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Question

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

Vedclass · Accepted Answer

exactly three values of  $\;\lambda $

Vedclass · Answer

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

$\Delta=\left|\begin{array}{lll}{a_{1}} & {b_{1}} & {c_{1}} \ {a_{2}} & {b_{2}} & {c_{2}} \ {a_{3}} & {b_{3}} & {c_{3}}\end{array}ight|$

$\Delta_{1}, \Delta_{2}, \Delta_{3}$ are obtained by replacing column $1,2,3$ of $\Delta$ by $\left(d_{1}, d_{2}, d_{3}ight)$ column

$\left|\begin{array}{ccc}{1} & {\lambda} & {-1} \ {\lambda} & {-1} & {-1} \ {1} & {1} & {-\lambda}\end{array}ight|=0$

$\Rightarrow 1(\lambda+1)-\lambda\left(1-\lambda^{2}ight)-1(1+\lambda)=0$

$\Rightarrow(1+\lambda)\left[\lambda^{2}-\lambdaight]=0$

$\Rightarrow \lambda=-1,0,1$

The system of linear equations $x + \lambda y - z = 0,\lambda x - y - z = 0\;,\;x + y - \lambda z = 0$ has a non-trivial solution for:

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