How do you solve #4r + 5s - 3t = 9#, #6r - 3s + 1t = 67#, #2r + 0s + 1t = 22#?

1 Answer
Mar 5, 2016

Answer:

#r=9, s=-3, t=4#

Explanation:

This system of variables is equal to
#[[4,5,-3],[6,-3,1],[2,0,1]]*[[r],[s],[t]]=[[9],[67],[22]]#
We can solve it in this way
#Delta=[[4,5,-3],[6,-3,1],[2,0,1]]=-12+10+0-(18+30+0)=-2-48=-50#
#Delta r=[[9,5,-3],[67,-3,1],[22,0,1]]=-27+110+0-(198+335+0)=83-533=-450#
#Delta s=[[4,9,-3],[6,67,1],[2,22,1]]=268+18-396-(-402+88+54)=-110+260=150#
#Delta t=[[4,5,9],[6,-3,67],[2,0,22]]=-264+670+0-(-54+660+0)=406-606=-200#

Finally

#r=(Delta r)/Delta=(-450)/-50=9#
#s=(Delta s)/Delta=150/-50=-3#
#t=(Delta t)/Delta=(-200)/-50=4#