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If $P=\left[\begin{array}{lll}1 & 2 & 1 \\ 1 & 3 & 1\end{array}\right] Q=P P^{T}$, then the value of the
determinant of $Q$ is
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determinant of $Q$ is
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Verified Answer
The correct answer is:
2
Given, $P=\left[\begin{array}{lll}1 & 2 & 1 \\ 1 & 3 & 1\end{array}\right]$
$$
\begin{array}{l}
\therefore \quad Q=P P^{T}=\left[\begin{array}{lll}
1 & 2 & 1 \\
1 & 3 & 1
\end{array}\right]\left[\begin{array}{ll}
1 & 1 \\
2 & 3 \\
1 & 1
\end{array}\right] \\
=\left[\begin{array}{ll}
1 \times 1+2 \times 2+1 \times 1 & 1 \times 1+2 \times 3+1 \times 1 \\
1 \times 1+3 \times 2+1 \times 1 & 1 \times 1+3 \times 3+1 \times 1
\end{array}\right] \\
=\left[\begin{array}{ll}
1+4+1 & 1+6+1 \\
1+6+1 & 1+9+1
\end{array}\right]=\left[\begin{array}{lr}
6 & 8 \\
8 & 11
\end{array}\right] \\
\therefore \text { The determintant of } Q=\left|\begin{array}{lr}
6 & 8 \\
8 & 11
\end{array}\right| \\
=66-64=2
\end{array}
$$
$$
\begin{array}{l}
\therefore \quad Q=P P^{T}=\left[\begin{array}{lll}
1 & 2 & 1 \\
1 & 3 & 1
\end{array}\right]\left[\begin{array}{ll}
1 & 1 \\
2 & 3 \\
1 & 1
\end{array}\right] \\
=\left[\begin{array}{ll}
1 \times 1+2 \times 2+1 \times 1 & 1 \times 1+2 \times 3+1 \times 1 \\
1 \times 1+3 \times 2+1 \times 1 & 1 \times 1+3 \times 3+1 \times 1
\end{array}\right] \\
=\left[\begin{array}{ll}
1+4+1 & 1+6+1 \\
1+6+1 & 1+9+1
\end{array}\right]=\left[\begin{array}{lr}
6 & 8 \\
8 & 11
\end{array}\right] \\
\therefore \text { The determintant of } Q=\left|\begin{array}{lr}
6 & 8 \\
8 & 11
\end{array}\right| \\
=66-64=2
\end{array}
$$
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