Question 784631
-3x+4y-z=-1
 5x-2y+3z=-1
 -x+3y-4z=-22
<pre>
Line up the letters and
show all coefficients:

-3x + 4y - 1z =  -1
 5x - 2y + 3z =  -1
-1x + 3y - 4z = -22

To solve using Cramer's rule:

There are 4 columns,

1. The column of x-coefficients {{{matrix(3,1,-3,5,-1)}}}

2. The column of y-coefficients {{{matrix(3,1,4,-2,3)}}}

3. The column of z-coefficients {{{matrix(3,1,-1,3,-4)}}}

4. The column of constants:     {{{red(matrix(3,1,-1,-1,-22))}}}

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,-3,4,-1,5,-2,3,-1,3,-4))}}}.

It has value {{{D=58}}}.  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(-1),4,-1,red(-1),-2,3,red(-22),3,-4))}}}.

It has value {{{D[x]=-232}}}.

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(-1),-1,5,red(-1),3,-1,red(-22),-4))}}}.

It has value {{{D[y]=-116}}}.

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,-3,4,red(-1),5,-2,red(-1),-1,3,red(-22)))}}}.

It has value {{{D[x]=-150}}}.

Now the formulas for x, y and z are

{{{x=D[x]/D=(-232)/(58)=-4}}}
{{{y=D[y]/D=(-116)/(58)=-2}}}
{{{x=D[z]/D=(290)/(58)=5}}}


Edwin</pre>