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

a

PPT

b

pp-1

c

qp

d

pp

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a

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Discussions (2)

dandano
7 years ago

When you find the transpose of p you discover that it is dsame as q. so pq=ppT

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