Question 824928
-x-3y+z=7
 3x +z=-1
 3y+z=5
<pre>
Put in 1 or -1 coefficients, and +0y for the missing term in
the 2nd equation.  Put 0x for the missing term in the third
equation:

{{{system(-1x-3y+1z=red(7),3x+0y+1z=red(-1),0x+3y+1z=red(5))}}}

Form the denominator determinant by using only the coefficients
to the left of the = sign.  You don't use the red numbers in D,
but you will in the others:

{{{D=abs(matrix(3,3,
-1,-3, 1,
3, 0, 1,
0, 3, 1))}}} = 21

[Do you know how to evaluate a 3x3 determinant?
If not, post again asking how to.  I will assume
you already know how.  That's where I got the 21.]

x is the FIRST unknown and so {{{D[x]}}} is just like {{{D}}}
except that the FIRST column is replaced by the three red numbers,
which I made up: 

{{{D[x]=abs(matrix(3,3,
red(7),-3, 1,
red(-1), 0, 1,
red(5), 3, 1))}}} = -42

y is the SECOND unknown and so {{{D[y]}}} is just like {{{D}}}
except that the SECOND column is replaced by the three red numbers: 


{{{D[y]=abs(matrix(3,3,
-1,red(7), 1,
3, red(-1), 1,
0, red(5), 1))}}} = 0

z is the THIRD unknown and so {{{D[z]}}} is just like {{{D}}}
except that the THIRD column is replaced by the three red numbers: 

{{{D[z]=abs(matrix(3,3,
-1,-3, red(7),
3, 0, red(-1),
0, 3, red(5)))}}} = 105

So:
{{{D=21}}}
{{{D[x]=-42}}}
{{{D[y]=0}}}
{{{D[z]=105}}}

Then the solutions are given by:

{{{x}}}={{{D[x]/D}}}={{{(-42)/21}}} = {{{-2}}}
{{{y}}}={{{D[y]/D}}}={{{(0)/21}}} = {{{"0"}}}
{{{z}}={{{D[z]/D}}}={{{(105)/21}}} = {{{5}}}

Edwin</pre>