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If the system of equations $x+y+z=1$, $x+2 y+4 z=k$ and $x+4 y+10 z=k^2$ is consistent, then $k$ is equal to
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Verified Answer
The correct answer is:
1, 2
Here, it is given that the given system of equations is consistent.
$$
\begin{aligned}
\therefore \Delta & =\left|\begin{array}{ccc}
1 & 1 & 1 \\
1 & 2 & 4 \\
1 & 4 & 10
\end{array}\right|=0 \\
\Delta_1 & =\left|\begin{array}{ccc}
1 & 1 & 1 \\
k & 2 & 4 \\
k^2 & 4 & 10
\end{array}\right|=2\left(k^2-3 k+2\right)=0 \\
\Delta_2 & =\left|\begin{array}{ccc}
1 & 1 & 1 \\
1 & k & 4 \\
1 & k^2 & 10
\end{array}\right|=3\left(k^2-3 k+2\right)=0 \\
\Delta_3 & =\left|\begin{array}{lll}
1 & 1 & 1 \\
1 & 2 & k \\
1 & 4 & k^2
\end{array}\right|=k^2-3 k+2=0
\end{aligned}
$$
Here, above all are having $k^2-3 k+2=0$
$$
\begin{array}{rlrl}
\Rightarrow & (k-1)(k-2) =0 \\
\Rightarrow & k =1,2
\end{array}
$$
$$
\begin{aligned}
\therefore \Delta & =\left|\begin{array}{ccc}
1 & 1 & 1 \\
1 & 2 & 4 \\
1 & 4 & 10
\end{array}\right|=0 \\
\Delta_1 & =\left|\begin{array}{ccc}
1 & 1 & 1 \\
k & 2 & 4 \\
k^2 & 4 & 10
\end{array}\right|=2\left(k^2-3 k+2\right)=0 \\
\Delta_2 & =\left|\begin{array}{ccc}
1 & 1 & 1 \\
1 & k & 4 \\
1 & k^2 & 10
\end{array}\right|=3\left(k^2-3 k+2\right)=0 \\
\Delta_3 & =\left|\begin{array}{lll}
1 & 1 & 1 \\
1 & 2 & k \\
1 & 4 & k^2
\end{array}\right|=k^2-3 k+2=0
\end{aligned}
$$
Here, above all are having $k^2-3 k+2=0$
$$
\begin{array}{rlrl}
\Rightarrow & (k-1)(k-2) =0 \\
\Rightarrow & k =1,2
\end{array}
$$
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