Solve the following equations by the method of inversion. (i) x + y = 1, y + z =5/3 , z +x=4/3 [ Find A inverse using row transformations.]
please solve this question and steps by steps
Answers
Answer:
a
Step-by-step explanation:
Answer:
For getting the inverse of the given matrix A by row elementary operations we may write the given matrix as A=IA
And, we know AA
−1
=I
Now ∣A∣=2×7+1×(−12)+3×(−1)=14−12−3=−1
=0
So, A
−1
is possible.
∵
⎣
⎢
⎢
⎡
2
−5
−3
−1
3
2
3
1
3
⎦
⎥
⎥
⎤
=
⎣
⎢
⎢
⎡
1
0
0
0
1
0
0
0
1
⎦
⎥
⎥
⎤
A
[R
2
→R
2
+R
1
]
⇒
⎣
⎢
⎢
⎡
2
−3
−3
−1
2
2
3
4
3
⎦
⎥
⎥
⎤
=
⎣
⎢
⎢
⎡
1
1
0
0
1
0
0
0
1
⎦
⎥
⎥
⎤
A
[
∵R
3
→
R
3
−
R
2
]
⇒
⎣
⎢
⎢
⎡
2
−3
0
−1
2
0
3
4
−1
⎦
⎥
⎥
⎤
=
⎣
⎢
⎢
⎡
1
1
−1
0
1
−1
0
0
1
⎦
⎥
⎥
⎤
A
[
R
1
→
R
1
+
R
2
]
⇒
⎣
⎢
⎢
⎡
−1
−3
0
1
2
0
7
4
−1
⎦
⎥
⎥
⎤
=
⎣
⎢
⎢
⎡
2
1
−1
1
1
−1
0
0
1
⎦
⎥
⎥
⎤
A
[
∵R
2
→
R
2
−
3R
1
]
⇒
⎣
⎢
⎢
⎡
−1
0
0
1
−1
0
7
−17
−1
⎦
⎥
⎥
⎤
=
⎣
⎢
⎢
⎡
2
−5
−1
1
−2
−1
0
0
1
⎦
⎥
⎥
⎤
A
[
∵R
1
→
andR
3
→
R
1
+
−1
R
2
×R
3
]
⇒
⎣
⎢
⎢
⎡
−1
0
0
0
−1
0
−10
−17
1
⎦
⎥
⎥
⎤
=
⎣
⎢
⎢
⎡
−3
−5
1
−1
−2
1
0
0
−1
⎦
⎥
⎥
⎤
A
[
∵R
1
→
andR
2
→
R
1
+
R
2
+
10R
3
17R
3
]
⇒
⎣
⎢
⎢
⎡
−1
0
0
0
−1
0
0
0
1
⎦
⎥
⎥
⎤
=
⎣
⎢
⎢
⎡
7
12
1
9
15
1
−10
−17
−1
⎦
⎥
⎥
⎤
A
[
∵R
1
→
andR
2
→
−1R
1
−1R
2
]
⇒
⎣
⎢
⎢
⎡
1
0
0
0
1
0
0
0
1
⎦
⎥
⎥
⎤
=
⎣
⎢
⎢
⎡
−7
−12
1
−9
−15
1
10
17
−1
⎦
⎥
⎥
⎤
A
As, I=A
−1
A=AA
−1
So, the inverse of A is A
−1
=
⎣
⎢
⎢
⎡
−7
−12
1
−9
−15
1
10
17
−1
⎦
⎥
⎥
⎤
Step-by-step explanation: