Solve the equation using matrices
3x+2y+z=16;
2x+3y+2z=23;
5x+2y+2z=21.
Answers
Answer:
Step-by-step explanation:
2x+3y+3z=5
x−2y+z=−4
3x−y−2z=3.
Represent it in matrix form
⎣
⎢
⎢
⎡
2
1
3
3
−2
−1
3
1
−2
⎦
⎥
⎥
⎤
⎣
⎢
⎢
⎡
x
y
z
⎦
⎥
⎥
⎤
=
⎣
⎢
⎢
⎡
5
−4
3
⎦
⎥
⎥
⎤
which is in the form of AX=B
A=
⎣
⎢
⎢
⎡
2
1
3
3
−2
−1
3
1
−2
⎦
⎥
⎥
⎤
∣A∣=10+15+15=40
=0
∴ A
−1
exists
To find adjoint of A
A
11
=5,A
12
=5,A
13
=5
A
21
=3,A
22
=−13,A
23
=11
A
31
=9,A
32
=1,A
33
=−7
Adj(A)=co-factor
⎣
⎢
⎢
⎡
5
3
9
5
−13
1
5
11
−7
⎦
⎥
⎥
⎤
=
⎣
⎢
⎢
⎡
5
5
5
3
−13
11
9
1
−7
⎦
⎥
⎥
⎤
A
−1
=
∣A∣
1
Adj(A)
=
40
1
⎣
⎢
⎢
⎡
5
5
5
3
−13
11
9
1
−7
⎦
⎥
⎥
⎤
X=A
−1
B
=
40
1
⎣
⎢
⎢
⎡
5
5
5
3
−13
11
9
1
−7
⎦
⎥
⎥
⎤
⎣
⎢
⎢
⎡
5
−4
3
⎦
⎥
⎥
⎤
X=
40
1
⎣
⎢
⎢
⎡
25−12+27
25+52+3
25−44−21
⎦
⎥
⎥
⎤
X=
40
1
⎣
⎢
⎢
⎡
40
80
−40
⎦
⎥
⎥
⎤
⎣
⎢
⎢
⎡
x
y
z
⎦
⎥
⎥
⎤
=
⎣
⎢
⎢
⎡
1
2
−1
⎦
⎥
⎥
⎤
Hence, x=1,y=2 and z=−1