solve the equation by matrix method x+y+z=3;y+z=1;x+z=3
Answers
Answer:
From the given set of equations we have,
⎣
⎢
⎢
⎡
1
2
3
1
3
−1
−1
1
−7
⎦
⎥
⎥
⎤
⎣
⎢
⎢
⎡
x
y
z
⎦
⎥
⎥
⎤
=
⎣
⎢
⎢
⎡
3
10
1
⎦
⎥
⎥
⎤
AX=B
X=A
−1
B
A
−1
=
∣A∣
1
adj(A)
∣A∣=
∣
∣
∣
∣
∣
∣
∣
∣
1
2
3
1
3
−1
−1
1
−7
∣
∣
∣
∣
∣
∣
∣
∣
=1(−21+1)−1(−14−3)−1(−2−9)
=8
=0
Next let us find the adjoint of C
C
11
=(−1)
1+1
∣
∣
∣
∣
∣
∣
3
−1
1
−7
∣
∣
∣
∣
∣
∣
=−20
C
12
=(−1)
1+2
∣
∣
∣
∣
∣
∣
2
3
1
−7
∣
∣
∣
∣
∣
∣
=17
C13=(−1)
1+3
∣
∣
∣
∣
∣
∣
2
3
3
−1
∣
∣
∣
∣
∣
∣
=−11
C
21
=(−1)
2+1
∣
∣
∣
∣
∣
∣
1
−1
−1
−7
∣
∣
∣
∣
∣
∣
=8
C
22
=(−1)
2+2
∣
∣
∣
∣
∣
∣
1
3
−1
−7
∣
∣
∣
∣
∣
∣
=−4
C
23
=(−1)
2+3
∣
∣
∣
∣
∣
∣
1
3
1
−1
∣
∣
∣
∣
∣
∣
=4
C
31
=(−1)
3+1
∣
∣
∣
∣
∣
∣
1
3
−1
1
∣
∣
∣
∣
∣
∣
=4
C
32
=(−1)
3+2
∣
∣
∣
∣
∣
∣
1
2
−1
1
∣
∣
∣
∣
∣
∣
=−3
C
33
=(−1)
3+3
∣
∣
∣
∣
∣
∣
1
2
1
3
∣
∣
∣
∣
∣
∣
=1
C
A
=
⎝
⎜
⎜
⎛
−20
8
4
17
−4
−3
−11
4
1
⎠
⎟
⎟
⎞
adj(A)=C
A
T
=
⎝
⎜
⎜
⎛
−20
17
−11
8
−4
4
4
−3
1
⎠
⎟
⎟
⎞
A
−1
=
8
1
⎝
⎜
⎜
⎛
−20
17
−11
8
−4
4
4
−3
1
⎠
⎟
⎟
⎞
X=A
−1
B
=
8
1
⎝
⎜
⎜
⎛
−20
17
−11
8
−4
4
4
−3
1
⎠
⎟
⎟
⎞
⎣
⎢
⎢
⎡
3
10
1
⎦
⎥
⎥
⎤
=
8
1
⎣
⎢
⎢
⎡
24
8
8
⎦
⎥
⎥
⎤
⎣
⎢
⎢
⎡
x
y
z
⎦
⎥
⎥
⎤
=
⎣
⎢
⎢
⎡
3
1
1
⎦
⎥
⎥
⎤
$$
x=3,y=1,z=1
