How do you solve #5x - 5y + 10z = -11#, #10x + 5y - 5z = 1# and #15x - 15y -10z = -1# using matrices?

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Answer 1 · 2018-08-02T07:22:27

The solution is #((x),(y),(z))=((-2/5),(1/5),(-4/5))#

Perform the Gauss-Jordan Elimination on the augmented matrix

#A=((5,-5,10,|,-11),(10,5,-5,|,1),(15,-15,-10,|,-1))#

Make the pivot in the first column and first row

#R1larr((R1)/5)#

#((1,-1,2,|,-2.2),(10,5,-5,|,1),(15,-15,-10,|,-1))#

Eliminate the first column

#R2larr(R2-10R1)# and #R3larr(R3-15R1)#

#((1,-1,2,|,-2.2),(0,15,-25,|,23),(0,0,-40,|,32))#

Make the pivot in the second column and second row

#R2larr((R2)/15)#

#((1,-1,2,|,-2.2),(0,1,-5/3,|,23/15),(0,0,-40,|,32))#

Eliminate the second column

#R1larr(R1+R2)#

#((1,0,1/3,|,-2/3),(0,1,-5/3,|,23/15),(0,0,-40,|,32))#

Make the pivot in the third column and third row

#R3larr((R3)/-40)#

#((1,0,1/3,|,-2/3),(0,1,-5/3,|,23/15),(0,0,1,|,-4/5))#

Eliminate the third column

#R1larr(R1-1/3R3)#, and #R2larr(R2+5/3R3)#,

#((1,0,0,|,-2/5),(0,1,0,|,1/5),(0,0,1,|,-4/5))#

The solution is

#((x),(y),(z))=((-2/5),(1/5),(-4/5))#