Consider the system of equations `x-2y+3z=-1, x-3y+4z=1` and `-x+y-2z=k` Statement 1: The system of equation has no solution for `k!=3` and Statement 2: The determinant `|[1,3,-1], [-1,-2,k] , [1,4,1]| !=0` for `k!=0`

Options

A. Statement I is true, Statement II is also true, Statement II is the correct explanation of Statement I
B. Statement I is true, Statement II is also true, Statement II is not the correct explanation of Statement I
C. Statement I is true, Statement II is false
D. Statement I is true,Statement II is true

Answer

A

Explanation

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):}].`