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  • Can someone help me with this, - Matrices and Determinants - JEE Main-4

Let S be the set of all real values of k for which the system of linear equations

x+y+z=2

2x+y-z=3

3x+2y+kz=4

has a unique solution. Then S is :

 

  • Option 1)

    an empty set

  • Option 2)

    equal to {0}

  • Option 3)

    equal to R

  • Option 4)

    equal to R −{0}

 

Answers (2)

best_answer

As we learned 

 

Cramer's rule for solving system of linear equations -

When   \Delta \neq 0

x=\Delta _{1}/\Delta , y=\Delta _{2}/\Delta , z=\Delta _{3}/\Delta

- wherein

 

 

x+y+z=2

2x+y-z=3

3x+2y+kz=4

\bigtriangleup =\begin{vmatrix} 1 &1 &1 \\ 2&1 &-1 \\ 3 & 2 & k \end{vmatrix}\Rightarrow \left ( k+2 \right )+\left ( -3-2k \right )+1\left ( 4-3 \right )=0

\Rightarrow -k=0

\Rightarrow k=0

So it has unique solution for all x\equiv R-\left \{ 0 \right \}

 


Option 1)

an empty set

Option 2)

equal to {0}

Option 3)

equal to R

Option 4)

equal to R −{0}

Posted by

Himanshu

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