p = \(\begin{vmatrix} x & 3 & 0 \\ 2 & y & 3\\ 4 & 2 & 4 \end{vmatrix}\)

Q = \(\begin{vmatrix} x & 2 & z \\ 3 & y & 2\\ 0 & 3 & z \end{vmatrix}\)
PQ is equivalent to

PP^T

pp^-1

Correct answer is A

p = \(\begin{vmatrix} 0 & 3 & 0 \\ 2 & 1 & 3\\ 4 & 2 & 2 \end{vmatrix}\)

Q = \(\begin{vmatrix} 0 & 2 & 4 \\ 3 & 1 & 2\\ 0 & 3 & 2 \end{vmatrix}\) = p^T

pq = pp^T

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p = \(\begin{vmatrix} x & 3 & 0 \\ 2 & y & 3\\ 4 & 2 & 4 \end{vmatrix}\)

Q = \(\begin{vmatrix} x & 2 & z \\ 3 & y & 2\\ 0 & 3 & z \end{vmatrix}\)
PQ is equivalent to

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p = \(\begin{vmatrix} x & 3 & 0 \\ 2 & y & 3\\ 4 & 2 & 4 \end{vmatrix}\) Q = \(\begin{vmatrix} x & 2 & z \\ 3 & y & 2\\ 0 & 3 & z \end{vmatrix}\) PQ is equivalent to

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p = \(\begin{vmatrix} x & 3 & 0 \\ 2 & y & 3\\ 4 & 2 & 4 \end{vmatrix}\)

Q = \(\begin{vmatrix} x & 2 & z \\ 3 & y & 2\\ 0 & 3 & z \end{vmatrix}\)
PQ is equivalent to