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  • For the system of linear equations <img alt="\mathrm{2x + 4y + 2az = b}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B2x%20&plus;%204y%20&plus;%202az%20%3D%20b%7D

For the system of linear equations
\mathrm{2x + 4y + 2az = b}
\mathrm{x + 2y + 3z = 4}
\mathrm{2x -5y + 2z = 8}
which of the following is NOT correct?

Option: 1

It has infinitely many solutions if \mathrm{a = 3, b = 8}


Option: 2

It has unique solution if \mathrm{a = b=8}


Option: 3

It has unique solution if \mathrm{a = b = 6}


Option: 4

It has infinitely many solutions if \mathrm{a = 3, b=6}


Answers (1)

best_answer


$$ \begin{aligned} & \Delta=\left|\begin{array}{ccc} 2 & 4 & 2 \mathrm{a} \\ 1 & 2 & 3 \\ 2 & -5 & 2 \end{array}\right|=18(3-\mathrm{a}) \\\end{aligned}

$$ \begin{aligned} & \Delta_{\mathrm{x}}=\left|\begin{array}{ccc} \mathrm{b} & 4 & 2 \mathrm{a} \\ 4 & 2 & 3 \\ 8 & -5 & 2 \end{array}\right|=(64+19 \mathrm{~b}-72 \mathrm{a})\\\end{aligned}

For unique solution \Delta=0
\Rightarrow \mathrm{a} \neq 3$ and $\mathrm{b} \in \mathrm{R}
For infinitely many solution :

\begin{aligned} & \Delta=\Delta_x=\Delta_y=\Delta_z=0 \\ & \Rightarrow a=3 \quad \because \Delta=0 \\ & \text { and } \mathrm{b}=8 \quad \because \Delta_{\mathrm{x}}=0 \end{aligned}
 

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Rishi

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