find the non trivial solution for the following equations 2w+3x-y-z=0.
4w-6x-2y+2z=0
-6w+12x+3y-4z=0
Answers
Solution
verified
Verified by Toppr
Given system of equation-
3x+y+z=10
2x−y−z=0
x−y+2z=1
Let A=
⎣
⎢
⎢
⎡
3
2
1
1
−1
−1
1
−1
2
⎦
⎥
⎥
⎤
,B=
⎣
⎢
⎢
⎡
10
0
−1
⎦
⎥
⎥
⎤
,X=
⎣
⎢
⎢
⎡
x
y
z
⎦
⎥
⎥
⎤
Now,
AX=B
⎣
⎢
⎢
⎡
3
2
1
1
−1
−1
1
−1
2
⎦
⎥
⎥
⎤
⎣
⎢
⎢
⎡
X
Y
Z
⎦
⎥
⎥
⎤
,X=
⎣
⎢
⎢
⎡
10
0
−1
⎦
⎥
⎥
⎤
∣A∣=
∣
∣
∣
∣
∣
∣
∣
∣
3
2
1
1
−1
−1
1
−1
2
∣
∣
∣
∣
∣
∣
∣
∣
=3(−2−1)−1(4−(−1))+1(−2−(−1))=−9−5−1=−15
∵∣A∣=−15
=0
Thus, system of equation is consistent and has a unique solution.
AX=B
⇒X=A
−1
B
Now,
A
−1
=
∣A∣
adj.(A)
adj.(a)=
⎣
⎢
⎢
⎡
A
11
A
21
A
31
A
12
A
22
A
32
A
13
A
23
A
33
⎦
⎥
⎥
⎤
A=
⎣
⎢
⎢
⎡
3
2
1
1
−1
−1
1
−1
2
⎦
⎥
⎥
⎤
A
11
=−2−1=−3,A
12
=−[4−(−1)]=−5A
13
=−2−(−1)=−1
A
21
=−[2−(−1)]=−3,A
22
=6−1=5A
23
=−[−3−(1)]=4
A
31
=−1−(−1)=0,A
32
=−[−3−2]=5A
33
=−3−2=−5
Therefore,
adj(A)=
⎣
⎢
⎢
⎡
−3
−3
0
−5
5
5
−1
4
−5
⎦
⎥
⎥
⎤
′
=
⎣
⎢
⎢
⎡
−3
−5
−1
−3
5
4
0
5
−5
⎦
⎥
⎥
⎤
∴A
−1
=
−15
1
⎣
⎢
⎢
⎡
−3
−5
−1
−3
5
4
0
5
−5
⎦
⎥
⎥
⎤
=
⎣
⎢
⎢
⎡
5
1
3
1
15
1
5
1
−
3
1
15
−4
0
−
3
1
3
1
⎦
⎥
⎥
⎤
Therefore,
⎣
⎢
⎢
⎡
x
y
z
⎦
⎥
⎥
⎤
=
⎣
⎢
⎢
⎡
5
1
3
1
15
1
5
1
−
3
1
15
−4
0
−
3
1
3
1
⎦
⎥
⎥
⎤
⎣
⎢
⎢
⎡
10
0
−1
⎦
⎥
⎥
⎤
⎣
⎢
⎢
⎡
x
y
Z
⎦
⎥
⎥
⎤
=
⎣
⎢
⎢
⎢
⎢
⎢
⎡
2+0+0
3
10
+0+
3
1
3
2
+0−
3
1
⎦
⎥
⎥
⎥
⎥
⎥
⎤
=
⎣
⎢
⎢
⎢
⎢
⎢
⎡
2
3
11
3
1
⎦
⎥
⎥
⎥
⎥
⎥
⎤
Hence value of x,y and z are 2,
3
11
and
3
1
respectively.