5
Question
Using elementary transformation, find the inverse of the matrix ⎡⎢⎣2−332233−22⎤⎥⎦.
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Solution
⎡⎢⎣2−332233−22⎤⎥⎦A−1=⎡⎢⎣100010001⎤⎥⎦R1→R3⎡⎢⎣3−222232−33⎤⎥⎦A−1=⎡⎢⎣001010100⎤⎥⎦R1−R2⎡⎢⎣1−4−12232−33⎤⎥⎦A−1=⎡⎢⎣0−11010100⎤⎥⎦R2−2R1, R3−2R1⎡⎢⎣1−4−10105055⎤⎥⎦A−1=⎡⎢⎣0−1103−212−2⎤⎥⎦R2/10⎡⎢
⎢
⎢⎣1−4−10112055⎤⎥
⎥
⎥⎦A−1=⎡⎢⎣0−1103/10−2/1012−2⎤⎥⎦R1+4R2,R3−5R2⎡⎢
⎢
⎢
⎢
⎢⎣10101120052⎤⎥
⎥
⎥
⎥
⎥⎦A−1=⎡⎢
⎢
⎢
⎢
⎢
⎢
⎢⎣015150310−210112−1⎤⎥
⎥
⎥
⎥
⎥
⎥
⎥⎦R3×25,R1−R3,R1−13R3,⎡⎢⎣100010001⎤⎥⎦A−1=⎡⎢
⎢
⎢
⎢
⎢
⎢
⎢⎣−25035−151502515−25⎤⎥
⎥
⎥
⎥
⎥
⎥
⎥⎦A−1=15⎡⎢⎣−203−11021−2⎤⎥⎦.
⎡⎢⎣2−332233−22⎤⎥⎦A−1=⎡⎢⎣100010001⎤⎥⎦
R1→R3
⎡⎢⎣3−222232−33⎤⎥⎦A−1=⎡⎢⎣001010100⎤⎥⎦
R1−R2
⎡⎢⎣1−4−12232−33⎤⎥⎦A−1=⎡⎢⎣0−11010100⎤⎥⎦
R2−2R1, R3−2R1
⎡⎢⎣1−4−10105055⎤⎥⎦A−1=⎡⎢⎣0−1103−212−2⎤⎥⎦
R2/10
⎡⎢
⎢
⎢⎣1−4−10112055⎤⎥
⎥
⎥⎦A−1=⎡⎢⎣0−1103/10−2/1012−2⎤⎥⎦
R1+4R2,R3−5R2
⎡⎢
⎢
⎢
⎢
⎢⎣10101120052⎤⎥
⎥
⎥
⎥
⎥⎦A−1=⎡⎢
⎢
⎢
⎢
⎢
⎢
⎢⎣015150310−210112−1⎤⎥
⎥
⎥
⎥
⎥
⎥
⎥⎦
R3×25,R1−R3,R1−13R3,
⎡⎢⎣100010001⎤⎥⎦A−1=⎡⎢
⎢
⎢
⎢
⎢
⎢
⎢⎣−25035−151502515−25⎤⎥
⎥
⎥
⎥
⎥
⎥
⎥⎦
A−1=15⎡⎢⎣−203−11021−2⎤⎥⎦.
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