Question 6 [10 points]
Consider the following system of linear equations:
11+x2-x3 = -2
X1+4x2-4x3 = -14
-3x1-6X7+7x3 19
Let A be the coefficient matrix and X the solution matrix to the system. Solve the system by first computing A-l and then using it to find X.
You can resize a matrix (when appropriate) by clicking and dragging the bottom-right corner of the matrix. Find A inverse and X.
Answers
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Answer:
A=
⎣
⎢
⎢
⎡
8
2
1
4
1
2
3
1
2
⎦
⎥
⎥
⎤
AA
−1
=I
⎣
⎢
⎢
⎡
8
2
1
4
1
2
3
1
2
⎦
⎥
⎥
⎤
A
−1
=
⎣
⎢
⎢
⎡
1
0
0
0
1
0
0
0
1
⎦
⎥
⎥
⎤
R
1
→R
1
−4R
2
⎣
⎢
⎢
⎡
0
2
1
0
1
2
−1
1
2
⎦
⎥
⎥
⎤
A
−1
=
⎣
⎢
⎢
⎡
1
0
0
−4
1
0
0
0
1
⎦
⎥
⎥
⎤
R
1
→R
1
×−1
⎣
⎢
⎢
⎡
0
2
1
0
1
2
1
1
2
⎦
⎥
⎥
⎤
A
−1
=
⎣
⎢
⎢
⎡
−1
0
0
4
1
0
0
0
1
⎦
⎥
⎥
⎤
C
1
⟺C
3
⎣
⎢
⎢
⎡
1
1
2
0
1
2
0
2
1
⎦
⎥
⎥
⎤
A
−1
=
⎣
⎢
⎢
⎡
0
0
1
4
1
0
−1
0
0
⎦
⎥
⎥
⎤
R
2
→R
2
−R
1
⎣
⎢
⎢
⎡
1
0
2
0
1
2
0
2
1
⎦
⎥
⎥
⎤
A
−1
=
⎣
⎢
⎢
⎡
0
0
1
4
−3
0
−1
1
0
⎦
⎥
⎥
⎤
R
3
→R
3
−2R
1
⎣
⎢
⎢
⎡
1
0
0
0
1
2
0
2
1
⎦
⎥
⎥
⎤
A
−1
=
⎣
⎢
⎢
⎡
0
0
1
4
−3
−8
−1
1
2
⎦
⎥
⎥
⎤
C
3
→C
3
−2C
2
⎣
⎢
⎢
⎡
1
0
0
0
1
2
0
0
−3
⎦
⎥
⎥
⎤
A
−1
=
⎣
⎢
⎢
⎡
0
0
1
4
−3
−8
−9
7
18
⎦
⎥
⎥
⎤
R
3
→R
3
−2R
2
⎣
⎢
⎢
⎡
1
0
0
0
1
0
0
0
−3
⎦
⎥
⎥
⎤
A
−1
=
⎣
⎢
⎢
⎡
0
0
1
4
−3
−2
−9
7
4
⎦
⎥
⎥
⎤
R
3
→R
3
×
3
−1
⎣
⎢
⎢
⎡
1
0
0
0
1
0
0
0
1
⎦
⎥
⎥
⎤
A
−1
=
⎣
⎢
⎢
⎢
⎡
0
0
3
−1
4
−3
3
2
−9
7
3
−4
⎦
⎥
⎥
⎥
⎤
A
−1
=
⎣
⎢
⎢
⎢
⎡
0
0
3
−1
4
−3
3
2
−9
7
3
−4
⎦
⎥
⎥
⎥
⎤
⎣
⎢
⎢
⎡
x
y
z
⎦
⎥
⎥
⎤
=
⎣
⎢
⎢
⎢
⎡
0
0
3
−1
4
−3
3
2
−9
7
3
−4
⎦
⎥
⎥
⎥
⎤
⎣
⎢
⎢
⎡
19
5
7
⎦
⎥
⎥
⎤
⎣
⎢
⎢
⎡
x
y
z
⎦
⎥
⎥
⎤
=
⎣
⎢
⎢
⎢
⎡
20−63
−15+49
3
−19+10−28
⎦
⎥
⎥
⎥
⎤
⎣
⎢
⎢
⎡
x
y
z
⎦
⎥
⎥
⎤
=
⎣
⎢
⎢
⎢
⎡
−43
34
3
−37
⎦
⎥
⎥
⎥
⎤