Question 1143266
<pre>
3x + 2y +  z =  5
2x + 3y +  z =  1
2x +  y + 3z = 11

solve using cramers rule
 
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Write in all the 1 and -1 coefficients:

{{{system(3x+2y+1z=red(6),
2x+3y+1z=red(3),
2x+1y+3z=red(-4))}}}

Cramer's rule:
 
There are 4 columns,
 
1. The column of x-coefficients {{{matrix(3,1,3,2,2)}}}
 
2. The column of y-coefficients {{{matrix(3,1,2,3,1)}}}
 
3. The column of z-coefficients {{{matrix(3,1,1,1,3)}}} 
 
4. The column of constants:     {{{red(matrix(3,1,5,1,11))}}}
 
There are four determinants:
 
1. The determinant {{{D}}} consists of just the three columns
of x, y, and z coefficients. in that order, but does not
contain the column of constants.
 
{{{D=abs(matrix(3,3,3,2,1,2,3,1,1,1,3))}}}. 
 
It has value {{{D=12}}}.  I'm assuming you know how to find the
value of a 3x3 determinant, for that's a subject all by itself.
If you don't know how, post again asking how. 
 
2. The determinant {{{D[x]}}} is like the determinant {{{D}}}
except that the column of x-coefficients is replaced by the
column of constants.  {{{D[x]}}} does not contain the column 
of x-coefficients.
 
{{{D[x]=abs(matrix(3,3,red(5),2,1,red(1),3,1,red(11),1,3))}}}.
 
It has value {{{D[x]=24}}}.
 
3. The determinant {{{D[y]}}} is like the determinant {{{D}}}
except that the column of y-coefficients is replaced by the
column of constants.  {{{D[y]}}} does not contain the column 
of y-coefficients.
 
{{{D[y]=abs(matrix(3,3,3,red(5),1,2,red(1),1,22,red(11),3))}}}.
 
It has value {{{D[y]=24}}}.
 
4. The determinant {{{D[z]}}} is like the determinant {{{D}}}
except that the column of z-coefficients is replaced by the
column of constants.  {{{D[z]}}} does not contain the column 
of z-coefficients.
 
{{{D[z]=abs(matrix(3,3,3,2,red(5),2,3,red(1),2,1,red(11)))}}}.
 
It has value {{{D[x]=36}}}.
 
Now the formulas for x, y and z are
 
{{{x=D[x]/D[""]=24/12=2}}}
{{{y=D[y]/D=(-24)/12=-2}}}
{{{x=D[z]/D=36/12=3}}}
 
Edwin</pre>