Question 33382
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SEE BELOW MY COMMENTS AND SUGGESTIONS AND THE EXAMPLE AT THE END...........
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
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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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