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Question 33608: solve each system
4x+8y+z=2
x+7y-3z=-14
2x-3y+2z=23
Answer by venugopalramana(3286)   (Show Source):
You can put this solution on YOUR website! 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...
Quadratic_Equations/20884: 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
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