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For the system of equations
\mathrm{x + y + z = 6}
\mathrm{x + 2y + \alpha z = 10}
\mathrm{x + 3y + 5z = \beta ,} which one of the following is NOT true :

Option: 1

System has a unique solution for \mathrm{\alpha =3,\beta \neq 14}
 


Option: 2

System has a unique solution for \mathrm{ \alpha = 3, \beta = 14.}
 


Option: 3

System has no solution for \mathrm{\alpha = 3, \beta = 24.}
 


Option: 4

System has infinitely many solutions for \mathrm{\alpha = 3, \beta = 14.}


Answers (1)

best_answer

\begin{aligned} & \Delta=\left|\begin{array}{ccc} 1 & 1 & 1 \\ 1 & 2 & \alpha \\ 1 & 3 & 5 \end{array}\right| \\ & =(10-3 \alpha)-(5-\alpha)+(3-2) \\ & =6-2 \alpha \end{aligned}

$$ \begin{aligned} & \Delta \mathrm{x}=\left|\begin{array}{ccc} 6 & 1 & 1 \\ 10 & 2 & \alpha \\ \beta & 3 & 5 \end{array}\right| \\ & =6(10-3 \alpha)-(50-\alpha 13)+(30-2 \beta) \\ & =40-18 \alpha+\alpha \beta-2 \beta \\\end{aligned}

\begin{aligned}& \Delta y=\left|\begin{array}{ccc} 1 & 6 & 1 \\ 1 & 10 & \alpha \\ 1 & \beta & 5 \end{array}\right| \\ & =(50-\alpha \beta)-6(5-\alpha)+(\beta-10) \\ & =10+6 \alpha+\beta-\alpha \beta \\\end{aligned}

\begin{aligned}& \Delta \mathrm{z}=\left|\begin{array}{ccc} 1 & 1 & 6 \\ 1 & 2 & 10 \\ 1 & 3 & \beta \end{array}\right| \\ & =(2 \beta-30)-(\beta-10)+6(1) \\ & =\beta-14 \end{aligned}
for Infinite solution \begin{array}{ll} \Delta=0, & \Delta_{\mathrm{x}}=\Delta_{\mathrm{y}}=\Delta_{\mathrm{z}}=0 \\ \underline{\alpha=3}, & \underline{\beta=14} \end{array}

For unique solution \alpha \neq 3

Correct Answer is option 1

Posted by

Pankaj Sanodiya

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