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Question 40783: use cramers rule to solve
5x-4y+z=17
x+y+z=4
-10x+8y-2z=11
Answer by venugopalramana(3286)   (Show Source):
You can put this solution on YOUR website! SEE THE FOLLOWING AND TRY
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I need help with usings Cramer's Rule. The question
is: Solve the following system of equations using
Cramer's Rule.
2x + 4y - 3z = 2
2x - 2y + 3z = 3
3x - 4y + 5z = 4
I understand that I have to first have to set up the
determinants. Which are
D = 2 4 -3 Answer Column is 2
2 -2 3 3
3 -4 5 4
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OK
BUT LET US INTRODUCE SOME NOMENCLATURE TO MAKE IT
EASY TO REMEMBER.
FIRST MAKE D= DETERMINANT OF COEFFICIENTS.THAT IS
WRITE COEFFICIENTS OF X,Y AND Z IN THE SAME ORDER IN
ALL 3 EQNS.AS SHOWN BELOW. WE PUT THEM APART TO WRITE
COLUMN NUMBERS.
COLUMN I(X)....COL.II(Y)...COL.III(Z)
2,................4,.........-3..............I ROW
2,...............-2,..........3...............II ROW
3,...............-4,..........5..................III
ROW
OK ...NOW WE SHALL FIND ITS VALUE...WE DO IT IN A
STEP-WISE PROCESS AS GIVEN BELOW..THERE ARE 3 ROWS AND
3 COLUMNS.WE CAN FIND ITS VALUE USING ANY ONE ROW OR
COLUMN.LET US USE FIRST COLUMN TO EXPAND IN THIS
EXAMPLE.FIRST COLUMN HAS 3 ELEMENTS...2,2,AND 3.LET US
CALL THEM E11,E21 AND E31 TO SHOW THAT THEY ARE FROM
ROW1,COL.1;ROW2,COL.1;AND ROW3,COL.1.THE FORMULA FOR
VALUE OF DETERMINANT IS
D=(E11)*(S11)*(M11)+(E21)*(S21)*(M21)+(E31)*(S31)*(M31)
..I
THAT IS WE MULTIPLY EACH ELEMENT WITH ITS CO-FACTOR
AND ADD ALL THE RESULTS.COFACTOR OF AN ELEMENT IS
PRODUCT OF 2 NUMBERS...ONE IS FOR SIGN(S11,S21,S31)
AND ANOTHER FOR VALUE(M11,M21,M31).
THE NUMBER FOR SIGN IS (-1)^(ROW NUMBER+COLUMN
NUMBER)OF THE ELEMENT.HERE THE FIRST ELEMENT 2 IS FROM
ROW1 AND COL.1.SO ITS SIGN NUMBER IS
(-1)^(1+1)=(-1)^2=+1=S11
THE NUMBER FOR VALUE IS CALLED THE MINOR OF THE
ELEMENT.IT IS OBTAINED BY DELETING THE ROW AND COLUMN
CONTAINING THE ELEMENT.OUR ELEMENT IS 2 FROM ROW1 AND
COLUMN 1, AS GIVEN ABOVE.SO REMOVE THE ROW1 AND COLUMN
1.SO WE GET A 2 BY 2 DETERMINANT.IT IS
-2,3
-4,5
LET US CALL THIS EQUAL TO M11 TO INDICATE IT IS MINOR
OF ELEMENT 11 (E11)THAT IS ROW 1 AND COLUMN 1.
SO THE FIRST ELEMENT GIVES US ITS CONTRIBUTION
AS...(2)*(+1)(M11)...WE SHALL FIND THE VALUE OF M11 IN
THE SAME WAY AS ABOVE.BUT LET US SKIP THIS FOR A
MOMENT AND COMPLETE OUR JOB.
NOW TAKING THE NEXT NUMBER 2 AGAIN WHICH IS FROM ROW 2
AND COL.1.SO ITS
SIGN =S21=(-1)^(2+1)=(-1)^3=-1
MINOR IS = M21 GIVEN BY
4,-3
-4,5
SO CONTRIBUTION OF E21 IS
(2)(-1)(M21)
SIMILARLY CONTRIBUTION OF E31 IS
(3)*{(-1)^(3+1)}*(M31)=(3)*(1)*(M31) WHERE M31 IS THE
2 BY 2 DETERMINANT
4,-3
-2,3
SO NOW WE GOT THE 3 BY 3 DETERMINANT CONVERTED TO 2 BY
2 DETERMINANT.THE SAME METHOD WE CAN USE TO EVALUATE
ANY ORDER DETERMINANT BY REDUCING IT TO A LOWER ORDER
DETERMINANT IN EACH STEP.4 BY 4 TO 3 BY 3 TO 2 BY 2
WHICH IS THE END AS WE SHOW BELOW.
LET US FINISH THIS NOW BY ONE EXAMPLE OF 2 BY 2
DETERMINANT SAY M11
-2,3
-4,5
M11=BY THE SAME METHOD
.=(-2)(1)(5)+(-4)(-1)(3)=-10+12=2
SO BY THIS WAY WE CAN FIND M21 AND M31 AND HENCE FIND
D BY THE ABOVE FORMULA-I...VIZ..
D=(E11)*(S11)*(M11)+(E21)*(S21)*(M21)+(E31)*(S31)*(M31)
..I
NEXT MAKE DX
..BY REPLACING COEFFICIENTS OF X IN D
WITH CONSTANT TERMS
SO DX IS
COLUMN I(CONSTANTS)....COL.II(Y)...COL.III(Z)
2,.
...............4,.........-3..............I
ROW
3,
...............-2,..........3...............II
ROW
4,...
............-4,..........5..................III
ROW
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D of x = 2 4 -3
3 -2 3
4 -4 5
SO YOURS IS OK TOO
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SIMILARLY DY IS
COLUMN I(X)....COL.II(CONSTANT)...COL.III(Z)
2,
.............
2,.........-3..............I
ROW
2,...
.........
3,..........3...............II
ROW
3,........
.......4,..........5..................III
ROW --------------------------------
D of y = 2 2 -3
2 3 3
3 4 5
SO YOURS IS OK
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SIMILARLY DZ IS
COLUMN I(X)....COL.II(Y)...COL.III(CONSTANT)
2,...........
.....4,......
.2..............I ROW
2,.........
......-2,..........3...............II
ROW
3,......
......-4,.......
4..................III
ROW
-----------------------
D of Z = 2 4 2
2 -2 3
3 -4 4
SO YOURS IS OK TOO
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Now my problem is how do I evaluate the determinants?
I think that I understand how to set it up. I'm just
not sure what to do after that. Any help you can give
me would be greatly appreciated. I'm not necessarily
looking for the entire question, just an explanation
of what to do next. I've looked over the web for hours
and none of the Algebra websites explains how to do
this step. Thanks for the help and have a great night.
Jonna
HOPE NOW YOUR PROBLEM IS SOLVED AS FAR AS EVALUATING
THE DETERMINANTS IS COCERNED.LET US FINISH BY GIVING
FINAL FORMULAE TO FIND X,Y AND Z
X=DX/D
..;
..Y=DY/D
..;
AND
.Z=DZ/D
.II
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NOW SEE THE FOLLOWING ADDITIONAL PROBLEMS WORKED
EARLIER.
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Matrices-and-determiminant/17820: Please help I have
no idea how to do this problem: Use Cramer's Rule to
solve each system.
1. 2x+y=4
3x-y=6
2. 2x+3y+ z= 5
x+y-2z= -2
-3x +z=-7
1 solutions
Answer 8586 by venugopalramana(1088) About Me on
2005-10-31 03:43:14 (Show Source):
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
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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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