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  • If the system of linear equations                 <img alt="\begin{aligned} &2 x+y-z=7 \\ &x-3 y+2 z=1

If the system of linear equations
                \begin{aligned} &2 x+y-z=7 \\ &x-3 y+2 z=1 \\ &x+4 y+\delta z=\mathrm{k}, \text { where } \delta, \mathrm{k} \in \mathbf{R} \end{aligned}
has infinitely many solutions, then \delta+\mathrm{k} is equal to :

Option: 1

-3


Option: 2

3


Option: 3

6


Option: 4

9


Answers (1)

best_answer

For infinite solutions

\mathrm{\Delta=\left|\begin{array}{ccc} 2 & 1 & -1 \\ 1 & -3 & 2 \\ 1 & 4 & \delta\end{array}\right|=0}

\mathrm{\Rightarrow 2(-3 \delta-8)-1(\delta-2)-1(4+3)=0} \\

\mathrm{\Rightarrow-6 \delta-16-\delta+2-7=0} \\

\mathrm{\Rightarrow-7 \delta=16+7-2 }\\

\mathrm{\Rightarrow-7 \delta=21}

\mathrm{\Rightarrow \delta=-3}

Also

\mathrm{\Delta_{3}=0} \\

\mathrm{\left|\begin{array}{ccc} 2 & 1 & 7 \\ 1 & -3 & 1 \\ 1 & 4 & k \end{array}\right|=0 }

\mathrm{\Rightarrow 2(-3 k -4)-1(k -1)+7(4+3)=0} \\

\mathrm{\Rightarrow-6 k-8-k+1+49=0} \\

\mathrm{\Rightarrow-7 k=-42} \\

\mathrm{\Rightarrow k=6} \\

\mathrm{\therefore \delta+k=3}

Hence the answer is option 2

Posted by

Rakesh

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