For getting the inverse of the given matrix A by row elementary operations we may write the given matrix as A=IA
And, we know AA−1=I
Now |A|=2×7+1×(−12)+3×(−1)=14−12−3=−1≠0
So, A−1 is possible.
∵⎡⎢⎣2−13−531−323⎤⎥⎦=⎡⎢⎣100010001⎤⎥⎦A
[R2→R2+R1]
⇒⎡⎢⎣2−13−324−323⎤⎥⎦=⎡⎢⎣100110001⎤⎥⎦A
[∵R3→R3−R2]
⇒⎡⎢⎣2−13−32400−1⎤⎥⎦=⎡⎢⎣100110−1−11⎤⎥⎦A
[R1→R1+R2]
⇒⎡⎢⎣−117−32400−1⎤⎥⎦=⎡⎢⎣210110−1−11⎤⎥⎦A
[∵R2→R2−3R1]
⇒⎡⎢⎣−1170−1−1700−1⎤⎥⎦=⎡⎢⎣210−5−20−1−11⎤⎥⎦A
[∵R1→R1+R2andR3→−1×R3]
⇒⎡⎢⎣−10−100−1−17001⎤⎥⎦=⎡⎢⎣−3−10−5−2011−1⎤⎥⎦A
[∵R1→R1+10R3andR2→R2+17R3]
⇒⎡⎢⎣−1000−10001⎤⎥⎦=⎡⎢⎣79−101215−1711−1⎤⎥⎦A
[∵R1→−1R1andR2→−1R2]
⇒⎡⎢⎣100010001⎤⎥⎦=⎡⎢⎣−7−910−12−151711−1⎤⎥⎦A
As, I=A−1A=AA−1
So, the inverse of A is A−1=⎡⎢⎣−7−910−12−151711−1⎤⎥⎦