SOLUTION: 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

Algebra ->  Matrices-and-determiminant -> SOLUTION: 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       Log On


   



Question 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

Answer by venugopalramana(3286) About Me  (Show Source):
You can put this solution on YOUR website!
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%282%2C2%2C2%2C1%2C3%2C-1%29=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%282%2C2%2C4%2C1%2C6%2C-1%29=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%282%2C2%2C2%2C4%2C3%2C6%29=2*6-3*4=12=12=0
..now cramers rule says that
%28x%2FCX%29=%28y%2FCY%29=%281%2FC%29..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 ,
%28x%2FCX%29=%28y%2FCY%29=%28z%2FCZ%29=%281%2FC%29..
but the procedure is same ..
2x+3y+ z= 5
x+y-2z= -2
-3x +z=-7 ...
...just to give you the idea
C=matrix%283%2C3%2C2%2C3%2C1%2C1%2C1%2C-2%2C-3%2C0%2C1%29..and
CZ=matrix%283%2C3%2C2%2C3%2C5%2C1%2C1%2C-2%2C-3%2C0%2C7%29..etc..hope you can work out the rest