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  • Help me understand! - Matrices and Determinants - JEE Main-3

The number of values of k for which the  system of linear equations,
(K+2)x +10 y = k  

kx + (k+3)y= k-1
has no solution, is :

  • Option 1)

    1

  • Option 2)

    2

  • Option 3)

    3

  • Option 4)

    infinitely many

 

Answers (2)

best_answer

As we have learned

Cramer's rule for solving system of linear equations -

When \Delta =0 and atleast one of   \Delta_{1},\Delta _{2} and \Delta _{3}  is non-zero , system of equations has no solution

- wherein

a_{1}x+b_{1}y+c_{1}z=d_{1}

a_{2}x+b_{2}y+c_{2}z=d_{2}

a_{3}x+b_{3}y+c_{3}z=d_{3}

and 

\Delta =\begin{vmatrix} a_{1} &b_{1} &c_{1} \\ a_{2} & b_{2} &c_{2} \\ a_{3}&b _{3} & c_{3} \end{vmatrix}

 

 for no solutions \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}}\neq \frac{c_{1}}{c_{2}}

\frac{K+2}{K} = \frac{10}{K+3}\neq \frac{K}{K-1}

k^{2}-5k+6=0

k=2,3

only solution is K= 3

 

 

 


Option 1)

1

This is correct 

Option 2)

2

This is incorrect 

Option 3)

3

This is incorrect 

Option 4)

infinitely many

This is incorrect 

Posted by

Himanshu

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