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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)

best_answer

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

Posted by

Aadil

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