Let $$\mathrm{D}_{\mathrm{k}}=\left|\begin{array}{ccc}1 & 2 k & 2 k-1 \\ n & n^{2}+n+2 & n^{2} \\ n & n^{2}+n & n^{2}+n+2\end{array}\right|$$. If $\sum_\limits{k=1}^{n} \mathrm{D}_{\mathrm{k}}=96$, then $n$ is equal to _____________.
Answer (integer)
6
Solution
$$
\begin{aligned}
& \sum_{k=1}^n D_k=\left|\begin{array}{ccc}
\sum 1 & 2 \sum k & 2 \sum k-\sum 1 \\
n & n^2+n+2 & n^2 \\
n & n^2+n & n^2+n+2
\end{array}\right| \\\\
& =\left|\begin{array}{ccc}
n & n(n+1) & n^2 \\
n & n^2+n+2 & n^2 \\
n & n^2+n & n^2+n+2
\end{array}\right| \\\\
& =\left|\begin{array}{ccc}
0 & -2 & 0 \\
0 & 2 & -n-2 \\
n & n^2+n & n^2+n+2
\end{array}\right| \\\\
& =2((-n)(-n-2))=96 \\\\
& \Rightarrow n^2+2 n=48 \\\\
& \Rightarrow n=6,-8 \\\\
& \Rightarrow n=6
\end{aligned}
$$
About this question
Subject: Mathematics · Chapter: Matrices and Determinants · Topic: Types of Matrices and Operations
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