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  • For the system of linear equations <img alt="\alpha x+y+z=1, \quad x+\alpha y+z=1, \quad x+y+\alpha z=\beta" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Calpha%20x&plus;y&pl

For the system of linear equations \alpha x+y+z=1, \quad x+\alpha y+z=1, \quad x+y+\alpha z=\beta  which one of the following statements is NOT correct ?

Option: 1

It has infinitely many solutions if \alpha =2 and \beta =-1  
 


Option: 2

It has no solution if \alpha =-2 and \beta =1
 


Option: 3

x+y+z=\frac{3}{4} if \alpha =2 and \beta =1


Option: 4

It has infinitely many solutions if \alpha =1 and \beta =1


Answers (1)

best_answer

\begin{aligned} & \left|\begin{array}{lll} \alpha & 1 & 1 \\ 1 & \alpha & 1 \\ 1 & 1 & \alpha \end{array}\right|=0 \\ & \alpha\left(\alpha^2-1\right)-1(\alpha-1)+1(1-\alpha)=0 \\ & \alpha^3-3 \alpha+2=0 \end{aligned}

$$ \begin{aligned} & \alpha^2(\alpha-1)+\alpha(\alpha-1)-2(\alpha-1)=0 \\ & (\alpha-1)\left(\alpha^2+\alpha-2\right)=0 \\ & \alpha=1, \alpha=-2,1 \end{aligned} $$ For \: \alpha=1, \beta=1

\left.\begin{array}{l} x+y+z=1 \\ x+y+z=b \end{array}\right\} \text { infinite solution }

for \alpha =2,\beta =1

\begin{aligned} &\Delta=4\\ &\begin{aligned} & \Delta_1=\left|\begin{array}{lll} 1 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{array}\right|=3-1-1 \quad \Rightarrow x=\frac{1}{4} \\ & \Delta_2=\left|\begin{array}{lll} 2 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 2 \end{array}\right|=2-1=1 \quad \Rightarrow y=\frac{1}{4} \\ & \Delta_3=\left|\begin{array}{lll} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 1 \end{array}\right|=2-1=1 \quad \Rightarrow z=\frac{1}{4} \\ & \end{aligned} \end{aligned}

For \alpha=2 \Rightarrow \text { unique solution }

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Anam Khan

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