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  • Please help! The system of linear equations, x+y+z=2, 2x+y-z=3, 3x+2y+Kz=4 has a unique solution if?

The system of linear equations, x+y+z=2, 2x+y-z=3, 3x+2y+Kz=4 has a unique solution if?

  • Option 1)

    K\neq 0

  • Option 2)

    -1< K< 1

  • Option 3)

    -2<K<2

  • Option 4)

    K=0

 

Answers (1)

For unique solution D\neq 0

\begin{vmatrix} 1 & 1& 1\\ 2& 1& -1\\ 3& 2 & K \end{vmatrix}\neq 0

\begin{vmatrix} 0 & 0& 1\\ 1& 2& -1\\ 1& 2-K & K \end{vmatrix}\neq 0

\therefore (2-K)-2\neq 0\\ \therefore K\neq 0


Option 1)

K\neq 0

This option is correct 

Option 2)

-1< K< 1

This option is incorrect 

Option 3)

-2<K<2

This option is incorrect 

Option 4)

K=0

This option is incorrect 

Posted by

Vakul

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