Systems of 3 linear equations?

Question

9x+4y-z=0
3x-2y+4z=6
6x-8y-3z=3

Answer 1

equation (2)*2
6x-4y+8z=12
9x+4y-z=0
adding
15x+7z=12 (4)
(1)*2
18x+8y-2z=0
6x-8y-3z=3
adding
24x-5z=3 (5)
(4)*5 75x+35z=60
(5)*7 168x+35z=21
adding 243x=81
x=1/3
substituting
8-5z=3
-5z=-5
z=1
sub
3+4y-1=0
4y=-2
y=-1/2
solution set{1/3,-1/2,1}

Answer 2

to describe precisely why there is not any answer to this one calls for rather more advantageous arithmetic, yet enable me clarify a number of the words so that you'll be able to understand not absolutely "i will't confirm it," yet WHY there is not any answer to this one. It has to do with something stated as linear independence. If a gadget of n equations and n unknowns is linearly self sustaining, then whatever the constants after the equals signal are, there's a answer (and as an advantage, in the experience that they are linearly self sustaining and all 3 constants after the equals signal are 0, then the answer's continually x = 0, y = 0, z = 0, and so on.). What linear independence skill is that you may't write any of the left halves of the equations in words of multiplying and including different left halves of equations contained in the gadget. it rather is, if one equation is a million.7 situations between the others, then the gadget isn't linearly self sustaining. If one equation is a similar as 4 situations one equation minus 2 situations yet another equation, then it isn't linearly self sustaining. in the experience that your gadget isn't linearly self sustaining, then there are 2 opportunities -- both there are not any thoughts (as on your case), or there are an unlimited kind of thoughts (as an party, in the experience that your gadget had a million, -a million and three because the constants on the right hand area, then any values of x, y, and z ought to paintings as thoughts see you later as they fulfill x - y + 2z = a million, that can be an complete airplane answer). In extra reachable words, in case you try fixing with the help of eliminating technique and also you eventually end up with all zeroes on the left 0.5 of one equation and something it really is not a nil contained in the right, then there is not any answer.

Answer 3

Two ways to do this one: brute force or via matrices.
Brute force: solve the first equation for z and put it into eqns 2 and 3; now you can solve two simultaneous equations for x and y. The just plug back in eqn 1 to get z.
Elegant way:
A is a 3x3 matrix:
9 4 -1
3 -2 4
6 -8 -3
B is a column matrix
x
y
z
And so is C
0
6
3
So that AB=C, and multiplying by the inverse of A
B = AinverseC

Answer 4

wow this is funny I am learning this in class!

I dont know if this is right at all. But its almost midnight...

*a) 9x+4y-z=0
b) 3x-2y+4z=6
c) 6x-8y-3z=3

a+b
9x+4y-z=0
3x-2y+4z=6
-----------------
make the z's cancel out

4(9x+4y-z=0)
3x-2y+4z=6
distribute
---------------------
add together a+b after distrubuting
36x+16y-4z=0
3x-2y+4z=6
______________
~~39x+14y=6~~ (remember z's canceled out)

Then do
a+c
9x+4y-z=0
6x-8y-3z=3
-----------------
cancel out the z's again
-3(9x+4y-z=0)
6x-8y-3z=3
--------------------
distrubute and cancel out the z's

then add the rows.
27x-12y=-3
6x-8y=3
________________
~~-21x-20y=0~~
===========

Take the a+b answer and the a+c answer and combine them and cancel out one of the varaibles::

39x+14y=6
-21x-20y=0

Cancel out the y's cause they have the least common multiple.
distrubute until the y's cancel out.

20(39x+14y=6)
14(-21x-20y=0)
--------------------
780x+280y=120
-294x-280y=0
--------------------
y's cancel out
486x=0
___ ___
486 486

then divide both sides by 486 to get the x by itself
x=0

then put 0 in for x
-21(0)-20y=0
-20y=0
__ __
-20 -20
divide both sides by -20 to get y by itself

y=0

then put x and y into one of the original equations

3(0)-2(0)+4z=6
0+0+4z=6
4z=6
__ __
4 4

divide both sides by 4 to get z by itself

z= 6/4 which reduces to
z= 3/2

final ANSWER:: (0,0,3/2)


sorry if its wrong.

Answer 5

Yep, that's what it is.