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If $A=\left[\begin{array}{lll}1 & 0 & 1 \\ 2 & 1 & 0 \\ 3 & 2 & 1\end{array}\right]$, then det $A$ is equal to
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1779 Upvotes
Verified Answer
The correct answer is:
$2$
We have,
$$
\begin{aligned}
|A| & =\left|\begin{array}{lll}
1 & 0 & 1 \\
2 & 1 & 0 \\
3 & 2 & 1
\end{array}\right|=1(1-0)+0+1(4-3) \\
& =1+1=2
\end{aligned}
$$
$$
\begin{aligned}
|A| & =\left|\begin{array}{lll}
1 & 0 & 1 \\
2 & 1 & 0 \\
3 & 2 & 1
\end{array}\right|=1(1-0)+0+1(4-3) \\
& =1+1=2
\end{aligned}
$$
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