If P = begin{bmatrix} 1 & 2 & 1 1 & 3 & 1 en | vedclass

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(A) Given, $P = \begin{bmatrix} 1 & 2 & 1 \ 1 & 3 & 1 \end{bmatrix}$.We need to find $Q = P P^{T}$.$P^{T} = \begin{bmatrix} 1 & 1 \ 2 & 3 \ 1 & 1 \

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(A) Given, $P = \begin{bmatrix} 1 & 2 & 1 \ 1 & 3 & 1 \end{bmatrix}$.We need to find $Q = P P^{T}$.$P^{T} = \begin{bmatrix} 1 & 1 \ 2 & 3 \ 1 & 1 \end{bmatrix}$.$Q = \begin{bmatrix} 1 & 2 & 1 \ 1 & 3 & 1 \end{bmatrix} \begin{bmatrix} 1 & 1 \ 2 & 3 \ 1 & 1 \end{bmatrix}$.$Q = \begin{bmatrix} (1 	imes 1 + 2 	imes 2 + 1 	imes 1) & (1 	imes 1 + 2 	imes 3 + 1 	imes 1) \ (1 	imes 1 + 3 	imes 2 + 1 	imes 1) & (1 	imes 1 + 3 	imes 3 + 1 	imes 1) \end{bmatrix}$.$Q = \begin{bmatrix} (1 + 4 + 1) & (1 + 6 + 1) \ (1 + 6 + 1) & (1 + 9 + 1) \end{bmatrix} = \begin{bmatrix} 6 & 8 \ 8 & 11 \end{bmatrix}$.Now, the determinant of $Q$ is $|Q| = (6 	imes 11) - (8 	imes 8)$.$|Q| = 66 - 64 = 2$.

If $P = \begin{bmatrix} 1 & 2 & 1 \\ 1 & 3 & 1 \end{bmatrix}$ and $Q = P P^{T}$, then the value of the determinant of $Q$ is:

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