Cramer's Rule Using Determinants to solve systems of equations

1. Vector Calculus And
Linear Algebra
Presented by :
Abdul Sattar

2. CRAMER'S RULE
(Using Determinants to solve systems of equations)

3. INTRODUCTION
 Suppose that we have a square system
with n equation in the same number of
variables ( ). Then the
solution of the system has the
following cases.
1) If the system has non-zero
coefficient determinant D=det(A),
then the system has unique solution
and this solution is of the form
nxxx ,......., 21
,,.......,,
D
D
x
D
D
x
D
D
x n
n  2
2
1
1
iD
th
i nbbb ,.....,, 21

4. a) if at least one of Di is non-zero then
the system has no solution
b) if all Di‘s are zero, then the system
has infinite number of solutions.
In this case,
if the given system is homogeneous;
that is, right hand side is zero
then we have ffollowing possibilities of
its solution.
2) If the system has zero coefficient
determinant D=det(A) , then we have
two possibilities as discussed below.

6. EXAMPLE : 1
Find the solution of the system
Solution :
In matrix form, the given system of
equations can be written as Ax=b,
where
74
63
52



zyx
zyx
zyx

































7
6
5
,,
411
113
121
b
z
y
x
xA

7. Here , matrix A is a square matrix of order
3, so Cramer’s rule can be applied Now,
411
113
121
||)det(  AAD
)4(1)11(2)5(1 
023 
)13(1)112(2)14(1 
Therefore, the system has unique solution.
For finding unique solution, let us first find
D1,D2 and D3 it can be easily verified that
)76(1)724(2)14(5
417
116
125
1 D
46
)13(1)17(2)5(5



8. EXAMPLE : 2
Find the solution of the system
7242
532
32



zyx
zyx
zyx
Solution :
In matrix form, the given system of
equations can be written as Ax=b,
where

































7
5
3
,,
242
312
121
b
z
y
x
xA

9. )621(1)112(5)724(1
471
163
151
2 D
23
)15()11(517


)13(5)621(2)67(1
711
613
521
3 D
23
)4(5)15(213


Therefore, the system has unique solution.
1
23
23
,1
23
23
,2
23
46 321










D
D
z
D
D
y
D
D
x

10. Here , matrix A is a square matrix of order
3, so Cramer’s rule can be applied Now,
)28(1)64(2)122(1
242
312
121
||  AD
6)2(210 
0
Therefore, either the system has no
solution or infinite number of solution. Let
us check for it.
)720(1)2110(2)122(3
247
315
123
1 D
13)11(2)10(3 
05 
Therefore, the system has no solution as at
least one Di, i=1,2,3 is nonzero.

11. EXAMPLE : 3
Find the solution of the system
24987
15654
632



zyx
zyx
zyx
Solution :
In matrix form, the given system of
equations can be written as Ax=b,
where

































24
15
6
,,
987
654
321
b
z
y
x
xA

12. Here , matrix A is a square matrix of order
3, so Cramer’s rule can be applied Now,
)3532(3)4236(2)4845(1
987
654
321
||  AD
)3(3)6(23 
0
Also,
)120120(3)144135(2)4845(6
9824
6515
326
1 D
0
)0(3)9(2)3(6



13. )10596(3)4236(6)144135(1
9247
6154
361
2 D
0
)9(3)6(69


)3532(6)10596(2)120120(1
2487
1554
621
3 D
0
)3(6)9(20


Therefore, the system has infinitle number
of solutions.
Now, 0385
54
21


14. Therefore, þ(A)=2
Omitting m-r = 3-2 = 1
Considering n-r = 3-2 = 1 variable as
arbitary, the remaining system becomes
xzy
xzy
41565
632


Where x is arbitary
Now ,
31512
65
32
D
96)415(3)636(
6415
36
1 


 xxx
x
x
D
xxx
x
x
D 3)6(5)415(2
4155
62
1 




15. Therefore ,
x
x
D
D
zx
x
D
D
y 






3
3
,23
3
96 21
Let x = k , where k is arbitary , then the
infinite number of solutions of the given
system is kzkykx  ,23,
where k is an arbitary constant.