
matrix is a table of elements. These elements can be just about
anything, usually the same class of objects. Addition, subtraction,
multiplication, division and multiplication by a constant are defined
for matrices. ( These operations will be discussed in a later course
and are not important for their application solving linear systems of
equations.)
Here
are some matrices:
The
first matrix has 2 rows and 2 columns: row one contains the elements
1 and 2, row two contains the elements 3 and 4
and 2
columns: column 1 contains the elements 1 and 3 and column 2 contains
the elements 2 and 4.
The
second matrix has only one row.
The
third matrix has only one column.
The
fourth matrix has two rows and three columns. This matrix is of
interest for us to solve the previous system of linear equations.
In
general terms it's customary to write the elements in a matrix for
reference like so:
a_{ij}_{
}where i is the row number and j is the column number, both
starting from 1.
for
example:
So
in the example matrices above, consider the first matrix:
a_{11
}= 1 a_{12 }= 2 a_{21 }= 3 a_{22 }= 4
Every
matrix has a value associated with it called the determinant. It is
found by multiplying crossing diagonals in a certain way. First it is
represented like a matrix but instead of using parenthesis, vertical
bars are used like so:
The
determinate is found like so: ( a_{11 }• a_{22 })
– ( a_{21 } • a_{12 })
An
easy way to remember how to calculate a determinant is visualizing
the previous diagram. Note the direction of the arrows and the plus
and minus signs.
Start
at 1, multiply the elements along the diagonal starting at 1.
Multiply the elements along the second diagonal starting at 2 and
subtract that from the first product.
This
method of solving linear equations is known as Cramer's Rule.
Now
let's solve the previous system of equations.
First
let's write our two equations in terms of a matrix, but before we do
this let's simplify and reduce the equations as much as possible.
Doing this will make the necessary calculations easier to do.
original:
3x
+ 4y = 20
2x
 10y = 8
The
second equation can have a 2 factored out and we get:
3x
+ 4y = 20
x
– 5y = 4
Now
we're ready to go.
This matrix is the coefficients of
the variables.
Now
evaluate the determinant:
(3)
( 5 )  (1)( 4) = – 15 – 4 = – 19
19
is the denominator of the solutions for x and y. If this number is
ever 0, then there is no solution to the system of equations.
Now
the solutions for x and y also have similar determinants:
First
x: replace the x column with the b column:
we
have (20)(5)  (4)(4) = 100 16 = 116
Since
this is the numerator for x we have x = 116/19 = 116/19
Now
for y:
Numerator
for y is (3)(4)  (1)(20)
12
 20 = 8
So
y = 8 / 19 = 8/19
Solution
is (116/19, 8/18)
Example:
solve
5x
+ 4y = 42
3x
+ y = 12
now
mentally visualize the 3 determinants
the
denominator is: (5)(1)  (3)(4) = 5 + 12 = 17
the
numerator for x is (42)(1)  (12)(4) = 42 + 48 = 90
then
numerator for y is (5)(12)  (3)(42) = 60 +126 = 66
solution
x = 90/17, y = 66/17
check:
5(90/17) + 4(66/17) = 714/17 = 42
Another
example:
2x
– 5y = 16
x
+ 4y = 20
denominator:
(2)(4) – (1)(5) = 8 + 5 = 13
x:
(4)(16)  (20)(5) = 64 + 100 = 164
y:
(2)(20)  (1)(16) = 40 16 = 24
solution:
x = 164/13 y = 24/13
check
2^{nd} eq: 164/13 + 4*24/13 = (164 + 96)/13 = 260 / 13 = 20
Here's
an alternate visualization (equivalent to before) to remember how to
calculate the numerators for x and y.
If
you look closely to the way we crossed out columns, what remains is
the y or x column and the b column. In effect the b column replaces
the column of the variable of interest. So what remains is:
for
x:
for
y:
If
you carry out the multiplications and additions on these determinants
as normal, you will find the same formulas given earlier.
Important
note here: Do not memorize the formulas!!! Use visualization instead!
Advanced:
Solve a linear system of equations
of 3 variables, x, y, and z.
The previous method will work
(theoretically this is true for any number of variables, but
practically becomes cumbersome and better techniques are available.)
Solve:
4x + y + z = 0
3x  y + z = 2
x + y + 2z =  1
matrix:
determinant:
for the denominator we must find the
value of the determinant:
The quickest way to evaluate this 3
by 3 determinant is to use Sarrus' rule. We'll copy the first two
columns in the determinant past the last column to create 6 diagonals
with 3 elements apiece then multiply them and add or subtract for the
result.
Notice, in dark black are the
columns of the original determinant. The first two columns are
repeated in green as columns 4 and 5. We've 'augmented' this
determinant so that we can create diagonals with 3 elements each. Now
we need to multiple the elements for each diagonal and add the red
diagonals together and subtract the blue diagonals from that sum.
(Zero products will be shown in this example only for clarity.)
(4)(1)(2) + (1)(1)(1) + (1)(3)(1) 
(1)(1)(1)  (1)(1)(4)  (2)(3)(1)
8 + 1 + 3 + 1  4  6 = 13 ( which
is the denominator)
Now x: (first column is replaced
with the b column)
(0)(1)(2) + (1)(1)(1) + (1)(2)(1)
 (1)(1)(1)  (1)(1)(0)  (2)(2)(1)
0 1  2  1  0 + 4 = 0
Now y:
(4)(2)(2) + (0)(1)(1) + (1)(3)(1)
 (1)(2)(1)  (1)(1)(4)  (2)(3)(0)
16 + 0  3 + 2 + 4  0 = 13
Now z:
(4)(1)(1) + (1)(2)(1) + (0)(3)(1)
 (1)(1)(0)  (1)(2)(4)  (1)(3)(1)
4  2 + 0  0 + 8 + 3 = 13
x = 0, y = 13/13 = 1, z = 13/13 =
1
(x,y,z) = (0, 1, 1)
Verify that this solution does solve
the three original equations.
Below are the plots of these three
planes, showing how each pair intersect in a line then how all three
intersect in one point.
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