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Consider the system of equations x-2y+3z...

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`

A

Statement 1 is true, Statement 2 is true, Statement 2 is a correct explanation for Statement 1

B

Statement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation for Statement 1

C

Statement 1 is true, Statement 2 is False

D

Statement 1 is False, Statement 2 is true

Text Solution

Verified by Experts

The correct Answer is:
A
We have,
`D = |(1,-2,3),(-1,1,-2),(1,-3,4)|`
`rArr D = |(1,-2,3),(0,-1,1),(0,-1,1)| = 0 " " [("Applying " R_(2) rarr R_(2) - R_(1)),(R_(3) rarr R_(3) - R_(1))]`
and `D_(2) = |(1,-1,3),(-1,k,-2),(1,1,4)| = - |(1,3,-1),(-1,-2,k),(1,4,1)| " " ["Applying " C_(2) hArr C_(3)]`
`rArr D_(2) = - |(1,3,-1),(0,1,k -1),(0,1,2)| = k -3 " " [("Applying " R_(2) rarr R_(2) - R_(1)),(R_(3) rarr R_(3) - R_(1))]`
Clearly, `D_(2) != 0 " for " k != 3` and the determinant is not zero for `k != 3`.
Hence, both the statement are true and statement 2 is a correct explanation for statement 1
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