Solve equations by matrix method. X+y+z =6. 2x5y+5z=27. 2x+5y11z=45
Answers
Answer:
Given system of equations
x−y+2z=7
3x+4y−5z=−5
2x−y+3z=12
This can be written as
AX=B
where A=
⎣
⎢
⎢
⎡
1
3
2
−1
4
−1
2
−5
3
⎦
⎥
⎥
⎤
,X=
⎣
⎢
⎢
⎡
x
y
z
⎦
⎥
⎥
⎤
,B=
⎣
⎢
⎢
⎡
7
−5
12
⎦
⎥
⎥
⎤
Here, ∣A∣=1(12−5)+1(9+10)+2(−3−8)
⇒∣A∣=7+19−22=4
Since, ∣A∣
=0
Hence, the system of equations is consistent and has a unique solution given by X==A
−1
B
A
−1
=
∣A∣
adjA
and adjA=C
T
C
11
=(−1)
1+1
∣
∣
∣
∣
∣
∣
4
−1
−5
3
∣
∣
∣
∣
∣
∣
⇒C
11
=12−5=7
C
12
=(−1)
1+2
∣
∣
∣
∣
∣
∣
3
2
−5
3
∣
∣
∣
∣
∣
∣
⇒C
12
=−(9+10)=−19
C
13
=(−1)
1+3
∣
∣
∣
∣
∣
∣
3
2
4
−1
∣
∣
∣
∣
∣
∣
⇒C
13
=−3−8=−11
C
21
=(−1)
2+1
∣
∣
∣
∣
∣
∣
−1
−1
2
3
∣
∣
∣
∣
∣
∣
⇒C
21
=−(−3+2)=1
C
22
=(−1)
2+2
∣
∣
∣
∣
∣
∣
1
2
2
3
∣
∣
∣
∣
∣
∣
⇒C
22
=3−4=−1
C
23
=(−1)
2+3
∣
∣
∣
∣
∣
∣
1
2
−1
−1
∣
∣
∣
∣
∣
∣
⇒C
23
=−(−1+2)=−1
C
31
=(−1)
3+1
∣
∣
∣
∣
∣
∣
−1
4
2
−5
∣
∣
∣
∣
∣
∣
⇒C
31
=5−8=−3
C
32
=(−1)
3+2
∣
∣
∣
∣
∣
∣
1
3
2
−5
∣
∣
∣
∣
∣
∣
⇒C
32
=−(−5−6)=11
C
33
=(−1)
3+3
∣
∣
∣
∣
∣
∣
1
3
−1
4
∣
∣
∣
∣
∣
∣
⇒C
33
=4+3=7
Hence, the co-factor matrix is C=
⎣
⎢
⎢
⎡
7
1
−3
−19
−1
11
−11
−1
7
⎦
⎥
⎥
⎤
⇒adjA=C
T
=
⎣
⎢
⎢
⎡
7
−19
−11
1
−1
−1
−3
11
7
⎦
⎥
⎥
⎤
⇒A
−1
=
∣A∣
adjA
=
4
1
⎣
⎢
⎢
⎡
7
−19
−11
1
−1
−1
−3
11
7
⎦
⎥
⎥
⎤
Solution is given by
⎣
⎢
⎢
⎡
x
y
z
⎦
⎥
⎥
⎤
=
4
1
⎣
⎢
⎢
⎡
7
−19
−11
1
−1
−1
−3
11
7
⎦
⎥
⎥
⎤
⎣
⎢
⎢
⎡
7
−5
12
⎦
⎥
⎥
⎤
⎣
⎢
⎢
⎡
x
y
z
⎦
⎥
⎥
⎤
=
4
1
⎣
⎢
⎢
⎡
49−−5−36
−133+5+132
−77+5+84
⎦
⎥
⎥
⎤
⎣
⎢
⎢
⎡
x
y
z
⎦
⎥
⎥
⎤
=
4
1
⎣
⎢
⎢
⎡
8
4
12
⎦
⎥
⎥
⎤
⎣
⎢
⎢
⎡
x
y
z
⎦
⎥
⎥
⎤
=
⎣
⎢
⎢
⎡
2
1
3
⎦
⎥
⎥
⎤
Hence, x=2,y=1,z=3