1
Question
Using elementary transformations, find the inverse of the followng matrix.
⎡⎢⎣13−2−30−5250⎤⎥⎦
Using elementary transformations, find the inverse of the followng matrix.
⎡⎢⎣13−2−30−5250⎤⎥⎦
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Solution
Let A=⎡⎢⎣13−2−30−5250⎤⎥⎦. We know that A=IA,
∴⎡⎢⎣13−2−30−5250⎤⎥⎦=⎡⎢⎣13−2−39−110−14⎤⎥⎦=⎡⎢⎣100310201⎤⎥⎦A
(Using R2→R2+3R1 and R3→R3−2R1)
⇒⎡⎢
⎢⎣13−201−1190−14⎤⎥
⎥⎦=⎡⎢
⎢⎣10013190−201⎤⎥
⎥⎦A (Using R2→19R2)
⇒⎡⎢
⎢
⎢⎣13−201−11900259⎤⎥
⎥
⎥⎦=⎡⎢
⎢
⎢⎣10013190−53191⎤⎥
⎥
⎥⎦A (Using R3→R3+R2)
⇒⎡⎢
⎢⎣13−201−119001⎤⎥
⎥⎦=⎡⎢
⎢
⎢⎣10013190−35425925⎤⎥
⎥
⎥⎦A (Using R3→(925)R3)
⇒⎡⎢
⎢
⎢⎣−152251825−254251125−35125925⎤⎥
⎥
⎥⎦A
(Using R2→R2+(119)R3 and R1→R1+2R3)
⇒⎡⎢⎣100010001⎤⎥⎦=⎡⎢
⎢
⎢⎣1−25−35−254251125−35125925⎤⎥
⎥
⎥⎦A (Using R1→R1−3R2)
Hence, A−1=⎡⎢
⎢
⎢⎣1−25−35−254251125−351595⎤⎥
⎥
⎥⎦=125⎡⎢⎣25−10−15−10411−1519⎤⎥⎦(∵AA−1=I)
Let A=⎡⎢⎣13−2−30−5250⎤⎥⎦. We know that A=IA,
∴⎡⎢⎣13−2−30−5250⎤⎥⎦=⎡⎢⎣13−2−39−110−14⎤⎥⎦=⎡⎢⎣100310201⎤⎥⎦A
(Using R2→R2+3R1 and R3→R3−2R1)
⇒⎡⎢
⎢⎣13−201−1190−14⎤⎥
⎥⎦=⎡⎢
⎢⎣10013190−201⎤⎥
⎥⎦A (Using R2→19R2)
⇒⎡⎢
⎢
⎢⎣13−201−11900259⎤⎥
⎥
⎥⎦=⎡⎢
⎢
⎢⎣10013190−53191⎤⎥
⎥
⎥⎦A (Using R3→R3+R2)
⇒⎡⎢
⎢⎣13−201−119001⎤⎥
⎥⎦=⎡⎢
⎢
⎢⎣10013190−35425925⎤⎥
⎥
⎥⎦A (Using R3→(925)R3)
⇒⎡⎢
⎢
⎢⎣−152251825−254251125−35125925⎤⎥
⎥
⎥⎦A
(Using R2→R2+(119)R3 and R1→R1+2R3)
⇒⎡⎢⎣100010001⎤⎥⎦=⎡⎢
⎢
⎢⎣1−25−35−254251125−35125925⎤⎥
⎥
⎥⎦A (Using R1→R1−3R2)
Hence, A−1=⎡⎢
⎢
⎢⎣1−25−35−254251125−351595⎤⎥
⎥
⎥⎦=125⎡⎢⎣25−10−15−10411−1519⎤⎥⎦(∵AA−1=I)
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