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If $\left[\begin{array}{lll}1 & x & 1\end{array}\right]\left[\begin{array}{lll}1 & 3 & 2 \\ 0 & 5 & 1 \\ 0 & 3 & 2\end{array}\right]\left[\begin{array}{c}x \\ 1 \\ -2\end{array}\right]=0$, then $x$ is
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2224 Upvotes
Verified Answer
The correct answer is:
$\frac{1}{2}$
We have
$$
\begin{array}{l}
\left[\begin{array}{lll}
1 & \mathrm{x} & 1
\end{array}\right]\left[\begin{array}{lll}
1 & 3 & 2 \\
0 & 5 & 1 \\
0 & 3 & 2
\end{array}\right]\left[\begin{array}{c}
\mathrm{x} \\
1 \\
-2
\end{array}\right]=0 \\
\Rightarrow\left[\begin{array}{ll}
1 & 5 \mathrm{x}+6 & \mathrm{x}+4
\end{array}\right]\left[\begin{array}{c}
\mathrm{x} \\
1 \\
-2
\end{array}\right]=0 \\
\Rightarrow \mathrm{x}+5 \mathrm{x}+6-2 \mathrm{x}-8=0 \Rightarrow 4 \mathrm{x}-2=0 \Rightarrow \mathrm{x}=\frac{1}{2}
\end{array}
$$
$$
\begin{array}{l}
\left[\begin{array}{lll}
1 & \mathrm{x} & 1
\end{array}\right]\left[\begin{array}{lll}
1 & 3 & 2 \\
0 & 5 & 1 \\
0 & 3 & 2
\end{array}\right]\left[\begin{array}{c}
\mathrm{x} \\
1 \\
-2
\end{array}\right]=0 \\
\Rightarrow\left[\begin{array}{ll}
1 & 5 \mathrm{x}+6 & \mathrm{x}+4
\end{array}\right]\left[\begin{array}{c}
\mathrm{x} \\
1 \\
-2
\end{array}\right]=0 \\
\Rightarrow \mathrm{x}+5 \mathrm{x}+6-2 \mathrm{x}-8=0 \Rightarrow 4 \mathrm{x}-2=0 \Rightarrow \mathrm{x}=\frac{1}{2}
\end{array}
$$
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