Solve the following equations with the help of of Rank method a-2y+z-W+1=0 32-22+36 +4=0 5n-4y+W+3=0
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Solve the system of equations, using matrix method
2x+3y+3z=5,x−2y+z=−4,3x−y−2z=3
Medium
Solution
verified
Verified by Toppr
Given system of equations
2x+3y+3z=5
x−2y+z=−4
3x−y−2z=3
This can be written as
AX=B
where A=
⎣
⎢
⎢
⎡
2
1
3
3
−2
−1
3
1
−2
⎦
⎥
⎥
⎤
,X=
⎣
⎢
⎢
⎡
x
y
z
⎦
⎥
⎥
⎤
,B=
⎣
⎢
⎢
⎡
5
−4
3
⎦
⎥
⎥
⎤
Here, ∣A∣=2(4+1)−3(−2−3)+3(−1+6)
⇒∣A∣=10+15+15=40
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
∣
∣
∣
∣
∣
∣
−2
−1
1
−2
∣
∣
∣
∣
∣
∣
⇒C
11
=4+1=5
C
12
=(−1)
1+2
∣
∣
∣
∣
∣
∣
1
3
1
−2
∣
∣
∣
∣
∣
∣
⇒C
12
=−(−2−3)=5
C
13
=(−1)
1+3
∣
∣
∣
∣
∣
∣
1
3
−2
−1
∣
∣
∣
∣
∣
∣
⇒C
13
=−1+6=5
C
21
=(−1)
2+1
∣
∣
∣
∣
∣
∣
3
−1
3
−2
∣
∣
∣
∣
∣
∣
⇒C
21
=−(−6+3)=3
C
22
=(−1)
2+2
∣
∣
∣
∣
∣
∣
2
3
3
−2
∣
∣
∣
∣
∣
∣
⇒C
22
=−4−9=−13
C
23
=(−1)
2+3
∣
∣
∣
∣
∣
∣
2
3
3
−1
∣
∣
∣
∣
∣
∣
⇒C
23
=−(−2−9)=11
C
31
=(−1)
3+1
∣
∣
∣
∣
∣
∣
3
−2
3
1
∣
∣
∣
∣
∣
∣
⇒C
31
=3+6=9
C
32
=(−1)
3+2
∣
∣
∣
∣
∣
∣
2
1
3
1
∣
∣
∣
∣
∣
∣
⇒C
32
=−(2−3)=1
C
33
=(−1)
3+3
∣
∣
∣
∣
∣
∣
2
1
3
−2
∣
∣
∣
∣
∣
∣
⇒C
33
=−4−3=−7
Hence, the co-factor matrix is C=
⎣
⎢
⎢
⎡
5
3
9
5
−13
1
5
11
−7
⎦
⎥
⎥
⎤
⇒adjA=C
T
=
⎣
⎢
⎢
⎡
5
5
5
3
−13
11
9
1
−7
⎦
⎥
⎥
⎤
⇒A
−1
=
∣A∣
adjA
=
40
1
⎣
⎢
⎢
⎡
5
5
5
3
−13
11
9
1
−7
⎦
⎥
⎥
⎤
Solution is given by
⎣
⎢
⎢
⎡
x
y
z
⎦
⎥
⎥
⎤
=
40
1
⎣
⎢
⎢
⎡
5
5
5
3
−13
11
9
1
−7
⎦
⎥
⎥
⎤
⎣
⎢
⎢
⎡
5
−4
3
⎦
⎥
⎥
⎤
⎣
⎢
⎢
⎡
x
y
z
⎦
⎥
⎥
⎤
=
40
1
⎣
⎢
⎢
⎡
25−12+27
25+52+3
25−44−21
⎦
⎥
⎥
⎤
⎣
⎢
⎢
⎡
x
y
z
⎦
⎥
⎥
⎤
=
40
1
⎣
⎢
⎢
⎡
40
80
−40
⎦
⎥
⎥
⎤
⎣
⎢
⎢
⎡
x
y
z
⎦
⎥
⎥
⎤
=
⎣
⎢
⎢
⎡
1
2
−1
⎦
⎥
⎥
⎤
Hence, x=1,y=2,z=