Question 974842
Solve the system: 2x+3y+6z=2
                  -x+y+z=0
I tried by the equation {{{ X=A^(-1)B }}}
Please help me solve it..
<pre>
This system has more unknowns than equations, so it is labeled an
"under-determined system", so it has infinitely many solutions.
The coefficient matrix is not a square matrix, thus it has no
inverse.  If you've studied left-inverse and right-inverses,
you may be able to get the general solution for the infinite
set of solutions that way, but you can get that easier just by 
ordinary methods:


{{{system(2x+3y+6z=2,
-x+y+z=0)}}}

Multiply 2nd equation by 2

{{{system(2x+3y+6z=2,
-2x+2y+2z=0)}}}

Add them

{{{5y+8z = 2}}}
{{{5y = 2-8z}}}
{{{y = (2-8z)/5}}}

Substitute in

{{{-x+y+z=0}}}
{{{-x+(2-8z)/5+z=0}}}
{{{-5x+(2-8z)+5z=0}}}
{{{-5x+2-8z+5z=0}}}
{{{-5x+2-3z=0}}}
{{{-5x=-2+3z}}}
{{{5x=2-3z}}}
{{{x=(2-3z)/5}}}

There are infinitely many solutions.  Just substitute arbitrary 
values for z in the expressions above.  Here, I'll set up a 
computer program to generate some of them:

If you choose z=-21, then substituting, {{{x=(2-3(-21))/5 = (2-(-63))/5=(65)/5=13}}}
Also, {{{y=(2-8(-21))/5 = (2-(-168))/5=(170)/5=34}}}

If you choose z=-16, then substituting, {{{x=(2-3(-16))/5 = (2-(-48))/5=(50)/5=10}}}
Also, {{{y=(2-8(-16))/5 = (2-(-128))/5=(130)/5=26}}}

If you choose z=-11, then substituting, {{{x=(2-3(-11))/5 = (2-(-33))/5=(35)/5=7}}}
Also, {{{y=(2-8(-11))/5 = (2-(-88))/5=(90)/5=18}}}

If you choose z=-6, then substituting, {{{x=(2-3(-6))/5 = (2-(-18))/5=(20)/5=4}}}
Also, {{{y=(2-8(-6))/5 = (2-(-48))/5=(50)/5=10}}}

If you choose z=-1, then substituting, {{{x=(2-3(-1))/5 = (2-(-3))/5=(5)/5=1}}}
Also, {{{y=(2-8(-1))/5 = (2-(-8))/5=(10)/5=2}}}

If you choose z=4, then substituting, {{{x=(2-3(4))/5 = (2-(12))/5=(-10)/5=-2}}}
Also, {{{y=(2-8(4))/5 = (2-(32))/5=(-30)/5=-6}}}

If you choose z=9, then substituting, {{{x=(2-3(9))/5 = (2-(27))/5=(-25)/5=-5}}}
Also, {{{y=(2-8(9))/5 = (2-(72))/5=(-70)/5=-14}}}

If you choose z=14, then substituting, {{{x=(2-3(14))/5 = (2-(42))/5=(-40)/5=-8}}}
Also, {{{y=(2-8(14))/5 = (2-(112))/5=(-110)/5=-22}}}

If you choose z=19, then substituting, {{{x=(2-3(19))/5 = (2-(57))/5=(-55)/5=-11}}}
Also, {{{y=(2-8(19))/5 = (2-(152))/5=(-150)/5=-30}}}
 
I chose those value because they give integer solutions, however,
you can get fractional solutions by choosing other values for z.

Edwin</pre>