Question 143429
Determine the augmented matrix:p+s+c=600; 12p+10s+8c=5900; 0-s+c=175 
Solve for the matrix and please show all work. 
<pre><b>

{{{  p+  s+ c =   600}}}
{{{12p+10s+8c = 5900}}}
{{{0  -s+ c =  175}}} 

Solve for the matrix and please show all work. 

Put in all the {{{1}}} coefficients and make the {{{0}}} a {{{0p}}}

{{{1p+1s+1c=600}}}
{{{12p+10s+8c=5900}}}
{{{0p-1s+1c=175}}}

Now erase all the letters and you have

{{{drawing(110,50,-10,10,-10,10, line(3.8,10,3.8,-10), 
(matrix(3,4,1,1,1,600,12,10,8,5900,0,-1,1,175))
)}}}

The idea is to get three {{{0}}}'s in the lower left corner:
Multiply the top row through by -12

{{{drawing(200,50,-10,10,-10,10, line(1.5,10,1.5,-10), 
(matrix(3,4,-12,-12,-12,-7200,12,10,8,5900,0,-1,1,175))
)}}}

Add the top row to the middle row:

{{{drawing(200,50,-10,10,-10,10, line(1.5,10,1.5,-10), 
(matrix(3,4,-12,-12,-12,-7200,0,-2,-4,-1300,0,-1,1,175))
)}}}

Restore the top row

{{{drawing(200,50,-10,10,-10,10, line(-2,10,-2,-10), 
(matrix(3,4,1,1,1,600,0,-2,-4,-1300,0,-1,1,175))
)}}}

Multiply the bottom row by -2

{{{drawing(200,50,-10,10,-10,10, line(-2,10,-2,-10), 
(matrix(3,4,1,1,1,600,0,-2,-4,-1300,0,2,-2,-350))
)}}}
Add the middle row to the bottom row:

{{{drawing(200,50,-10,10,-10,10, line(-2,10,-2,-10), 
(matrix(3,4,1,1,1,600,0,-2,-4,-1300,0,0,-6,-1650))
)}}}

Now we've gotten three {{{0}}}'s in the bottom left.
So we make the matrix back into 3 equations:

{{{1p+1s+1c=600}}}
{{{0p-2s-4c=-1300}}}
{{{0p+0s-6c=-1650}}}

Drop the 0 terms and make the 1's invisible:

{{{p+s+c=600}}}
{{{-2s-4c=-1300}}}
{{{-6c=-1650}}}

Solve from the bottom to top.
Solve the bottom equation for {{{c}}}

{{{-6c=-1650}}}
{{{(-6c)/(-6)=(-1650)/(-6)}}}
{{{c=275}}}

Plug that into the second equation:

{{{-2s-4c=-1300}}}
{{{-2s-4(275)=-1300}}}
{{{-2s-1100=-1300}}}
{{{-2s=-200}}}
{{{(-2s)/(-2)=(-200)/(-2)}}}
{{{s=100}}}

Plug that and {{{c=275}}} into the first
equation:

{{{p+s+c=600}}}
{{{p+100+275=600}}}
{{{p+375=600}}}
{{{p=225}}}

So

({{{p}}},{{{s}}},{{{c}}}) = ({{{225}}},{{{100}}},{{{275}}})

Edwin</pre>