SOLUTION: Use the matrix capabilities of a graphing utility to write the augmented matrix corresponding to the system of equations in reduced row-echelon form. Then solve the system. If the

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Question 1120019: Use the matrix capabilities of a graphing utility to write the augmented matrix corresponding to the system of equations in reduced row-echelon form. Then solve the system. If the system is dependent, express x, y, and z in terms of the parameter a.
3x+3y+12z=3
x+y+4z=1
2x+5y+20z=8
-x+2y+8z=5
**NOTE: I've determined the determinate is zero and tried entering "no solution" but that is incorrect. I've also figured -1,2,0,0
Found 3 solutions by Alan3354, greenestamps, ikleyn:
Answer by Alan3354(69443)   (Show Source): You can put this solution on YOUR website!
Use the matrix capabilities of a graphing utility to write the augmented matrix corresponding to the system of equations in reduced row-echelon form. Then solve the system. If the system is dependent, express x, y, and z in terms of the parameter a.
3x+3y+12z=3
x+y+4z=1
2x+5y+20z=8
-x+2y+8z=5
**NOTE: I've determined the determinate [sic] is zero and tried entering "no solution" but that is incorrect. I've also figured -1,2,0,0
===============
determinant, not determinate
------------
If Det=0, there are an infinite # of solutions, no unique solution.
--------
Not the same as "no solution"
Answer by greenestamps(13200)   (Show Source): You can put this solution on YOUR website!


The system has only three variables; the answer must be an ordered set of three numbers, not four.

Quick inspection shows that the first equation is three times the second. That means the first equation is superfluous; it also means the determinant will be zero.

You say you have "figured -1,2,0,0". I will guess that you got that using matrices. But what you show is only part of the story.

As pointed out earlier, what you really found was -1,2,0, since there are only three variables.

Using only the last three of the four equations, the matrix you end up with is
  1  0  0 -1
  0  1  4  2
  0  0  0  0


The first row says

x = -1

The second row says

y+4z = 2

The third row says 0 = 0, which means the system is dependent, with an infinite family of solutions.

So solve the equation from the second row for one of the variables and use that to define the set of solutions using the parameter a.

y + 4z = 2
y = 2-4z

Then the solution set in parametric form is

x = -1
y = 2-4a
z = a
Answer by ikleyn(52802)   (Show Source): You can put this solution on YOUR website!
.
Use the matrix capabilities of a graphing utility to write the augmented matrix corresponding to the system of equations
in reduced row-echelon form. Then solve the system. If the system is dependent, express x, y, and z in terms of the parameter a.
3x+3y+12z=3
x+y+4z=1
2x+5y+20z=8
-x+2y+8z=5
**NOTE: I've determined the determinate is zero and tried entering "no solution" but that is incorrect. I've also figured -1,2,0,0
~~~~~~~~~~~~~~~~~~~~~~ 

You have the system of 4 equations for only 3 unknown.

So, let's look in it more attentively.


Then you see that the first equation is exactly the second, multiplied by 3.


So, these two equations represent, actually, one single equation and do contribute nothing more to remaining equations.


Therefore, we can consider and focus on the EQUIVALENT system, which contains equations (2), (3) and (4) ONLY. It is


 x +  y +  4z = 1    (2)
2x + 5y + 20z = 8    (3)
-x + 2y +  8z = 5    (4)


Its determinant is 0.   (For quick calculations, I used the free of charge online solver  
                        https://www.algebra.com/algebra/homework/Matrices-and-determiminant/determinant-of-3x3-matrix.solver  of this site).


Now, let's apply the Gauss elimination. As the first step, let's eliminate "x".

For it, multiply eq(1) by 2 and then subract from eq(2). 
        Also, add eq(1) to eq(3).  You will get


 x +  y +  4z = 1    (5)    (same as (2) )
0x + 3y + 12z = 6    (6)    (instead of (3) )
0x + 3y + 12z = 6    (7)    (instead of (4) )


Next step, eliminate "y". For it, subtract eq(6) from eq(7). You will get


 x +  y +  4z = 1    (8)
0x + 3y + 12z = 6    (9)
0x + 0y +  0z = 0    (10)


Now you actually have the system of 2 equations for 3 unknowns


 x +  y +  4z = 1    (11)
0x +  y +  4z = 2    (12)


You can make one more/next step by subtracting eq(12) from eq(11). It will give you

  x           = -1    (13)
      y +  4z =  2    (14)


Now take  a = z  as  a parameter.  Then from eq(14),  y = 2-4a.


Thus the original system has infinitely many solutions  (x,y,z) = (-1,2-4a,a) for any value of real number "a".


Answer.  The system is DEPENDENT and has infinitely many solutions of the form  (x,y,z) = (-1,2-4a,a), where "a" is the parameter (any real number).

Solved.


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