<PRE><B>Question 1129997</B>
&lt;br&gt;
Your &quot;answer&quot; makes no sense; the answer should show the values of x, y, and z that satisfy all three equations.&lt;br&gt;
There are always an endless number of different ways to solve a system of equations using Gaussian elimination.  I will show one.&lt;br&gt;
The beginning matrix is this:&lt;br&gt;
{{{matrix(3,4,0.1,-0.2,0.3,4,0.5,-0.1,0.4,16,0.7,-0.2,0.3,16)}}}&lt;br&gt;
The first thing I would do is multiply each row (i.e., each equation) by 10 to get rid of the decimals, since working with integers is much easier.&lt;br&gt;
{{{matrix(3,4,1,-2,3,40,5,-1,4,160,7,-2,3,160)}}}&lt;br&gt;
Now before I plunge into the standard techniques for Gaussian elimination, I&#39;m going to look at the equations to see if there is some major simplification that can be made.  And I see that the coefficients of y and z in both the first and third equations are the same (-2 and 3); that means I can get an equation in x only by subtracting the first equation from the third.&lt;br&gt;
So my first step (after converting the matrix to all integers) will be to replace row 3 with (row 3 minus row 1):&lt;br&gt;
{{{matrix(3,4,1,-2,3,40,5,-1,4,160,6,0,0,120)}}}&lt;br&gt;
Next I will combine two steps: dividing row 3 by 6, and making it the first row (since it will be in the form I want for the first row of the final matrix):&lt;br&gt;
{{{matrix(3,4,1,0,0,20,1,-2,3,40,5,-1,4,160)}}}&lt;br&gt;
Next use row 1 to get 0&#39;s in the first column of rows 2 and 3: replace row 2 with (row 1 minus row 2) and replace row 3 with (row 3 minus 5 times row 1):&lt;br&gt;
{{{matrix(3,4,1,0,0,20,0,2,-3,-20,0,-1,4,60)}}}&lt;br&gt;
The first column of my matrix is in its final form; next I want to get a &quot;1&quot; in row 2 column 2.  I can do that by replacing row 2 with (row 2 plus row 3):&lt;br&gt;
{{{matrix(3,4,1,0,0,20,0,1,1,40,0,-1,4,60)}}}&lt;br&gt;
Next, to finish column 2, I need a &quot;0&quot; in row 3 column 2; I can do that by replacing row 3 with (row 3 plus row 2):&lt;br&gt;
{{{matrix(3,4,1,0,0,20,0,1,1,40,0,0,5,100)}}}&lt;br&gt;
Column 2 is finished; next I need a &quot;1&quot; in row 3 column 3.  I can get that by dividing row 3 by 5:&lt;br&gt;
{{{matrix(3,4,1,0,0,20,0,1,1,40,0,0,1,20)}}}&lt;br&gt;
And last I need a &quot;0&quot; in row 2 column 3; I get that by replacing row 2 with (row 2 minus row 3):&lt;br&gt;
{{{matrix(3,4,1,0,0,20,0,1,0,20,0,0,1,20)}}}&lt;br&gt;
The matrix is in the desired final form; it shows the solution to the system is&lt;br&gt;
x=20; y=20; z=20</PRE>