How do you solve 3a + 6b + c + 4d = 23a+6b+c+4d=2, a - 2b + 3c - d = -3a2b+3cd=3, 2a - 4c + d= -152a4c+d=15 and a + 8b + 2c + d = 6a+8b+2c+d=6 using matrices?

1 Answer
Dec 15, 2017

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))