Consider the system of linear equations

x_{1}+2x_{2}+x_{3}=3

2x_{1}+3x_{2}+x_{3}=3

3x_{1}+5x_{2}+2x_{3}=1

The system has

  • Option 1)

    infinite number of solutions

  • Option 2)

    exactly 3 solutions

  • Option 3)

    a unique solution

  • Option 4)

    no solution

 

Answers (2)

As we learnt in 

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}

 

 x_{1}+2x_{2}+x_{3}=3

2x_{1}+3x_{2}+x_{3}=3

3x_{1}+5x_{2}+2x_{3}=1

\therefore \begin{vmatrix} 1&2&1\\2&3&1\\ 3&5&2\end{vmatrix}

= > 1\left ( 6-5 \right )-2\left ( 4-3 \right )+1\left ( 10-9\right )

= > 1-2\left ( 1\right )+1\left ( 1\right )

= > 1-2+1= 0

Since\, \, \, \, D=0 \, \, \, so \, \, that \, \, no\, \, solution

 If \, \, \, \, D_{1},D_{2},D_{3}\, \, \, having\, \, \, atleast\, \, \, one\, \, \, non\, \, \, zero.

 

 


Option 1)

infinite number of solutions

Incorrect Option

 

Option 2)

exactly 3 solutions

Incorrect Option

 

Option 3)

a unique solution

Incorrect Option

 

Option 4)

no solution

Correct Option

 

N neha

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