linear algebra
Cramer's rule
system of linear equations


Given


 




{





a

1


x
+

b

1


y
+

c

1


z


=



d

1









a

2


x
+

b

2


y
+

c

2


z


=



d

2









a

3


x
+

b

3


y
+

c

3


z


=



d

3












{\displaystyle \left\{{\begin{matrix}a_{1}x+b_{1}y+c_{1}z&={\color {red}d_{1}}\\a_{2}x+b_{2}y+c_{2}z&={\color {red}d_{2}}\\a_{3}x+b_{3}y+c_{3}z&={\color {red}d_{3}}\end{matrix}}\right.}




which in matrix format is



 





[




a

1





b

1





c

1







a

2





b

2





c

2







a

3





b

3





c

3





]




[



x




y




z



]


=


[






d

1











d

2











d

3







]


.


{\displaystyle {\begin{bmatrix}a_{1}&b_{1}&c_{1}\\a_{2}&b_{2}&c_{2}\\a_{3}&b_{3}&c_{3}\end{bmatrix}}{\begin{bmatrix}x\\y\\z\end{bmatrix}}={\begin{bmatrix}{\color {red}d_{1}}\\{\color {red}d_{2}}\\{\color {red}d_{3}}\end{bmatrix}}.}




Then the values of 



x
,
y


{\displaystyle x,y}

 and 



z


{\displaystyle z}

 can be found as follows:







x
=



|






d

1







b

1





c

1









d

2







b

2





c

2









d

3







b

3





c

3





|


|




a

1





b

1





c

1







a

2





b

2





c

2







a

3





b

3





c

3





|



,

y
=



|




a

1







d

1







c

1







a

2







d

2







c

2







a

3







d

3







c

3





|


|




a

1





b

1





c

1







a

2





b

2





c

2







a

3





b

3





c

3





|



,

 and 

z
=



|




a

1





b

1







d

1









a

2





b

2







d

2









a

3





b

3







d

3







|


|




a

1





b

1





c

1







a

2





b

2





c

2







a

3





b

3





c

3





|



.


{\displaystyle x={\frac {\begin{vmatrix}{\color {red}d_{1}}&b_{1}&c_{1}\\{\color {red}d_{2}}&b_{2}&c_{2}\\{\color {red}d_{3}}&b_{3}&c_{3}\end{vmatrix}}{\begin{vmatrix}a_{1}&b_{1}&c_{1}\\a_{2}&b_{2}&c_{2}\\a_{3}&b_{3}&c_{3}\end{vmatrix}}},\quad y={\frac {\begin{vmatrix}a_{1}&{\color {red}d_{1}}&c_{1}\\a_{2}&{\color {red}d_{2}}&c_{2}\\a_{3}&{\color {red}d_{3}}&c_{3}\end{vmatrix}}{\begin{vmatrix}a_{1}&b_{1}&c_{1}\\a_{2}&b_{2}&c_{2}\\a_{3}&b_{3}&c_{3}\end{vmatrix}}},{\text{ and }}z={\frac {\begin{vmatrix}a_{1}&b_{1}&{\color {red}d_{1}}\\a_{2}&b_{2}&{\color {red}d_{2}}\\a_{3}&b_{3}&{\color {red}d_{3}}\end{vmatrix}}{\begin{vmatrix}a_{1}&b_{1}&c_{1}\\a_{2}&b_{2}&c_{2}\\a_{3}&b_{3}&c_{3}\end{vmatrix}}}.}






Task
Given the following system of equations:









{



2
w
−
x
+
5
y
+
z
=
−
3




3
w
+
2
x
+
2
y
−
6
z
=
−
32




w
+
3
x
+
3
y
−
z
=
−
47




5
w
−
2
x
−
3
y
+
3
z
=
49








{\displaystyle {\begin{cases}2w-x+5y+z=-3\\3w+2x+2y-6z=-32\\w+3x+3y-z=-47\\5w-2x-3y+3z=49\\\end{cases}}}




solve for 



w


{\displaystyle w}

, 



x


{\displaystyle x}

, 



y


{\displaystyle y}

 and 



z


{\displaystyle z}

, using Cramer's rule.




crule[m_, b_] := Module[{d = Det[m], a},
  Table[a = m; a[[All, k]] = b; Det[a]/d, {k, Length[m]}]]
 
crule[{
  {2, -1, 5, 1},
  {3, 2, 2, -6},
  {1, 3, 3, -1},
  {5, -2, -3, 3}
 } , {-3, -32, -47, 49}]