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If the system of simultaneous linear equations $3 x-4 y+k z+13=0, x+2 y-z-9=0$ and $k x-y+3 z+7=0$ has a unique solution $x=\alpha, y=\beta, z=\gamma$ for $k \neq m$ and $2 \beta-\gamma=8$, then $\alpha+m=$
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The correct answer is:
-2
$\begin{aligned} & \text { }\left|\begin{array}{ccc}3 & -4 & k \\ 1 & 2 & -1 \\ k & -1 & 3\end{array}\right| \neq 0 \\ & \Rightarrow \quad 3(6-1)+4(3+k)+k(-1-2 k) \neq 0 \\ & \Rightarrow 15+12+4 k-k-2 k^2 \neq 0 \\ & \Rightarrow \quad 2 k^2-3 k-27 \neq 0 \Rightarrow 2 k^2-9 k+6 k-27 \neq 0 \\ & \Rightarrow k \neq \frac{9}{2},-3\end{aligned}$
On putting $x=\alpha, y=\beta, z=\gamma$ inequation
$$
\begin{gathered}
x+2 y-z-9=0 \\
\alpha+2 \beta-\gamma-9=0 \\
\alpha+8-9=0 [It is given that 2 \beta-\gamma=8 ]
\\
\alpha=1
\end{gathered}
$$
As $k \neq m \Rightarrow m=9 / 2-3$
$$
\alpha+m=1+\frac{9}{2}=\frac{11}{2} \text { or } \alpha+m=1-3=-2
$$
On putting $x=\alpha, y=\beta, z=\gamma$ inequation
$$
\begin{gathered}
x+2 y-z-9=0 \\
\alpha+2 \beta-\gamma-9=0 \\
\alpha+8-9=0 [It is given that 2 \beta-\gamma=8 ]
\\
\alpha=1
\end{gathered}
$$
As $k \neq m \Rightarrow m=9 / 2-3$
$$
\alpha+m=1+\frac{9}{2}=\frac{11}{2} \text { or } \alpha+m=1-3=-2
$$
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