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| Hallhair |
 Posted: Wed Feb 09, 2005 12:08 pm |
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i can't figure out how to do 4x4 augmented matrices like
3a+5b-2c+d=3
7a-3b+4c-3d=15
4a+b-c+5d=-2
2a-4b+6c-d=11
if anyone can help me with this great i just can seem to get it i get
3x3 and 2x2's but i can't get 4x4's thanks |
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| Robin Chapman |
| Posted: Wed Feb 09, 2005 12:15 pm |
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Hallhair wrote:
[quote:ef1218988b]i can't figure out how to do 4x4 augmented matrices like
3a+5b-2c+d=3
7a-3b+4c-3d=15
4a+b-c+5d=-2
2a-4b+6c-d=11
if anyone can help me with this great i just can seem to get it i get
3x3 and 2x2's but i can't get 4x4's thanks
[/quote:ef1218988b]
The augmented matrix is
(3 5 -2 1 3)
(7 -3 4 -3 15)
(4 1 -1 5 -2)
(2 -4 6 -1 11).
If I had to solve these equations, I'd reduce the above matrix
to echelon form by elementary row operations and then read
off the solution(s).
--
Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html
"Elegance is an algorithm"
Iain M. Banks, _The Algebraist_ |
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| Hallhair |
| Posted: Thu Feb 10, 2005 11:57 am |
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Robin Chapman <rjc@ivorynospamtower.freeserve.co.uk> wrote in message news:<cudgfa$mr6$1@south.jnrs.ja.net>...
[quote:2d22af4361]Hallhair wrote:
i can't figure out how to do 4x4 augmented matrices like
3a+5b-2c+d=3
7a-3b+4c-3d=15
4a+b-c+5d=-2
2a-4b+6c-d=11
if anyone can help me with this great i just can seem to get it i get
3x3 and 2x2's but i can't get 4x4's thanks
The augmented matrix is
(3 5 -2 1 3)
(7 -3 4 -3 15)
(4 1 -1 5 -2)
(2 -4 6 -1 11).
If I had to solve these equations, I'd reduce the above matrix
to echelon form by elementary row operations and then read
off the solution(s).
[/quote:2d22af4361]
k so i know you would do that but how can somone please show me step
by step or somthing so i can understand this becsaue i don't get it. |
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| Robin Chapman |
| Posted: Thu Feb 10, 2005 12:15 pm |
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Guest
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Hallhair wrote:
[quote:63029d6ce5]Robin Chapman <rjc@ivorynospamtower.freeserve.co.uk> wrote in message
news:<cudgfa$mr6$1@south.jnrs.ja.net>...
Hallhair wrote:
i can't figure out how to do 4x4 augmented matrices like
3a+5b-2c+d=3
7a-3b+4c-3d=15
4a+b-c+5d=-2
2a-4b+6c-d=11
if anyone can help me with this great i just can seem to get it i get
3x3 and 2x2's but i can't get 4x4's thanks
The augmented matrix is
(3 5 -2 1 3)
(7 -3 4 -3 15)
(4 1 -1 5 -2)
(2 -4 6 -1 11).
If I had to solve these equations, I'd reduce the above matrix
to echelon form by elementary row operations and then read
off the solution(s).
k so i know you would do that but how can somone please show me step
by step or somthing so i can understand this becsaue i don't get it.
[/quote:63029d6ce5]
There's a lot of steps! Have you done row reduction for smaller matrices:
it's the same procedure. To start row reduction on the above matrix
you would replace
row 2 by row 2 - (7/3) row 1,
row 3 by row 3 - (4/3) row 1 and
row 4 by row 4 - (2/3) row 1.
That would put zeroes below the top left entry 3. (I'm not doing this
arithmetic for you!)
Can you do that?
--
Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html
"Elegance is an algorithm"
Iain M. Banks, _The Algebraist_ |
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| Gerry Myerson |
| Posted: Thu Feb 10, 2005 5:15 pm |
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In article <cug4sa$gih$1@south.jnrs.ja.net>,
Robin Chapman <rjc@ivorynospamtower.freeserve.co.uk> wrote:
[quote:d5496ba027]Hallhair wrote:
Robin Chapman <rjc@ivorynospamtower.freeserve.co.uk> wrote in message
news:<cudgfa$mr6$1@south.jnrs.ja.net>...
Hallhair wrote:
i can't figure out how to do 4x4 augmented matrices like
3a+5b-2c+d=3
7a-3b+4c-3d=15
4a+b-c+5d=-2
2a-4b+6c-d=11
if anyone can help me with this great i just can seem to get it i get
3x3 and 2x2's but i can't get 4x4's thanks
The augmented matrix is
(3 5 -2 1 3)
(7 -3 4 -3 15)
(4 1 -1 5 -2)
(2 -4 6 -1 11).
If I had to solve these equations, I'd reduce the above matrix
to echelon form by elementary row operations and then read
off the solution(s).
k so i know you would do that but how can somone please show me step
by step or somthing so i can understand this becsaue i don't get it.
There's a lot of steps! Have you done row reduction for smaller matrices:
it's the same procedure. To start row reduction on the above matrix
you would replace
row 2 by row 2 - (7/3) row 1,
row 3 by row 3 - (4/3) row 1 and
row 4 by row 4 - (2/3) row 1.
That would put zeroes below the top left entry 3. (I'm not doing this
arithmetic for you!)
Can you do that?
[/quote:d5496ba027]
Doing it by hand, I'd be more inclined to avoid fractions
(or at least delay the inevitable) by using the 1 in row 1
as a pivot instead: add 3 times row 1 into row 2,
subtract 5 times row 1 from row 3, add row 1 into row 4.
--
Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) |
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| Hallhair |
| Posted: Fri Feb 11, 2005 11:17 am |
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Guest
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Robin Chapman <rjc@ivorynospamtower.freeserve.co.uk> wrote in message news:<cug4sa$gih$1@south.jnrs.ja.net>...
[quote:7e33b48437]Hallhair wrote:
Robin Chapman <rjc@ivorynospamtower.freeserve.co.uk> wrote in message
news:<cudgfa$mr6$1@south.jnrs.ja.net>...
Hallhair wrote:
i can't figure out how to do 4x4 augmented matrices like
3a+5b-2c+d=3
7a-3b+4c-3d=15
4a+b-c+5d=-2
2a-4b+6c-d=11
if anyone can help me with this great i just can seem to get it i get
3x3 and 2x2's but i can't get 4x4's thanks
The augmented matrix is
(3 5 -2 1 3)
(7 -3 4 -3 15)
(4 1 -1 5 -2)
(2 -4 6 -1 11).
If I had to solve these equations, I'd reduce the above matrix
to echelon form by elementary row operations and then read
off the solution(s).
k so i know you would do that but how can somone please show me step
by step or somthing so i can understand this becsaue i don't get it.
There's a lot of steps! Have you done row reduction for smaller matrices:
it's the same procedure. To start row reduction on the above matrix
you would replace
row 2 by row 2 - (7/3) row 1,
row 3 by row 3 - (4/3) row 1 and
row 4 by row 4 - (2/3) row 1.
That would put zeroes below the top left entry 3. (I'm not doing this
arithmetic for you!)
Can you do that?
[/quote:7e33b48437]
can somone please do the 16 steps for me or show me a pattern
i know a 4x4 would be for example (2 4)
(5 5) i know you have to first make
the bottom left corner 0 the make the right bot corner 1 then the 4
to a 0 then the 2 to a 1 and i know on a 3x3 like (4 5 6 )
(7 8 9 )
(1 2 3 )
you would first make the 1 a 0 then you would make the 7 a 0 then
make the 2 a 0 then make the 3 a 1 then make the 9 a 0 then make the 8
a 1 then make the 6 a 0 then make the 5 a 0 then make the 4 a 1
so since the pattern changes for 2x2 to 3x3 what is the pattern for a
4x4?? |
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| Robin Chapman |
| Posted: Fri Feb 11, 2005 11:22 am |
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Guest
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Hallhair wrote:
[quote:87403ec026]Robin Chapman <rjc@ivorynospamtower.freeserve.co.uk> wrote in message
news:<cug4sa$gih$1@south.jnrs.ja.net>...
Hallhair wrote:
Robin Chapman <rjc@ivorynospamtower.freeserve.co.uk> wrote in message
news:<cudgfa$mr6$1@south.jnrs.ja.net>...
Hallhair wrote:
i can't figure out how to do 4x4 augmented matrices like
3a+5b-2c+d=3
7a-3b+4c-3d=15
4a+b-c+5d=-2
2a-4b+6c-d=11
if anyone can help me with this great i just can seem to get it i
get 3x3 and 2x2's but i can't get 4x4's thanks
The augmented matrix is
(3 5 -2 1 3)
(7 -3 4 -3 15)
(4 1 -1 5 -2)
(2 -4 6 -1 11).
If I had to solve these equations, I'd reduce the above matrix
to echelon form by elementary row operations and then read
off the solution(s).
k so i know you would do that but how can somone please show me step
by step or somthing so i can understand this becsaue i don't get it.
There's a lot of steps! Have you done row reduction for smaller matrices:
it's the same procedure. To start row reduction on the above matrix
you would replace
row 2 by row 2 - (7/3) row 1,
row 3 by row 3 - (4/3) row 1 and
row 4 by row 4 - (2/3) row 1.
That would put zeroes below the top left entry 3. (I'm not doing this
arithmetic for you!)
Can you do that?
can somone please do the 16 steps for me or show me a pattern
i know a 4x4 would be for example (2 4)
(5 5) i know you have to first make
the bottom left corner 0 the make the right bot corner 1 then the 4
to a 0 then the 2 to a 1 and i know on a 3x3 like (4 5 6 )
(7 8 9 )
(1 2 3 )
you would first make the 1 a 0 then you would make the 7 a 0 then
make the 2 a 0 then make the 3 a 1 then make the 9 a 0 then make the 8
a 1 then make the 6 a 0 then make the 5 a 0 then make the 4 a 1
so since the pattern changes for 2x2 to 3x3 what is the pattern for a
4x4??
[/quote:87403ec026]
For the present example you would aim to get a pattern
(* * * * *)
(0 * * * *)
(0 0 * * *)
(0 0 0 * *)
--
Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html
"Elegance is an algorithm"
Iain M. Banks, _The Algebraist_ |
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| Hallhair |
| Posted: Sun Feb 20, 2005 10:34 pm |
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Guest
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k i understand how you have to get
(****)
(0***)
(00**)
(000*)
but how do you get the 0's?? i know that you have to add to get
them but please do my example i gave befor so i can understand. |
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| The Qurqirish Dragon |
| Posted: Mon Feb 21, 2005 11:26 am |
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Guest
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Hallhair wrote:
[quote:ccc91dde49]k i understand how you have to get
(****)
(0***)
(00**)
(000*)
but how do you get the 0's?? i know that you have to add to
get
them but please do my example i gave befor so i can understand.
[/quote:ccc91dde49]
Let's try a different tack on this.
You said you know how to solve 2x2 and 3x3 matrices this way:
2x2:
(a b | c)
(d e | f)
you first get element d to be 0, then change e to 1. Next you change b
to 0, and then get a to be 1. Whatever is left in positions c and f is
your answer.
3x3:
(a b c | j)
(d e f | k)
(g h i | l)
You first get d and g to be 0, then h to be 0, and i to be 1 yielding:
(a b c | j)
(0 x x | x)
(0 0 1 | x)
where x is some number that may not be the same as the original
position.
you then work upwards to get f and c to be 0, and then get e to be 1,
yielding:
(a b o | x)
(0 1 0 | x)
(0 0 1 | x)
Now you get b to be 0, and get a to be 1. Whatever is in the positions
originally labelled j,k,l gives the solution.
The idea is to get zeroes in one entire column, other than a specific
element. The you do the same with another column, ignoring the row(s)
in which you didn't get a 0 before. repeat until one row has only one
non-0 element (if a row becomes all 0s, then the system is
underdefined, or inconsistant, depending on the value of the augmented
column)
Now, take the row with only one non-0, and divide the row to get a 1
there, and use this to get 0s in all the other rows, in this column.
This will give you another row with only one 0. repeat with this one,
and so on.
This is pretty much what the others have said, but you still seem
confused. SO, try this:
4x4:
(a b c d | s)
(e f g h | t)
(i j k m | u)
(n p q r | v)
I skipped letters l and o, to avoid confusion with 1 and 0.
Consider instead this sub-matrix:
(a b c | d | s)
(e f g | h | t)
(i j k | m | u)
Solve this just as you would a 3x3, but with two augmented rows. You
will be left with this:
(1 0 0 x | x)
(0 1 0 x | x)
(0 0 1 x | x)
(n p q r | v)
again, an x indicates a number which may not be the same as that
originally in that position.
Since you said you can do 3x3 matices, I will not explain how to get
here.
Now, subtract n times row 1 from row 4, p times row 2 from what is
left, and q times row 3 from what is left after that. This will leave
you with:
(1 0 0 x | x)
(0 1 0 x | x)
(0 0 1 x | x)
(0 0 0 x | x)
Now get a 1 in position r by dividing, and use this to get 0s in
positions d,h, and m. What is left in the last column is your answer.
By following this reasoning, you can solve a 5x5 by looking at the
first 4 rows, and solving that 4x4, and then get zeroes in the first 4
positions of row 5, and then use the fifth element in row 5 to get
zeroes above.
This is a bit longer (but not much more so) than doing normal
gauss-jordan elimination, and a bit longer than forward-elimination /
back-substitution, but it will work. |
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| Guest |
| Posted: Mon Feb 21, 2005 12:28 pm |
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You solve this just as you did in a 3x3 marix, by using the elementary
row operations. Get A(11) to be one and then get A(21), A(31) and
A(41) to equal zero. Next step get A(22) to equal one and then get
A(32) and A(42) to equal zero. Next step get A(33) to equal one and
then get A(34) to equal zero. Finally get A(44) to equal one and the
last row will tell you what d equals. Now work backwards to find c,
using row 3. Then substitute values of c & d into row 2 and solve for
b. Last step substitute values of b, c, and d into row one and solve
for a. |
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| Guest |
| Posted: Tue Feb 22, 2005 3:23 pm |
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Another way to solve this system of equations would be to use Cramer's
Rule. Where you use the coefficients and constants of the equations
given and form 5 different array of numbers and then find the
determinant. You can use Excel to help you find the determinant.
You would call you first arrary of numbers D. In excell you would
place each coefficient in a seperate cell. Now in a seperate cell from
the function command choose the command MDETERM and then highlight your
array of numbers and this will give you the determinate of D. Such as;
3 5 -2 1
7 -3 4 -3
4 1 -1 5 830
2 -4 6 -1
Notice the determinant is 830.
Now lets substitute the constants (3, 15, -2, & 11) in for the first
column. Now againg select a new cell using the MDETERM command to find
the determinant. Such as;
3 5 -2 1
15 -3 4 -3
-2 1 -1 5 830
11 -4 6 -1
Notice the determinant is 830. Now we can solve for a, because a =
Da/D which is 830/830 which equals 1. a = 1
Now subtstitute the constants in for the second column and follow the
same procedure above using the MDETERM command. Such as;
3 3 -2 1
7 15 4 -3 830
4 -2 -1 5
2 11 6 -1
Notice the determinant is 830. Now we can find b, because b = Db/D
which is 830/830 which equals 1. b = 1
Now substitute the constants in for column three. Follow same
procedures as we did above. Such as;
3 5 3 1
7 -3 15 -3 1660
4 1 -2 5
2 -4 11 -1
Notice this gives us the determinant 1660. Now we can find c, because
c = Dc/D which is 1660/830 which equals 2. So c = 2.
Now substitute the constants in for column four. Such as;
3 5 -2 3
7 -3 4 15
4 1 -1 -2 -830
2 -4 6 11
Notice the determinant is -830, so now we can solve for d, because d =
Dd/D which is -830/830 which equals -1. d = -1.
Now we have the solution of the system of equation.
(1, 1, 2, -1). |
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