Correct Answer - A
The given system of equations can be expressed as
`[{:(1, -2, 3),(1, -3, 4), (-1," "1,-2):}][(x), (y), (z)]=[(1), (1), (k)]`
Applying `R_(2) to R_(2)-R_(1), R_(3) to R_(3) + R_(1)`
`~[{:(1, -2, 3),(0, -1, 1), (0,-1,1):}][(x), (y), (z)]=[(-1), (2), (k-1)]`
Applying `R_(3) to R_(3) -R_(2)`
`~[{:(1, -2, 3),(0, -1, 1), (0," "0,0):}][(x), (y), (z)]=[(-1), (2), (k-3)]`
When `k ne 3`, the given system of equation has no solution.
`rArr` I is true. Clearly, Statement II is also true as it is rearrangement of rows and columns of
`[{:(1, -2, 3),(1, -3, 4), (-1," "1,-2):}].`