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The system of equations $\alpha x+y+z=\alpha -1,$ $x+\alpha y+z=\alpha -1,$ $x+y+\alpha z=\alpha -1$ has no solutions , if $\alpha$ is

Option 1)
either –2 or 1

Option 2)
-2

Option 3)
1

Option 4)
not –2.

Answers (2)

As we learnt in

Cramer's rule for solving system of linear equations -

When $\Delta =0$ and $\Delta _{1}=\Delta _{2}=\Delta _{3}=0$ ,

then the system of equations has infinite solutions.

- 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

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

$\Delta _{1},\Delta _{2},\Delta _{3}$ are obtained by replacing column 1,2,3 of $\Delta$ by $\left ( d_{1},d_{2},d_{3} \right )$ column

$\alpha x+ y+z= \alpha -1$

$x+\alpha y+z= \alpha -1$

$x+y+\alpha z= \alpha -1$

$\therefore \begin{vmatrix} \alpha &1 &1 \\ 1& \alpha &\alpha \\ 1 &1 &\alpha \end{vmatrix}= 0$

$\therefore \alpha(\alpha^{2}-1)- (\alpha -1)+ (1-\alpha)= 0$

$\therefore (\alpha-1)(\alpha^{2} + \alpha -2)= 0$

$(\alpha-1)(\alpha+2) (\alpha-1)= 0$

$(\alpha-1)^{2}(\alpha+2) = 0$

$\alpha= -2$

Option 1)

either –2 or 1

Incorrect option

Option 2)

-2

Correct option

Option 3)

1

Incorrect option

Option 4)

not –2.

Incorrect option

Posted by

Vakul

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