Inverse of thefollowing matrix using elementary Row transformations would be
=∣∣ ∣∣131011361∣∣ ∣∣
⎡⎢⎣−5+323−2−1−3+31⎤⎥⎦
We know, for matrix A,
A = IA
I = PA
→P=A−1
=∣∣ ∣∣131011361∣∣ ∣∣
A = IA
Lets convert the LHS into a Identity Matrix and apply the same transformations on the identity matrix on the right. The final matrix obtained in the right will give the inverse.
∣∣ ∣∣131011361∣∣ ∣∣=∣∣ ∣∣100010001∣∣ ∣∣A.
R3→R3−3R1∣∣ ∣∣1310110−3−2∣∣ ∣∣=∣∣ ∣∣100010−301∣∣ ∣∣A.
R3→R3+3R2 then R1→R1−3R2∣∣ ∣∣10−2011001∣∣ ∣∣=∣∣ ∣∣1−30010−3+31∣∣ ∣∣A
R2→R2−R3 then R1→R1+2R3∣∣ ∣∣100010001∣∣ ∣∣=∣∣ ∣∣−5+323−2−1−3+31∣∣ ∣∣A
I=A−1AA−1=⎡⎢⎣−5+323−2−1−3+31⎤⎥⎦