Can someone help me with the system of linear equations?

Question

The system of linear equations

a(sub)1 + 2a(sub)2 +a(sub)3 - a(sub)4 =0
3a(sub)1 + 2a(sub)2 - 3a(sub)3 =0
-4a(sub)1 - 4a(sub)2 + 2a(sub)3 + a(sub)4 =0
2a(sub)1 -4a(sub)3 =0

Can you explain how to solve this equation?
When I looked up the solution manual, there was
no explanation for it. They just put the answer.
The answer was
a(sub)1 = 4
a(sub)2 = -3
a(sub)3 = 2
a(sub)4 = 0

Can someome please show me the steps of this equation?

EX) a(sub)1 means that there is a little 1(one) next to the "a".
The 1(one) has to be below the "a".

Answer 1

The steps were already explained by the first answerer, and the solution is correct. what I want to emphasize is that indeed a(sub)4=0 and as for the other three unknowns, the results are of the form:
a(sub)1=4x; a(sub)2=-3x; a(sub)3=2x, where x can take any value, unless specified otherwise in the exercise.
The solution the manual provides is the particular case when x=1.
That's all. Good luck!

Answer 2

need help's answer was coming along just fine. This system has an infinite number of solutions, provided the following:
z = 0
w = 2y
x = (-3/2)y

So the solution you have is 1 possible solution. Note that if I choose, for example, a y = 6, I get the following values that will satisfy the system: w = 12, x = -9, y = 6, z = 0. Go ahead and plug them in. They work.

Answer 3

For any system of linear equations greater than 2 eqs in 2 unknowns, the best method is using Cramer's rule, which derives from the matrix formulations of the equations. First form the determinant of coefficients. In your first equation, the coefficients are 1,2,1,-1; in the second they are 3,2,-3,0; then -4,-4,2,1 and 2,0,-4,0. Form the 4x4 determinant of these coefficients

1..2..2..-1
3..2..-3..0
-4.-4..2..1
2..0..-4..0

Find this value and call it D. To find any of the a1..a4, form a new
determinant from D by replacing the column corresponding to the coefficient of the variable you want by the column of equation values 0,0,0,0 For example for a1 you woule make a new determinant as follows:


0..2..2..-1
0..2..-3..0
0..-4..2..1
0..0..-4..0

(NOTE: you must have an error somewhere, because any determinant with a column of zeros is zero.)

Evaluate this determinant D1, then the value of a1 is D1/D.

Similarly for a2, a3 and a4

Answer 4

Just because i don't feel like typing all those subs, i'm going to rename asub1 = w, asub2 = x, asub3 = y, asub4 = z. Now..

1st eq: w+2x+y-z=0
2nd eq: 3w+2x-3y=0
3rd eq: -4w-4x+2y+z=0
4th eq: 2w-4y=0

from 4th eq we know w=2y

plugging into 1st eq: 2x+3y-z=0
plugging into 2nd eq: 2x+3y=0
plugging into 3rd eq: -4x-6y+z=0

multiply the revised 2nd eq by -1:
-2x-3y=0
add that to the revised first eq:
-2x+2x-3y+3y-z=0
so z = 0

then we are left with, from the revised 3rd eq: -4x-6y=0, and from the revised 2nd eq, 2x+3y=0, but these two equations give us the same information. so it seems like there is no one unique solution because the equations given were not independent... maybe i did something wrong...