<PRE><B>Question 1202812</B>
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You can&#39;t check your work by looking at my solution, because we can easily take very different paths to the answer.&lt;br&gt;
Here is one way to get to the solution to the system of equations.&lt;br&gt;
The original matrix, directly from the system of equations:&lt;br&gt;
{{{matrix(3,4,1,2,1,-2,-2,-3,-1,3,4,8,4,-8)}}}&lt;br&gt;
I can see immediately that this system is not going to have a unique solution, because the 3rd row of the matrix is exactly 4 times the 1st row.  One of the operations we can perform is to multiply one row by a constant, so the first thing I would do in this example is multiply the 3rd row by 1/4:&lt;br&gt;
{{{matrix(3,4,1,2,1,-2,-2,-3,-1,3,1,2,1,-2)}}}&lt;br&gt;
Now I am ready to start on the standard sequence of steps to find the solution.&lt;br&gt;
First step: get &quot;1&quot; in row 1 column 1 if it is not already there.  Row 1 column 1 is already 1, so we don&#39;t need to do anything here.&lt;br&gt;
Next step: Use the &quot;1&quot; in row 1 column 1 to get &quot;0&quot; in every other row in column 1 by multiplying the first row by a constant and adding it to another row.&lt;br&gt;
Get &quot;0&quot; in row 2 column 1 by multiplying row 1 by 2 and adding it to row 2; get a &quot;0&quot; in row 3 column 1 by multiplying row 1 by -1 and adding it to row 3.&lt;br&gt;
{{{matrix(3,4,1,2,1,-2,0,1,1,-1,0,0,0,0)}}}&lt;br&gt;
Note the row of all &quot;0&quot; in row 3 is because in the previous matrix rows 3 and 1 were identical.&lt;br&gt;
Next step: get &quot;1&quot; in row 2 column 2 if it is not already there.  Again that &quot;1&quot; is already there, so there is no work to do here.&lt;br&gt;
Next step: Use the &quot;1&quot; in row 2 column 2 to get &quot;0&quot; in every other row in column 2 by multiplying the second row by a constant and adding it to another row.&lt;br&gt;
Get a &quot;0&quot; in row 1 column 2 by multiplying row 2 by -2 and adding it to row 1.&lt;br&gt;
{{{matrix(3,4,1,0,-1,0,0,1,1,-1,0,0,0,0)}}}&lt;br&gt;
In a general problem, the next step would be to get a &quot;1&quot; in row 3 column 3.  But since the third row is all &quot;0&quot;, we have gone as far as we can performing operations on the matrix.&lt;br&gt;
The matrix we finish with tells us....&lt;br&gt;
row 1: x-z = 0
row 2: y+z = -1&lt;br&gt;
From this we get an infinite set of solutions using a parameter t:&lt;br&gt;
(1) x = t
(2) x-z = 0; t-z = 0; so z = t
(3) y+z = -1; y+t = -1; so y = -t-1&lt;br&gt;
The family of solutions is defined by (x,y,z) = (t,-t-1,t) where t is any constant.&lt;br&gt;
To check the solution, we substitute these parametric expressions for x, y, and z in each of the original equations.&lt;br&gt;
(1) x+2y+z = (t)+(-2t-2)+t = -2  correct
(2) -2x-3y-z = (-2t)+(3t+3)-t = 3  correct
(3) 4x+8y+4z = (4t)+(-8t-8)+(4t) = -8  correct&lt;br&gt;
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