1
Question
Find the inverse of ⎡⎢⎣3−1−220−13−50⎤⎥⎦ by using elementary row transformations.
Open in App
Solution
A=IAI=A−1AA=IA
⎡⎢⎣3−1−220−13−50⎤⎥⎦=⎡⎢⎣100010001⎤⎥⎦A R1→R1−R2
⎡⎢⎣1−1−120−13−50⎤⎥⎦=⎡⎢⎣1−10010001⎤⎥⎦A R2→R2−2R1R3→R3−3R1
⎡⎢⎣1−1−10210−23⎤⎥⎦=⎡⎢⎣1−10−230−331⎤⎥⎦A R3→R3+R2
⎡⎢⎣1−1−1021004⎤⎥⎦=⎡⎢⎣1−10−230−561⎤⎥⎦A R2→R2/2R3→R3/4
⎡⎢⎣1−1−1011/2001⎤⎥⎦=⎡⎢⎣1−10−13/20−5/46/41/4⎤⎥⎦A R1→R1+R2
⎡⎢
⎢
⎢⎣10−120112001⎤⎥
⎥
⎥⎦=⎡⎢
⎢
⎢⎣1120−1320−546414⎤⎥
⎥
⎥⎦A R1→R1+R3/2R2→R2−R3/2
⎡⎢⎣100010001⎤⎥⎦= ⎡⎢
⎢
⎢
⎢⎣−585418−38−34−18−546414⎤⎥
⎥
⎥
⎥⎦A
I=A−1A
⇒A−1=⎡⎢
⎢
⎢
⎢⎣−585418−38−34−18−546414⎤⎥
⎥
⎥
⎥⎦
I=A−1A
A=IA
⎡⎢⎣3−1−220−13−50⎤⎥⎦=⎡⎢⎣100010001⎤⎥⎦A R1→R1−R2
⎡⎢⎣1−1−120−13−50⎤⎥⎦=⎡⎢⎣1−10010001⎤⎥⎦A R2→R2−2R1R3→R3−3R1
⎡⎢⎣1−1−10210−23⎤⎥⎦=⎡⎢⎣1−10−230−331⎤⎥⎦A R3→R3+R2
⎡⎢⎣1−1−1021004⎤⎥⎦=⎡⎢⎣1−10−230−561⎤⎥⎦A R2→R2/2R3→R3/4
⎡⎢⎣1−1−1011/2001⎤⎥⎦=⎡⎢⎣1−10−13/20−5/46/41/4⎤⎥⎦A R1→R1+R2
⎡⎢
⎢
⎢⎣10−120112001⎤⎥
⎥
⎥⎦=⎡⎢
⎢
⎢⎣1120−1320−546414⎤⎥
⎥
⎥⎦A R1→R1+R3/2R2→R2−R3/2
⎡⎢⎣100010001⎤⎥⎦= ⎡⎢
⎢
⎢
⎢⎣−585418−38−34−18−546414⎤⎥
⎥
⎥
⎥⎦A
I=A−1A
⇒A−1=⎡⎢
⎢
⎢
⎢⎣−585418−38−34−18−546414⎤⎥
⎥
⎥
⎥⎦
Suggest Corrections
0
Similar questions
View More
Join BYJU'S Learning Program
Explore more
Join BYJU'S Learning Program
