How do you solve using gaussian elimination or gauss-jordan elimination, #x_1 + 3x_2 +x_3 + x_4= 3#, #2x_1- 2x_2 + x_3 + 2x_4 =8# and #3x_1 + x_2 + 2x_3 - x_4 =-1#?

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Answer 1 · 2018-01-07T09:37:09

#P={((-6-5p)/8, (2-p)/8, p, 3), p in RR}#

#([1,3,1,1,|,3],[2,-2,1,2,|,8],[3,1,2,-1,|,-1])~~([1,3,1,1,|,3],[0,-8,-1,0,|,2],[0,-8,-1,-4,|,-10])#
#R_2=R_2-2xxR_1#
#R_3=R_3-3xxR_1#
#R_3=R_3-R_2#

#([1,3,1,1,|,3],[0,-8,-1,0,|,2],[0,0,0,-4,|,-12])~~([1,3,1,1,|,3],[0,-8,-1,0,|,2],[0,0,0,1,|,3])#
#R_3=R_3xx-1/4#
#R_1=R_1-R_3#

#([1,3,1,0,|,0],[0,-8,-1,0,|,2],[0,0,0,1,|,3])~~([1,3,1,0,|,0],[0,1,1/8,0,|,1/4],[0,0,0,1,|,3])#
#R_2=R_2xx-1/8#
#R_1=R_1-3xxR_2#

#([1,0,5/8,0,|,-3/4],[0,1,1/8,0,|,1/4],[0,0,0,1,|,3])#

#x_3=pcolor(white)(QQQQQQQ)p in RR# (as a parameter)

#=>x_1=-3/4-5/8p=(-6-5p)/8#

#=>x_2=1/4-1/8p=(2-p)/8#

#=>x_4=3#

#P={((-6-5p)/8, (2-p)/8, p, 3), p in RR}#