using a matrix method solve for
2x + 3y =5
-5x + y = 13
Answers
Answer:
Given system of linear equations are
2x−3y+5z=11
3x+2y−4z=−5
x+y−2z=−3.
Represent it in matrix form
⎣
⎢
⎢
⎡
2
3
1
−3
2
1
5
−4
−2
⎦
⎥
⎥
⎤
⎣
⎢
⎢
⎡
x
y
z
⎦
⎥
⎥
⎤
=
⎣
⎢
⎢
⎡
11
−5
−3
⎦
⎥
⎥
⎤
which is in the form of AX=B
A=
⎣
⎢
⎢
⎡
2
3
1
−3
2
1
5
−4
−2
⎦
⎥
⎥
⎤
∣A∣=0−6+5=−1
=0
∴ A
−1
exists
To find adjoint of A
A
11
=0,A
12
=2,A
13
=1
A
21
=−1,A
22
=−9,A
23
=−6
A
31
=2,A
32
=23,A
33
=13
Adj(A)=co-factor
⎣
⎢
⎢
⎡
0
−1
2
2
−9
23
1
−6
13
⎦
⎥
⎥
⎤
=
⎣
⎢
⎢
⎡
0
2
1
−1
−9
−6
2
23
13
⎦
⎥
⎥
⎤
A
−1
=
∣A∣
1
Adj(A)
=
−1
1
⎣
⎢
⎢
⎡
0
2
1
−1
−9
−6
2
23
13
⎦
⎥
⎥
⎤
X=A
−1
B
=−
⎣
⎢
⎢
⎡
0
2
1
−1
−9
−6
2
23
13
⎦
⎥
⎥
⎤
⎣
⎢
⎢
⎡
11
−5
−3
⎦
⎥
⎥
⎤
X=−
⎣
⎢
⎢
⎡
0+5−6
22+45−69
11+25−39
⎦
⎥
⎥
⎤
X=−
⎣
⎢
⎢
⎡
−1
−2
−3
⎦
⎥
⎥
⎤
⎣
⎢
⎢
⎡
x
y
z
⎦
⎥
⎥
⎤
=
⎣
⎢
⎢
⎡
1
2
3
⎦
⎥
⎥
⎤
Hence, x=1,y=2 and z=3