Get Answers to all your Questions

header-bg qa

An ordered pair \left ( \alpha ,\beta \right ) for which the system of linear equations 

\left ( 1+\alpha \right )x+\beta y+z=2

\alpha x+\left ( 1+\beta \right )y+z=3

\alpha x+\beta y+2z=2

has a unique solution, is : 

  • Option 1)

    \left ( 2,4 \right )

     

     

     

  • Option 2)

    \left ( -4,2 \right )

  • Option 3)

    \left ( 1,-3 \right )

  • Option 4)

    \left ( -3,1\right )

Answers (1)

best_answer

 

Solution of a homogeneous system of linear equations -

Let Ax=0 

If A is non-singular then the system of equations will have a unique solution that is trivial soluion

-

 

 

Solution of a non-homogeneous system of linear equations by matrix method -

If A is a non-singular matrix then the system of equations given by Ax=b has a unique solution given by x=A^{-1}b

- wherein

 

\\(1 +\alpha)x + \beta y + z = 0 \\ \alpha x + (1+\beta )y + z =0 \\ \alpha x + \beta y +2z = 0

\\D = \begin{vmatrix} 1 + \alpha &\beta &1 \\ \alpha & 1 + \beta & 1 \\ \alpha & \beta & 2 \end{vmatrix}\\ C_1\rightarrow C_1 + C_2 + C_3

\\D =(\alpha + \beta + 2) \begin{vmatrix} 1 &\beta &1 \\ 1 & 1 + \beta & 1 \\ 1 & \beta & 2 \end{vmatrix}\\

R_2 \rightarrow R_2 - R_1 \;\;\;\;\;\; R_3 \rightarrow R_3 - R_1

\\D =(\alpha + \beta + 2) \begin{vmatrix} 1 &\beta &1 \\ 0 & 1 & 0 \\ 0 & 0 &1 \end{vmatrix}= \alpha +\beta +2

For a unique solution, \alpha + \beta \neq -2


Option 1)

\left ( 2,4 \right )

 

 

 

Option 2)

\left ( -4,2 \right )

Option 3)

\left ( 1,-3 \right )

Option 4)

\left ( -3,1\right )

Posted by

admin

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE