1
Question
Find the inverse of matrix by elementary row transformations.
A=⎡⎢⎣121−10221−3⎤⎥⎦
A=⎡⎢⎣121−10221−3⎤⎥⎦
Open in App
Solution
A=⎡⎢⎣121−10221−3⎤⎥⎦Now, for inverse of A,A=IA⎡⎢⎣121−10221−3⎤⎥⎦ = ⎡⎢⎣100010001⎤⎥⎦A⎡⎢⎣12102321−3⎤⎥⎦ = ⎡⎢⎣100110001⎤⎥⎦A (R2→R1+R2)⎡⎢⎣1210230−3−5⎤⎥⎦ = ⎡⎢⎣100110−201⎤⎥⎦A (R3→R3−2R1)
⎡⎢⎣10−20230−3−5⎤⎥⎦ = ⎡⎢⎣0−10110−201⎤⎥⎦A (R1→R1−R2)
⎡⎢
⎢
⎢⎣10−201320−3−5⎤⎥
⎥
⎥⎦ = ⎡⎢
⎢
⎢⎣0−1012120−201⎤⎥
⎥
⎥⎦A (R2→12R2)
⎡⎢
⎢
⎢
⎢
⎢⎣10−2013200−12⎤⎥
⎥
⎥
⎥
⎥⎦ = ⎡⎢
⎢
⎢
⎢
⎢⎣0−1012120−12321⎤⎥
⎥
⎥
⎥
⎥⎦A (R3→R3+3R2)⎡⎢
⎢
⎢
⎢
⎢⎣100013200−12⎤⎥
⎥
⎥
⎥
⎥⎦ = ⎡⎢
⎢
⎢
⎢
⎢⎣2−7−412120−12321⎤⎥
⎥
⎥
⎥
⎥⎦A (R1→R1−4R3)
⎡⎢
⎢
⎢⎣10001000−12⎤⎥
⎥
⎥⎦ = ⎡⎢
⎢
⎢⎣2−7−4−153−12321⎤⎥
⎥
⎥⎦A (R2→R2+3R3)
⎡⎢⎣100010001⎤⎥⎦ = ⎡⎢⎣2−7−4−1531−3−2⎤⎥⎦A (R3→−2R3)
Hence, ⎡⎢⎣2−7−4−1531−3−2⎤⎥⎦ is the inverse of the given matrix.
Now, for inverse of A,
A=IA
⎡⎢⎣121−10221−3⎤⎥⎦ = ⎡⎢⎣100010001⎤⎥⎦A
⎡⎢⎣12102321−3⎤⎥⎦ = ⎡⎢⎣100110001⎤⎥⎦A (R2→R1+R2)
⎡⎢⎣1210230−3−5⎤⎥⎦ = ⎡⎢⎣100110−201⎤⎥⎦A (R3→R3−2R1)
⎡⎢⎣10−20230−3−5⎤⎥⎦ = ⎡⎢⎣0−10110−201⎤⎥⎦A (R1→R1−R2)
⎡⎢
⎢
⎢⎣10−201320−3−5⎤⎥
⎥
⎥⎦ = ⎡⎢
⎢
⎢⎣0−1012120−201⎤⎥
⎥
⎥⎦A (R2→12R2)
⎡⎢
⎢
⎢
⎢
⎢⎣10−2013200−12⎤⎥
⎥
⎥
⎥
⎥⎦ = ⎡⎢
⎢
⎢
⎢
⎢⎣0−1012120−12321⎤⎥
⎥
⎥
⎥
⎥⎦A (R3→R3+3R2)
⎡⎢
⎢
⎢
⎢
⎢⎣100013200−12⎤⎥
⎥
⎥
⎥
⎥⎦ = ⎡⎢
⎢
⎢
⎢
⎢⎣2−7−412120−12321⎤⎥
⎥
⎥
⎥
⎥⎦A (R1→R1−4R3)
⎡⎢
⎢
⎢⎣10001000−12⎤⎥
⎥
⎥⎦ = ⎡⎢
⎢
⎢⎣2−7−4−153−12321⎤⎥
⎥
⎥⎦A (R2→R2+3R3)
⎡⎢⎣100010001⎤⎥⎦ = ⎡⎢⎣2−7−4−1531−3−2⎤⎥⎦A (R3→−2R3)
Hence,
⎡⎢⎣2−7−4−1531−3−2⎤⎥⎦ is the inverse of the given matrix.
Suggest Corrections
0
Similar questions
View More
Join BYJU'S Learning Program
Explore more
Join BYJU'S Learning Program
