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Question
Find the inverse of the matrices , if it exists.
⎡⎢⎣1−32305−2−50⎤⎥⎦
⎡⎢⎣1−32305−2−50⎤⎥⎦
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Solution
Let A=⎡⎢⎣1−32305−2−50⎤⎥⎦
We know that A=IA
∴⎡⎢⎣1−32305−2−50⎤⎥⎦=⎡⎢⎣100010001⎤⎥⎦A
Applying R2→R2+3R1 and R3→R3−2R1, we have:
⎡⎢⎣10039−1−2−114⎤⎥⎦=⎡⎢⎣13−2010001⎤⎥⎦A
Applying R1→R1+3R3 and R2→R2+8R3,
We have:
⎡⎢⎣10001−110214⎤⎥⎦=⎡⎢⎣−5−13−2010381⎤⎥⎦A
Applying R3→R3+R2 , we have:
⎡⎢⎣100010102125⎤⎥⎦=⎡⎢⎣−5−13−15011389⎤⎥⎦A
Applying R3→125R3, we have:
⎡⎢⎣10001010211⎤⎥⎦=⎡⎢
⎢⎣−5−13−350112538925⎤⎥
⎥⎦A
Applying R1→R1−10R3 and R2→R2−21R3, we have:
⎡⎢⎣100010001⎤⎥⎦=⎡⎢
⎢
⎢⎣1−25−35−25425125−351125925⎤⎥
⎥
⎥⎦A
∴A−1=⎡⎢
⎢
⎢⎣1−25−35−25425125−351125925⎤⎥
⎥
⎥⎦
We know that A=IA
∴⎡⎢⎣1−32305−2−50⎤⎥⎦=⎡⎢⎣100010001⎤⎥⎦A
Applying R2→R2+3R1 and R3→R3−2R1, we have:
⎡⎢⎣10039−1−2−114⎤⎥⎦=⎡⎢⎣13−2010001⎤⎥⎦A
Applying R1→R1+3R3 and R2→R2+8R3,
We have:
⎡⎢⎣10001−110214⎤⎥⎦=⎡⎢⎣−5−13−2010381⎤⎥⎦A
Applying R3→R3+R2 , we have:
⎡⎢⎣100010102125⎤⎥⎦=⎡⎢⎣−5−13−15011389⎤⎥⎦A
Applying R3→125R3, we have:
⎡⎢⎣10001010211⎤⎥⎦=⎡⎢ ⎢⎣−5−13−350112538925⎤⎥ ⎥⎦A
Applying R1→R1−10R3 and R2→R2−21R3, we have:
⎡⎢⎣100010001⎤⎥⎦=⎡⎢ ⎢ ⎢⎣1−25−35−25425125−351125925⎤⎥ ⎥ ⎥⎦A
∴A−1=⎡⎢ ⎢ ⎢⎣1−25−35−25425125−351125925⎤⎥ ⎥ ⎥⎦
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