Solve x y z x y z x y z 3,3 3,4 3 6 by Cramer’s method
Answers
Step-by-step explanation:
We have
x+y+z=6
x+2y+3z=14
x+4y+7z=30
The given system of equations in the matrix form are written as below:
⎣
⎢
⎢
⎡
1
1
1
1
2
4
1
3
7
⎦
⎥
⎥
⎤
⎣
⎢
⎢
⎡
x
y
z
⎦
⎥
⎥
⎤
=
⎣
⎢
⎢
⎡
6
14
30
⎦
⎥
⎥
⎤
AX=B...(1)
where
A=
⎣
⎢
⎢
⎡
1
1
1
1
2
4
1
3
7
⎦
⎥
⎥
⎤
,X=
⎣
⎢
⎢
⎡
x
y
z
⎦
⎥
⎥
⎤
and B=
⎣
⎢
⎢
⎡
6
14
30
⎦
⎥
⎥
⎤
∣A∣=1(14−12)−1(7−3)+1(4−2)
=2−4+2
=0
∴The equation either has no solution or an infinite number of solutions.
To decide about this, we proceed to find (Adj A)B.
Let
C be the matrix of cofactors of elements in ∣A∣ such that C=
∣
∣
∣
∣
∣
∣
∣
∣
C
11
C
21
C
31
C
12
C
22
C
32
C
13
C
23
C
33
∣
∣
∣
∣
∣
∣
∣
∣
Here
C
11
=
∣
∣
∣
∣
∣
∣
2
4
3
7
∣
∣
∣
∣
∣
∣
=2;C
12
=−
∣
∣
∣
∣
∣
∣
1
1
3
7
∣
∣
∣
∣
∣
∣
=−4;C
13
=
∣
∣
∣
∣
∣
∣
1
1
2
4
∣
∣
∣
∣
∣
∣
=2
C
21
=−
∣
∣
∣
∣
∣
∣
1
4
1
7
∣
∣
∣
∣
∣
∣
=−3;C
22
=
∣
∣
∣
∣
∣
∣
1
1
1
7
∣
∣
∣
∣
∣
∣
=6;C
23
=−
∣
∣
∣
∣
∣
∣
1
1
1
4
∣
∣
∣
∣
∣
∣
=−3 C
31
=
∣
∣
∣
∣
∣
∣
1
2
1
3
∣
∣
∣
∣
∣
∣
=1;C
32
=−
∣
∣
∣
∣
∣
∣
1
1
1
3
∣
∣
∣
∣
∣
∣
=−2;C
33
=
∣
∣
∣
∣
∣
∣
1
1
1
2
∣
∣
∣
∣
∣
∣
=1
C=
⎣
⎢
⎢
⎡
2
−3
1
−4
6
−2
2
−3
1
⎦
⎥
⎥
⎤
∴Adj A=
⎣
⎢
⎢
⎡
2
−4
2
−3
6
−3
1
−2
1
⎦
⎥
⎥
⎤
Then (Adj A)B=
⎣
⎢
⎢
⎡
2
−4
2
−3
6
−3
1
−2
1
⎦
⎥
⎥
⎤
⎣
⎢
⎢
⎡
6
14
30
⎦
⎥
⎥
⎤
=
⎣
⎢
⎢
⎡
0
0
0
⎦
⎥
⎥
⎤
=O
Hence, both conditions ∣A∣=0 and (Adj A)B=) are satisfied then the system of equations is consistent and has an infinite number of solutions.