TS EAMCET 04 May 2019 Shift 2 Solved Paper

Consider the matrix.
[

3	5	−1	4
2	1	3	−2
8	11	1	6
−7	−14	6	−14

]
Apply operation R1→R1−R2
[

1	4	−4	6
2	1	3	−2
8	11	1	6
−7	−14	6	−14

]
Apply operation R2→R2−2R1,R3→R3−8R1 and
R4→R4+7R1
[

1	4	−4	6
0	−7	11	−14
0	−21	33	−42
0	14	−22	28

]
Apply operation R3→R3−3R2
[

1	4	−4	6
0	−7	11	−14
0	0	0	0
0	14	−22	28

]
Apply operation R3↔R4
[

1	4	−4	6
0	−7	11	−14
0	14	−22	28
0	0	0	0

]
Apply operation R3→R3+2R2
[

1	4	−4	6
0	−7	11	−14
0	0	0	0
0	0	0	0

]
Therefore, the rank of the matrix is 2