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  • Consider the following system of equations <img alt="\mathrm{\begin{aligned} & \alpha x+2 y+z=1 \\ & 2 \alpha x+3 y+z=1 \\ & 3 x+\alpha y+2 z=\beta \end{aligned}}" src="https:

Consider the following system of equations

\mathrm{\begin{aligned} & \alpha x+2 y+z=1 \\ & 2 \alpha x+3 y+z=1 \\ & 3 x+\alpha y+2 z=\beta \end{aligned}}

for some \alpha ,\beta \epsilon \mathbb{R}. Then which of the following is NOT correct.

Option: 1

It has a solution if \mathrm{\alpha =-1 \: and\: \beta \neq 2} 


Option: 2

It has a solution for all \mathrm{\alpha \neq-1 \: and\: \beta =2}


Option: 3

It has no solution for \mathrm{\alpha=3 \: and\: \beta \neq2}


Option: 4

It has no solution for \mathrm{\alpha=-1 \: and\: for\: all\: \: \beta \varepsilon \mathbb{R}}


Answers (1)

best_answer

\begin{aligned} & \because \mathrm{D}=\left|\begin{array}{ccc} \alpha & 2 & 1 \\ 2 \alpha & 3 & 1 \\ 3 & \alpha & 2 \end{array}\right| \\ & D=\alpha(6-\alpha)+2(3-4 \alpha)+1\left(2 \alpha^2-9\right) \end{aligned}

\begin{aligned} & =6 \alpha-\alpha^2+6-8 \alpha+2 \alpha^2-9 \\ \end{aligned}

\begin{aligned} & \mathrm{D}=\alpha^2-2 \alpha-3 \end{aligned}
\mathrm{for\: no\, solution} , \mathrm{D}=0
\begin{aligned} \Rightarrow \quad & \alpha^2-2 \alpha-3=0 \\ & (\alpha+1)(\alpha-3)=0 \end{aligned}
\Rightarrow \quad \alpha=-1, \alpha=3
Now,
D_1=\left|\begin{array}{lll} 1 & 2 & 1 \\ 1 & 3 & 1 \\ \beta & \alpha & 2 \end{array}\right|, D_2=\left|\begin{array}{ccc} \alpha & 1 & 1 \\ 2 \alpha & 1 & 1 \\ 3 & \beta & 2 \end{array}\right| \text { and } D_3=\left|\begin{array}{ccc} \alpha & 2 & 1 \\ 2 \alpha & 3 & 1 \\ 3 & \alpha & \beta \end{array}\right|
if \alpha=-1 then
D_1=\left|\begin{array}{ccc} 1 & 2 & 1 \\ 1 & 3 & 1 \\ \beta & -1 & 2 \end{array}\right|, D_2=\left|\begin{array}{ccc} -1 & 1 & 1 \\ -2 & 1 & 1 \\ 3 & \beta & 2 \end{array}\right|, D_3=\left|\begin{array}{ccc} -1 & 2 & 1 \\ -2 & 3 & 1 \\ 3 & -1 & \beta \end{array}\right|
\mathrm{\Rightarrow \: only \: for \: \beta=2, D_1=0, D_2=0, D_3=0}
\therefore$ It has no solution if $\alpha=-1$ and $\beta \neq 2$ if $\alpha=3

if \alpha =3

\begin{aligned} & D_1=\left|\begin{array}{lll} 1 & 2 & 1 \\ 1 & 3 & 1 \\ \beta & 3 & 2 \end{array}\right|, D_2=\left|\begin{array}{lll} 3 & 1 & 1 \\ 6 & 1 & 1 \\ 3 & \beta & 2 \end{array}\right|, D_3=\left|\begin{array}{lll} 3 & 2 & 1 \\ 6 & 3 & 1 \\ 3 & 3 & \beta \end{array}\right| \\ & \Rightarrow \text { Only for } \beta=2, D_1=D_2=D_3=0 \\ & \Rightarrow \text { It has no solution for } \beta \neq 2 \\ & \therefore \text { It has no solution for } \alpha=3 \text { and for all } \beta \neq 2 \end{aligned}

Posted by

Ajit Kumar Dubey

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