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Question 58434: Can you please give me a step by step guide on how to solve 2x2 and 3x3 matrices using cramers rules! PLEASE PLEASE PLEASE!!!!! THANKS!
Answer by venugopalramana(3286)   (Show Source):
You can put this solution on YOUR website! Write a matrix equation
equivalent to the system of equations.
9x + 9y = -9
5x - 2y = 6
Cramer's rule.
6x + 4y = -4
y = -3x - 7
1 solutions
Answer 12176 by venugopalramana(1088) About Me on
2006-01-03 10:24:38 (Show Source):
Write a matrix equation equivalent to the system of
equations.
9x + 9y = -9
5x - 2y = 6
we write the matrix equation as (A)*(X)=(C)..where A
is the (2,2)matrix of coefficients namely,9,9,5 and -2
here.X is the (2,1) matrix of unknowns x and y and C
is the constants (2,1)matrix on the right side of the
eqn NAMELY -9 AND 6.you will find that from the rule
for equality of matrices,the above matrix eqn.in
effect means the same as that of the given equations.
(matrix(2,2,9,9,5,-2))*(matrix(2,1,x,y))=(matrix(2,1,-9,6))
Cramer's rule.
6x + 4y = -4
y = -3x - 7
GIVING BELOW EXAMPE OF CRAMERS RULE.
My question is.... I was wondering how to do the
cramer rule on a 3x3. I have found a bunch of examples
and stuff, but I want to know how in the world do you
find the determinants of the D, Dx, Dy.and Dz. If you
could just tell me how, that would be great.
1 solutions
Answer 9496 by venugopalramana(585) About Me on
2005-11-15 10:49:54 (Show Source):
SEE THE FOLLOWING AND COME BACK IF YOU HAVE
DIFFICULTY.HERE C,CX,CY,CZ REFER TO YOUR
D,DX,DY,DZ...JUST A DIFFERENCE IN NOMENCLATURE.I
SHOWED IN DETAIL A 2X2 DETERMINANT AND THEN IN BRIEF A
3X3 DETERMINANT
2x+y=4
3x-y=6
make a deteminant with coefficients of x (2,3)and
y(1,-1) in the 2 eqns.call it C.(Actually for a
determinant as you know ,the numbers are contained in
vertical bars at either end like |xx|,but in the
following the bars are omitted due to difficulty in
depiction.you may assume the bars are present)
C=matrix(2,2,2,1,3,-1)=2*(-1)-(1*3)=-5
..now use the constants (4,6)to replace coefficients
of x(2,3) in the above determinant C...call it CX..
CX=matrix(2,2,4,1,6,-1)=4*(-1)-1*6=-4-6=-10
..now use the constants (4,6)to replace coefficients
of y(1,-1) in the above determinant C...call it CY..
CY=matrix(2,2,2,4,3,6)=2*6-3*4=12=12=0
..now cramers rule says that
(x/CX)=(y/CY)=(1/C)..so we get
x/(-10)=y/0=1/-5
x=-10/-5=10/5=2
y=0/-5=0
************************************
so using the above method you can do the next problem
..here due to presence of 3 variables you will get
3rd.order determinants...4 in all...namely C,CX,CY and
CZ,the last formula also extends to include z ,
(x/CX)=(y/CY)=(z/CZ)=(1/C)..
but the procedure is same ..
2x+3y+ z= 5
x+y-2z= -2
-3x +z=-7 ...
...just to give you the idea
C=matrix(3,3,2,3,1,1,1,-2,-3,0,1)..and
CZ=matrix(3,3,2,3,5,1,1,-2,-3,0,7)..etc..hope you can
work out the rest
Matrices-and-determiminant/33609: solve each system
x+y-z= -1
4x-3y+2z= 16
2x-2y-3z=5
1 solutions
Answer 20008 by venugopalramana(1619) About Me on
2006-04-14 00:40:08 (Show Source):
Matrices-and-determiminant/21154: Can you help me to
solve this Matrix Transformation problem please?
x-2y+3z=3
2x+y+5z=8
3x-y-3z=-22
Thank You
1 solutions
Answer 12177 by venugopalramana(1088) About Me on
2006-01-03 10:32:42 (Show Source):
SEE THE FOLLOWING EXAMPLE ND TRY.IN CASE OF DIFFICULTY
PLEASE COME BACK
Hi, I'm in homeschooling and I'm having trouble with
matrices. I was wondering how to solve the problem
where you have to find the x,y, and z values in the
matrix:
[7 -7 5 | 9]
[9 5 -7 | -17]
[6 1 -7 | -2]
I'd appreciate the help. Thank you!
Caitlyn Reese
1 solutions
Answer 9969 by venugopalramana(585) About Me on
2005-11-28 07:14:11 (Show Source):
the 4 column heads represent x,y,z and constant term
in the matrix of system of eqns.
then each row gives us one eqn.like say row 1 gives us
that 7x-7y+5z=9..etc�
hence if we can make the matrix to become
1 0 0 ?
0 1 0 ??
0 0 1 ???
then from the explantion given above it means
1x=?.1y=?? And 1z=???
so we try to transform the matrix in to that form..by
the following steps.
in fact using the above explanation,you can see that
what we do at each step is just
divide each eqn. with a constant/add/subtract etc
which does not change the basic
eqn.for ex. dividing row 1 by 7 means change the given
eqn.7x-7y+5z=9 to x-y+5z/7=9/7
legend:- or1 means old row 1..nr1 means new row 1�r1
means the existing row 1 please note that no changes
are made in rows other than those mentioned at each
step.
start with given matrix �
7...... -7..... 5...... 9
9...... 5...... -7..... -17
6...... 1...... -7..... -2
step 1�we want to make 1st.row 1st.column as
1�.so�.nr1=or1/7...
1...... -1..... (5/7).. (9/7)
9...... 5...... -7..... -17
6...... 1...... -7..... -2
step 2..we want to make 2nd/3rd.rows,col.1 as
0...so...nr2=or2-9*r1........nr3=or3-6*r1
1...... -1..... (5/7).. (9/7)
0... 14..... (-7-9*5/7).... (-17-9*9/7)
0...... 7...... (-7-6*5/7)..... (-2-6*9/7)
step 3�we want to make 2nd.row.2nd.col.as
1..so..nr2=or2/14
1...... -1..... 5/7.... 9/7
0...... 1...... (-94/7)/14..... (-200/7)/14
0...... 7...... (-7-6*5/7)..... (-2-6*9/7)
step 4..we want to make 3rd.row.2nd.col.as
0�so�.nr3=or3-7*r2
1...... -1..... (5/7).. (9/7)
0...... 1...... (-94/7)/14..... (-200/7)/14
0 0 (-79/7)-7*(-94/98) (68/7)-7*(-200/98)
step 5�.we want to make 3rd.row.3rd.col.as
1�so�.nr3=or3/(-32/7)
1...... -1..... (5/7).. (9/7)
0...... 1...... (-94/98)... (-200/98)
0...... 0..... 1...... -1
step 6�we want to make 1st/2nd.row 3rd.col.as
0..so..nr1=or1-5*r3/7...nr2=or2+94*r3/98
1...... -1..... 0..... 2
0...... 1...... 0...... -3
0...... 0...... 1...... -1
step7�.we want to make 1st.row 2nd.col.as
0..so�.nr1=or1+r2
1...... 0...... 0...... -1
0...... 1...... 0...... -3
0...... 0...... 1...... -1
so x=-1.....y=-3.....and z=-1...you can check back
YOU CAN SEE THE FOLLOWING ADDITIONAL MATERIAL FOR
REFERENCE
How do I perform the next required row operation on
the following matrix and provide only the next table:
x y z
1 28 14 245
0 3 7 42
0 7 7 -38
1 solutions
Answer 9892 by venugopalramana(370) About Me on
2005-11-25 08:01:32 (Show Source):
trust you want to solve the equations for x,y and z
and you are at this stage now....assuming that
.....our objective is to finally get the matrix if
possible into the following form ....(i am using
....to seperate the numbers with suitable gaps..your
typing is giving raise to uneven gaps bringing a
little lack of clarity)
1.....0.....0.....x
0.....1.....0.....y
0.....0.....1.....z
now we have
1......28.....14.....245
0.......3......7......42
0.......7......7.....-38
new row2=old row2/3.......to get 1 as required in
row2.so we get...
1......28.....14.....245
0......3/3....7/3....42/3
0.......7......7.....-38
new row3=oldrow3-7*row2 to get 0 as required in row3
1......28.....14...........245
0.......1.....7/3...........14
0......7-7*1..7-7*7/3......-38-7*14
new row3 = old row3/(-28/3)..to get 1 as required in
row3
1......28.....14...................245
0.......1.....7/3...................14
0.......0....(-28/3)/(-28/3).....(-136)/(-28/3)
this gives us finally in the following form
1......28.....14............245
0.......1.....7/3...........14
0.......0......1............102/7
now we go back in the same way to get 0 in row2 and
row3
new row2=old row2-row3*7/3...and new row1=old
row1-row3*14...so we get
1......28......14-1*14.......245-(102/7)*14
0.......1.......7/3-(7/3)*1...14-(102/7)*(7/3)
0.......0.........1.............102/7
the above on simplification gives us
1.......28.......0..........41
0........1.......0..........-20
0........0.......1..........102/7
now finally we take new row1=old row1-28*row2
1.......28-28*1......0.......41-(-28*20)
0........1.......0...........-20
0........0.......1...........102/7
so the final answer is
1......0.......0.......601
0......1.......0.......-20
0......0.......1.......102/7
which tells us that
1*x+0*y+0*z=x=601
0*x+1*y+0*z=y=-20
0*x+0*y+1*z=z=102/7
note that each and every transformation we did above
can be interpreted as given in the last statement
given above...this i hope will give you the insight of
the process at every step.you can also substitute
these values of x,y and z in each and every matrix
above to see that they satify all the equations given
by the different matrices..in general each mtrix can
be taken as a set of simltanous equations in x,y and
z...they can be written as follows..take column 1 is
for x,column 2 is for y and column 3 is for z.so the
first matrix you gave
1......28.....14.....245
0.......3......7......42
0.......7......7.....-38
tells us that
1*x+28*y+14*z=245....etc...
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