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  • Let the system of linear equations  <img alt="\begin{aligned} & x+y+k z=2 \\ & 2 x+3 y-z=1 \\ & 3 x+4 y+2 z=k \end{aligned}" src="https://entrancecorner.oncodecogs.com/gi

Let the system of linear equations 

\begin{aligned} & x+y+k z=2 \\ & 2 x+3 y-z=1 \\ & 3 x+4 y+2 z=k \end{aligned}

have infinitely many solutions. Then the system

\begin{aligned} & (k+1) x+(2 k-1) y=7 \\ & (2 k+1) x+(k+5) y=10 \end{aligned}

has:

Option: 1

infinitely many solutions


Option: 2

unique solution satisfying x − y = 1


Option: 3

unique solution satisfying x + y = 1


Option: 4

no solution


Answers (1)

best_answer

\begin{aligned} & x+y+k z=2 \\ & 2 x+3 y-z=1 \\ & 3 x+4 y+2 z=k \end{aligned}

Have Infinitely many solution then

\begin{aligned} & \left|\begin{array}{ccc} 1 & 1 & \mathrm{k} \\ 2 & 3 & -1 \\ 3 & 4 & 2 \end{array}\right|=0 \\ & 1(10)-1(7)+\mathrm{k}(8-9)=0 \\ & \Rightarrow 10-7-\mathrm{k}=0 \\ & \Rightarrow \mathrm{k}=3 \end{aligned}

For k = 3

\begin{aligned} & 4 x+5 y=7 \\ & 7 x+8 y=10 \end{aligned}

has unique solution and solution is (–2, 3).
Hence solution is unique and satisfying x + y = 1

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