1
Question
Using elementary transformations, find the inverse of matrix⎡⎢⎣13−2−30−5250⎤⎥⎦, if it exists.
Using elementary transformations, find the inverse of matrix⎡⎢⎣13−2−30−5250⎤⎥⎦, if it exists.
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Solution
Let A=⎡⎢⎣13−2−30−5250⎤⎥⎦
We know that A=IA
⇒⎡⎢⎣13−2−30−5250⎤⎥⎦=⎡⎢⎣100010001⎤⎥⎦A
Applying R2→R2+3R1
⇒⎡⎢⎣13−2−3+3(1)0+3(3)−5+3(−2)250⎤⎥⎦=⎡⎢⎣1000+3(1)1+3(0)0+3(0)001⎤⎥⎦A
⇒⎡⎢⎣13−209−11250⎤⎥⎦=⎡⎢⎣100310001⎤⎥⎦A
Applying R3→R3−2R1
⇒⎡⎢⎣13−209−112−2(1)5−2(3)0−2(−2)⎤⎥⎦=⎡⎢⎣1003100−2(1)0−2(0)1−2(0)⎤⎥⎦A
⇒⎡⎢⎣13−209−110−14⎤⎥⎦=⎡⎢⎣100310−201⎤⎥⎦A
Applying R2→19R2
⇒⎡⎢
⎢
⎢⎣13−201−1190−14⎤⎥
⎥
⎥⎦=⎡⎢
⎢
⎢⎣10013190−201⎤⎥
⎥
⎥⎦A
Applying R3→R3+R2
⇒⎡⎢
⎢
⎢
⎢
⎢
⎢⎣13−201−1190+0−1+14+(−119)⎤⎥
⎥
⎥
⎥
⎥
⎥⎦=⎡⎢
⎢
⎢
⎢
⎢⎣10013190−2+130+191+0⎤⎥
⎥
⎥
⎥
⎥⎦A
⇒⎡⎢
⎢
⎢
⎢
⎢⎣13−201−11900259⎤⎥
⎥
⎥
⎥
⎥⎦=⎡⎢
⎢
⎢
⎢
⎢⎣10013190−53191⎤⎥
⎥
⎥
⎥
⎥⎦A
Applying R1→R1−3R2
⇒⎡⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢⎣1−3(0)3−3(1)−2−3(−119)01−11900253⎤⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥⎦=⎡⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢⎣1−3(13)0−3(19)0−3(0)13190−53191⎤⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥⎦A
⇒⎡⎢
⎢
⎢
⎢
⎢
⎢
⎢⎣105301−11900259⎤⎥
⎥
⎥
⎥
⎥
⎥
⎥⎦=⎡⎢
⎢
⎢
⎢
⎢
⎢
⎢⎣0−13013190−53191⎤⎥
⎥
⎥
⎥
⎥
⎥
⎥⎦A
Applying R3→925R3
⇒⎡⎢
⎢
⎢
⎢
⎢⎣105301−119001⎤⎥
⎥
⎥
⎥
⎥⎦=⎡⎢
⎢
⎢
⎢
⎢
⎢
⎢⎣0−13013190−35125925⎤⎥
⎥
⎥
⎥
⎥
⎥
⎥⎦A
Applying R2→R2+119R3
⇒⎡⎢
⎢
⎢
⎢
⎢⎣10530+119(0)1+119(0)−119+119(1)001⎤⎥
⎥
⎥
⎥
⎥⎦=⎡⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢⎣0−13013+119(−35)19+119(125)0+119(925)−35125925⎤⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥⎦A
⇒⎡⎢
⎢
⎢⎣1053010001⎤⎥
⎥
⎥⎦=⎡⎢
⎢
⎢
⎢
⎢
⎢
⎢⎣0−130−254251125−35125925⎤⎥
⎥
⎥
⎥
⎥
⎥
⎥⎦A
Applying R1→R1−53R3
⇒⎡⎢
⎢
⎢⎣1−53(0)0−53(0)53−53(1)010001⎤⎥
⎥
⎥⎦=⎡⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢⎣0−53(−35)−13−53(125)0−53(925)−254251125−35125925⎤⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥⎦A
⇒⎡⎢⎣100010001⎤⎥⎦=⎡⎢
⎢
⎢
⎢
⎢
⎢
⎢⎣55−615−35−254251125−35125925⎤⎥
⎥
⎥
⎥
⎥
⎥
⎥⎦A
⇒I=⎡⎢
⎢
⎢
⎢
⎢
⎢
⎢⎣55−615−35−254251125−35125925⎤⎥
⎥
⎥
⎥
⎥
⎥
⎥⎦A
⇒I=⎡⎢
⎢
⎢
⎢
⎢
⎢
⎢⎣1−25−35−254251125−35125925⎤⎥
⎥
⎥
⎥
⎥
⎥
⎥⎦A
This is similar to I=A−1A
Hence, A−1=⎡⎢
⎢
⎢
⎢
⎢
⎢
⎢⎣1−25−35−254251125−35125925⎤⎥
⎥
⎥
⎥
⎥
⎥
⎥⎦
We know that A=IA
⇒⎡⎢⎣13−2−30−5250⎤⎥⎦=⎡⎢⎣100010001⎤⎥⎦A
Applying R2→R2+3R1
⇒⎡⎢⎣13−2−3+3(1)0+3(3)−5+3(−2)250⎤⎥⎦=⎡⎢⎣1000+3(1)1+3(0)0+3(0)001⎤⎥⎦A
⇒⎡⎢⎣13−209−11250⎤⎥⎦=⎡⎢⎣100310001⎤⎥⎦A
Applying R3→R3−2R1
⇒⎡⎢⎣13−209−112−2(1)5−2(3)0−2(−2)⎤⎥⎦=⎡⎢⎣1003100−2(1)0−2(0)1−2(0)⎤⎥⎦A
⇒⎡⎢⎣13−209−110−14⎤⎥⎦=⎡⎢⎣100310−201⎤⎥⎦A
Applying R2→19R2
⇒⎡⎢ ⎢ ⎢⎣13−201−1190−14⎤⎥ ⎥ ⎥⎦=⎡⎢ ⎢ ⎢⎣10013190−201⎤⎥ ⎥ ⎥⎦A
Applying R3→R3+R2
⇒⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣13−201−1190+0−1+14+(−119)⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦=⎡⎢ ⎢ ⎢ ⎢ ⎢⎣10013190−2+130+191+0⎤⎥ ⎥ ⎥ ⎥ ⎥⎦A
⇒⎡⎢ ⎢ ⎢ ⎢ ⎢⎣13−201−11900259⎤⎥ ⎥ ⎥ ⎥ ⎥⎦=⎡⎢ ⎢ ⎢ ⎢ ⎢⎣10013190−53191⎤⎥ ⎥ ⎥ ⎥ ⎥⎦A
Applying R1→R1−3R2
⇒⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣1−3(0)3−3(1)−2−3(−119)01−11900253⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦=⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣1−3(13)0−3(19)0−3(0)13190−53191⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦A
⇒⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣105301−11900259⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦=⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣0−13013190−53191⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦A
Applying R3→925R3
⇒⎡⎢ ⎢ ⎢ ⎢ ⎢⎣105301−119001⎤⎥ ⎥ ⎥ ⎥ ⎥⎦=⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣0−13013190−35125925⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦A
Applying R2→R2+119R3
⇒⎡⎢ ⎢ ⎢ ⎢ ⎢⎣10530+119(0)1+119(0)−119+119(1)001⎤⎥ ⎥ ⎥ ⎥ ⎥⎦=⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣0−13013+119(−35)19+119(125)0+119(925)−35125925⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦A
⇒⎡⎢ ⎢ ⎢⎣1053010001⎤⎥ ⎥ ⎥⎦=⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣0−130−254251125−35125925⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦A
Applying R1→R1−53R3
⇒⎡⎢ ⎢ ⎢⎣1−53(0)0−53(0)53−53(1)010001⎤⎥ ⎥ ⎥⎦=⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣0−53(−35)−13−53(125)0−53(925)−254251125−35125925⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦A
⇒⎡⎢⎣100010001⎤⎥⎦=⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣55−615−35−254251125−35125925⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦A
⇒I=⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣55−615−35−254251125−35125925⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦A
⇒I=⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣1−25−35−254251125−35125925⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦A
This is similar to I=A−1A
Hence, A−1=⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣1−25−35−254251125−35125925⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦
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