How do you solve #w+4x+3y-11z=42#, #6w+9x+8y-9z=31#, #-5w+6x+3y+13z=2#, and #8w+3x-7y+6z=31# using matrices?

1 Answer
Oct 1, 2017

Answer:

The answer is #((w),(x),(y),(z))=((-12054/4889),(38342/4889),(-31301/4889),(-14357/4889))#

Explanation:

Perform the Gauss Jordan elimination on the augmented matrix

#A=((1,4,3,-11,|,42),(6,9,8,-9,|,31),(-5,6,3,13,|,2),(8,3,-7,6,|,31))#

Perform the row operations

#R2larrR2-6R1#, #R3larrR3+5R1#, #R4-8R1#

#A=((1,4,3,-11,|,42),(0,-15,-10,57,|,-221),(0,26,18,-42,|,212),(0,-29,-31,94,|,-305))#

#R2larr(R2)/(-15)#

#A=((1,4,3,-11,|,42),(0,1,2/3,-19/5,|,221/15),(0,26,18,-42,|,212),(0,-29,-31,94,|,-305))#

#R3larrR3-26R2#, #R4larrR4+29R2#,

#A=((1,4,3,-11,|,42),(0,1,2/3,-19/5,|,221/15),(0,0,2/3,284/5,|,-2566/15),(0,0,-35/3,-81/5,|,1834/15))#

#R3larr(R3)/(2/3)#

#A=((1,4,3,-11,|,42),(0,1,2/3,-19/5,|,221/15),(0,0,1,426/5,|,-1283/5),(0,0,-35/3,-81/5,|,1834/15))#

#R4larr(R4)*(-3/35)+R3#

#A=((1,4,3,-11,|,42),(0,1,2/3,-19/5,|,221/15),(0,0,1,426/5,|,-1283/5),(0,0,0,4889/5,|,-14357/5))#

#R4larr(R4)*(5/4889)#

#A=((1,4,3,-11,|,42),(0,1,2/3,-19/5,|,221/15),(0,0,1,426/5,|,-1283/5),(0,0,0,1,|,-14357/4889))#

#R3larr(R3)-(426/5)R4#

#A=((1,4,3,-11,|,42),(0,1,2/3,-19/5,|,221/15),(0,0,1,0,|,-31301/4889),(0,0,0,1,|,-14357/4889))#

#R2larr(R2)-(2/3)R3#, #R2larrR2+19/5R4#

#A=((1,4,3,-11,|,42),(0,1,0,0,|,38342/4889),(0,0,1,0,|,-31301/4889),(0,0,0,1,|,-14357/4889))#

#R1larr(R1)-(4)R2#, #R1larrR1-3R3#, #R1larrR1+11R4#

#A=((1,0,0,0,|,-12054/4889),(0,1,0,0,|,38342/4889),(0,0,1,0,|,-31301/4889),(0,0,0,1,|,-14357/4889))#