Consider the system of equations x-2 y+3 z=-1 ;-x+y-2 z=k ; x-3 y+4 z=1. STATEMENT-1: The system of equations has no solution for {\k} \neq 3 and STATEMENT -2: The determinant \left|\begin{array}{ccc}1 & 3 & -1 \ -1 & -2 & k \ 1 & 4 & 1|\end{array} \neq 0, for k \neq 3. Then

Correct Answer is : Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1

Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1

Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1

Statement-1 is True and Statement-2 is False

Statement-1 is False and Statement-2 is True

CUSAT 2019 BTech Solved Paper

Question : 44 of 250

Marks: +1, -0

Consider the system of equations x−2y+3z=−1;−x+y−2z=k;x−3y+4z=1. STATEMENT-1: The system of equations has no solution for k≠3 and
STATEMENT -2: The determinant |

1	3	−1
−1	−2	k
1	4	1

|≠0, for k≠3.

Then

Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1
Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1
Statement-1 is True and Statement-2 is False
Statement-1 is False and Statement-2 is True

Validate

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