How to solve the linear system with Cramer's Rule?

Consider the linear system:
#2x-3y+4z=20 2/3#
#x+2y-3z+13.5=0#
#-x-2y+5z=113/6#
Give two reasons for Cramer's Rule being applicable for solving this system.
Also use the rule to solve the linear system.

1 Answer
Jan 6, 2018

Answer:

#P={(1/2, -3, 8/3)}#

Explanation:

Give two reasons for Cramer's Rule being applicable for solving this system.

  1. #"Matrix"quadA in C^(nxxn)#
  2. #D=detA!=0#
  3. #P={(D_1/D, ..., D_n/D)}#

Assuming #20 2/3# is improper fraction. Rewriting given to the form:
#2x−3y+4z=62/3#
#x+2y−3z=-27/2#
#−x−2y+5z=113/6#

#detA=|[2,-3,4],[1,2,-3],[-1,-2,5]|=|[color(red)0,-7,10],[color(red)1,color(red)2,color(red)(-3)],[color(red)0,0,2]|#
#R_1=R_1-2xxR_2#
#R_3=R_3+R_2#

#detA=1xx(-1)^(1+2)xx|[-7,10],[0,2]|=-1(-14-10xx0)=color(red)(14=D)#


I will be using Sarrus' rule

#D_1=|[62/3,-3,4],[-27/2,2,-3],[113/6,-2,5]|=620/3+27xx4+(113*3)/2-(113xx4)/3-62xx2-(27xx15)/2=(168)/3-16-66/2=7#

#D_2=|[2,62/3,4],[1,-27/2,-3],[-1,113/6,5]|=-27xx5+(113xx4)/6+62-27xx2+113-(62xx5)/3=-14+(113xx4)/6-(620)/6=-42#

#D_3=|[2,-3,62/3],[1,2,-27/2],[-1,-2,113/6]|=(113xx2)/3-cancel((62xx2)/3)-(3xx27)/2+cancel((62xx2)/3)-2xx27+113/2=-54+226/3+16=(226-38xx3)/3=112/3#


#x=D_1/D=7/14=1/2#

#y=D_2/D=-42/14=-3#

#z=D_3/D=(112/3)/14=(112/14)/3=8/3#

#P={(1/2, -3, 8/3)}#