... resultingin:⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡100010001⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡ 33 32 3 1 23 222 1 13 121 1xxxxxxxxx=⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡−−−−1 13 13/ 43/ 10 33 /1 03/ 31Or, X=⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡ 33 32 3 1 23 222 1 13 121 1xxxxxxxxx=⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡−−−−1 13 13/ 43/ 10 33 /1 03/ 31Note ... in⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡100010001⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡ 33 32 3 1 23 222 1 13 121 1xxxxxxxxx=⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡−−−−1 13 13/ 43/ 10 33 /1 03/ 31Or, X=⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡ 33 32 3 1 23 222 1 13 121 1xxxxxxxxx=⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡−−−−1 13 13/ 43/ 10 33 /1 03/ 31The same process can be used ... by ( 2) and added to row 3 to replace row 3, resulting in:⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡−− 430 33 0 21 1⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡ 33 32 3 1 23 222 1 13 121 1xxxxxxxxx=⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡−−1 02 011001Row 2 is...