How do you solve #3a + 6b + c + 4d = 2#, #a - 2b + 3c - d = -3#, #2a - 4c + d= -15# and #a + 8b + 2c + d = 6# using matrices?

1 Answer
Dec 15, 2017

Answer:

The solution is #((a),(b),(c),(d))=((-5),(1/2),(2),(3))#

Explanation:

Perform the Gauss Jordan elimination on the augmented matrix

#A=((1,8,2,1,|,6),(1,-2,3,-1,|,-3),(3,6,1,4,|,2),(2,0,-4,1,|,-15))#

I have written the equations not in the sequence as in the question in order to get #1# as pivot.

Perform the folowing operations on the rows of the matrix

#R2larrR2-R1# ; #R3larrR3-3R1# ; #R4-2R1#

#A=((1,8,2,1,|,6),(0,-10,1,-2,|,-9),(0,-18,-5,1,|,-16),(0,-16,-8,-1,|,-27))#

#R3larr(R3)/(-10)#

#A=((1,8,2,1,|,6),(0,1,-1/10,1/5,|,9/10),(0,-18,-5,1,|,-16),(0,-16,-8,-1,|,-27))#

#R3larr(R3+18R2)# , #R4larr (R4+16R2)#

#A=((1,8,2,1,|,6),(0,1,-1/10,1/5,|,9/10),(0,0,-34/5,23/5,|,1/5),(0,0,-48/5,11/5,|,-63/5))#

#R3larr(R3xx(-5/34))#

#A=((1,8,2,1,|,6),(0,1,-1/10,1/5,|,9/10),(0,0,1,-23/34,|,-1/34),(0,0,-48/5,11/5,|,-63/5))#

#R4larr(R4+(48/5)*R3)#

#A=((1,8,2,1,|,6),(0,1,-1/10,1/5,|,9/10),(0,0,1,-23/34,|,-1/34),(0,0,0,-73/17,|,-219/17))#

#R4larr(R4*(-17/73))#

#A=((1,8,2,1,|,6),(0,1,-1/10,1/5,|,9/10),(0,0,1,-23/34,|,-1/34),(0,0,0,1,|,3))#

#R3larr(R3+23/34R4)#

#A=((1,8,2,1,|,6),(0,1,-1/10,1/5,|,9/10),(0,0,1,0,|,2),(0,0,0,1,|,3))#

#R2larr(R2+1/10R3)# , #R2larrR2-1/5R3#

#A=((1,8,2,1,|,6),(0,1,0,0,|,1/2),(0,0,1,0,|,2),(0,0,0,1,|,3))#

#R1larr(R1-8R2)# , #R1larrR1-2R3# ,; #R1larrR1-R4#

#A=((1,0,0,0,|,-5),(0,1,0,0,|,1/2),(0,0,1,0,|,2),(0,0,0,1,|,3))#