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.




 
(%i1) eqns: [ 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];
(%o1) [z + 5 y - x + 2 w = - 3, (- 6 z) + 2 y + 2 x + 3 w = - 32, 
                      (- z) + 3 y + 3 x + w = - 47, 3 z - 3 y - 2 x + 5 w = 49]
(%i2) A: augcoefmatrix (eqns, [w,x,y,z]);
                          [ 2  - 1   5    1    3   ]
                          [                        ]
                          [ 3   2    2   - 6   32  ]
(%o2)                     [                        ]
                          [ 1   3    3   - 1   47  ]
                          [                        ]
                          [ 5  - 2  - 3   3   - 49 ]
(%i3) C: coefmatrix(eqns, [w,x,y,z]);
                             [ 2  - 1   5    1  ]
                             [                  ]
                             [ 3   2    2   - 6 ]
(%o3)                        [                  ]
                             [ 1   3    3   - 1 ]
                             [                  ]
                             [ 5  - 2  - 3   3  ]
(%i4) c[n]:= (-1)^(n+1) * determinant (submatrix (A,n))/determinant (C);
                            n + 1
                       (- 1)      determinant(submatrix(A, n))
(%o4)            c  := ---------------------------------------
                  n                determinant(C)
(%i5) makelist (c[n],n,1,4);
(%o5)                          [2, - 12, - 4, 1]
(%i6) linsolve(eqns, [w,x,y,z]);
(%o6)                  [w = 2, x = - 12, y = - 4, z = 1]
 