SOLUTION: Is there a way to easily solve matrices, 3 system equations, and Cramer's rule
problems?
What are the short cuts or tricks to solving the above problems?
Question 21095: Is there a way to easily solve matrices, 3 system equations, and Cramer's rule
problems?
What are the short cuts or tricks to solving the above problems? Answer by venugopalramana(3286) (Show Source): You can put this solution on YOUR website!
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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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Kramers rule is used to solve simultaneous equations.In the working given below
let me use the notation to represent the determinant.However,in actual practice we should use |xxx|to indicate a determinant ,while (xxx)is used to represent a matrix.I shall check with the network authorities on how to represent determinants in their software.
In the present case there are 3 equations to solve for 3 unknowns.
The solution is given by Kramer's rule as follows.
x/=
y/=
z/=
1/
evaluating the determinants we get
x/(-3)=y/6=z/(-9)=1/(-3)
hence x=1
y=-2
z=3
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I have working on this problem for some time but I am still having a hard time working this one:
I am suppose to solve this system using elimination method
10x+6y+z=7 (1)
5x-9y-2z=3 (2)
15x-12y+2z=-5 (3)
20x-21y=-2 (4)
Please show me how to work this problem from this point.
Thank you
Good you have proceeded correctly and infact on the way to solving the problem by your self..you only need a little guidance on the path you should follow to solve the problem..o.k. ..let us see you have added equations 2 and 3 to get equation 3 ,which has accomplished elimination of one unknown z . The basic procedure is , if we start with 3 equations in 3 unknowns ,we try to eliminate one unknown taking one pair of equations at a time to get 2 new equations in 2 unknowns only.Then we take those 2 new equations to eliminate one another unknown to get one more new equation , but this time with one unknown only.This we can easily solve to find the unknown.Now , we travel backwards along the same path as we travelled to find the 2 other unknowns one after another by substituting the known values every time.Let us illustrate the procedure now with this example.Now that you have already got one new equation 4 from 2 and 3 to eliminate z., let us take equations 1 and 2 to eliminate the same unknown z.For this we observe the coefficients of z in the two equations which are 1 and -2 respectively.So we multiply equation 1 with 2 and add it to equation 2.
Eqn.1 * 2 gives us ...20x+12y+2z=14 .....(5)
Eqn.2 is .............5x-9y-2z = 3........(6)
Eqn.5 + Eqn.6 gives us .....25x+3y = 17....(7)
but from Eqn.4 we have .....20x-21y=-2......(4)..proceeding on the same basis ,we eliminate y from these 2 equations.
Eqn.7 * 7 gives us .........175x+21y=119....(8)
Eqn.8 + Eqn.4 gives us .....195x=117 ..or x= 117/195 = 39/65 = 3/5.....now substitute this value of x in eqn.4 to get y
y=(20*(3/5)+2)/21=14/21=2/3…….now substitute these values of x and y in eqn.1 to get z.
z=(7-10*(3/5)-6*(2/3))=-3………….. As a check ,you can substitute these values of x,y,and z in the 3 given equations to
verify that your answer is correct.