x+y-z=6 2x-3y+z=1 2x-4y+2z=1 solve by matrix method
Answers
Answer:
Here A=
⎣
⎢
⎢
⎡
2
1
3
1
3
1
1
−1
−2
⎦
⎥
⎥
⎤
∣A∣=
∣
∣
∣
∣
∣
∣
∣
∣
2
1
3
1
3
1
1
−1
−2
∣
∣
∣
∣
∣
∣
∣
∣
=2(−6+1)−1(−2+3)+1(1−9)
=−10−1−8=−19
=0
Let C
ij
be the factors of the elements a
ij
in A[a
ij
].Then
C
11
=(−1)
1+1
∣
∣
∣
∣
∣
∣
3
1
−1
−2
∣
∣
∣
∣
∣
∣
=−6+1=−5
C
12
=(−1)
1+2
∣
∣
∣
∣
∣
∣
1
3
−1
−2
∣
∣
∣
∣
∣
∣
=−(−2+3)=−1
C
13
=(−1)
1+3
∣
∣
∣
∣
∣
∣
1
3
3
1
∣
∣
∣
∣
∣
∣
=1−9=−8
C
21
=(−1)
2+1
∣
∣
∣
∣
∣
∣
1
1
1
−2
∣
∣
∣
∣
∣
∣
=−(−2−1)=3
C
22
=(−1)
2+2
∣
∣
∣
∣
∣
∣
2
3
1
−2
∣
∣
∣
∣
∣
∣
=−4−3=−7
C
23
=(−1)
2+3
∣
∣
∣
∣
∣
∣
2
3
1
1
∣
∣
∣
∣
∣
∣
=−(2−3)=1
C
31
=(−1)
3+1
∣
∣
∣
∣
∣
∣
1
3
1
−1
∣
∣
∣
∣
∣
∣
=−1−3=−4
C
32
=(−1)
3+2
∣
∣
∣
∣
∣
∣
2
1
1
−1
∣
∣
∣
∣
∣
∣
=−(−2−1)=3
C
33
=(−1)
3+3
∣
∣
∣
∣
∣
∣
2
1
1
3
∣
∣
∣
∣
∣
∣
=6−1=5
adjA=
⎣
⎢
⎢
⎡
−5
3
−4
−1
−7
3
−8
1
5
⎦
⎥
⎥
⎤
T
=
⎣
⎢
⎢
⎡
−5
−1
−8
3
−7
1
−4
3
5
⎦
⎥
⎥
⎤
A
−1
=
∣A∣
1
adjA
=
−19
1
⎣
⎢
⎢
⎡
−5
−1
−8
3
−7
1
−4
3
5
⎦
⎥
⎥
⎤
X=A
−1
B
⎣
⎢
⎢
⎡
x
y
z
⎦
⎥
⎥
⎤
=
−19
1
⎣
⎢
⎢
⎡
−5
−1
−8
3
−7
1
−4
3
5
⎦
⎥
⎥
⎤
⎣
⎢
⎢
⎡
2
5
6
⎦
⎥
⎥
⎤
⇒
⎣
⎢
⎢
⎡
x
y
z
⎦
⎥
⎥
⎤
=
−19
1
⎣
⎢
⎢
⎡
−19
−19
19
⎦
⎥
⎥
⎤
⇒
⎣
⎢
⎢
⎡
x
y
z
⎦
⎥
⎥
⎤
=
−19
1
⎣
⎢
⎢
⎡
−19
−19
19
⎦
⎥
⎥
⎤
⇒x=
−19
−19
,y=
−19
−19
and z=
−19
19
∴x=1,y=1 and z=−1
Step-by-step explanation:
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